answer：Okay, so I have this equation: ( x^2 - m y^2 + 2x + 2y = 0 ). The problem says that this equation represents two lines, and I need to find the value of ( m ). Hmm, okay. Let me think about how to approach this.First, I remember that if a quadratic equation represents two lines, it can be factored into the product of two linear equations. So, maybe I can try to factor this equation somehow. But before I jump into factoring, let me see if there's another way to check if it represents two lines.I recall that for a general quadratic equation ( Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0 ), it represents two lines if the discriminant ( B^2 - 4AC ) is greater than or equal to zero. Wait, is that right? Or is it something else? Let me think. Actually, I think the condition for the equation to represent two lines is that the determinant of the matrix associated with the quadratic form must be zero. That might be a more accurate condition.So, the equation is ( x^2 - m y^2 + 2x + 2y = 0 ). Let me write it in the standard form:( x^2 + 2x - m y^2 + 2y = 0 ).To check if it represents two lines, I need to ensure that the quadratic part can be factored into two linear terms. Alternatively, I can complete the squares for both ( x ) and ( y ) to see if it can be expressed as a difference of squares, which would factor into two lines.Let me try completing the square for ( x ) and ( y ).Starting with the ( x ) terms: ( x^2 + 2x ). To complete the square, I take half of the coefficient of ( x ), which is 1, square it to get 1, and add and subtract it.So, ( x^2 + 2x = (x + 1)^2 - 1 ).Now for the ( y ) terms: ( -m y^2 + 2y ). Let me factor out the ( -m ) first:( -m(y^2 - frac{2}{m} y) ).Now, inside the parentheses, I have ( y^2 - frac{2}{m} y ). To complete the square, take half of ( -frac{2}{m} ), which is ( -frac{1}{m} ), square it to get ( frac{1}{m^2} ), and add and subtract it inside the parentheses.So, ( y^2 - frac{2}{m} y = (y - frac{1}{m})^2 - frac{1}{m^2} ).Putting it back into the equation:( -m[(y - frac{1}{m})^2 - frac{1}{m^2}] = -m(y - frac{1}{m})^2 + frac{1}{m} ).Now, let's substitute these back into the original equation:( (x + 1)^2 - 1 - m(y - frac{1}{m})^2 + frac{1}{m} = 0 ).Simplify the constants:( (x + 1)^2 - m(y - frac{1}{m})^2 - 1 + frac{1}{m} = 0 ).So, ( (x + 1)^2 - m(y - frac{1}{m})^2 = 1 - frac{1}{m} ).Hmm, for this to represent two lines, the right-hand side should be zero because the equation should be a difference of squares equal to zero, which can then be factored into two linear equations. So, set the constant term equal to zero:( 1 - frac{1}{m} = 0 ).Solving for ( m ):( 1 = frac{1}{m} ) ( m = 1 ).Okay, so ( m ) must be 1. Let me check if this makes sense.Substituting ( m = 1 ) back into the equation:( (x + 1)^2 - (y - 1)^2 = 0 ).This can be factored as:( [(x + 1) - (y - 1)][(x + 1) + (y - 1)] = 0 ).Simplify each factor:First factor: ( x + 1 - y + 1 = x - y + 2 ).Second factor: ( x + 1 + y - 1 = x + y ).So, the equation becomes:( (x - y + 2)(x + y) = 0 ).Which indeed represents two lines: ( x - y + 2 = 0 ) and ( x + y = 0 ).Therefore, the value of ( m ) is 1.I think that makes sense. I completed the squares for both ( x ) and ( y ), set the constant term to zero to ensure it's a difference of squares, solved for ( m ), and verified by factoring. So, I'm confident that ( m = 1 ) is the correct answer.**Final Answer**The value of ( m ) is boxed{1}.

answer：Okay, so I need to find the remainder when 444 raised to the power of 444 is divided by 13. Hmm, that seems like a pretty big number, so calculating it directly isn't feasible. I remember something about modular arithmetic and maybe using Fermat's Little Theorem or something like that. Let me think through this step by step.First, I know that when dealing with exponents in modular arithmetic, it's helpful to simplify the base modulo the given number first. So, maybe I should start by finding what 444 is congruent to modulo 13. That should make things easier.Alright, let's calculate 444 divided by 13. I can do this by long division or maybe find a quicker way. Let me see, 13 times 34 is 442 because 13 times 30 is 390, and 13 times 4 is 52, so 390 + 52 is 442. So, 444 minus 442 is 2. That means 444 is congruent to 2 modulo 13. Wait, is that right? Let me double-check. 13 times 34 is indeed 442, and 444 minus 442 is 2. So yes, 444 ≡ 2 mod 13.Wait, hold on, earlier I thought 444 was congruent to 3 mod 13, but now I'm getting 2. Hmm, maybe I made a mistake earlier. Let me recalculate. 13 times 34 is 442, right? So 444 is 442 + 2, so 444 is 2 more than a multiple of 13. So, 444 ≡ 2 mod 13. Okay, so that corrects my earlier mistake. So, 444 is congruent to 2 mod 13, not 3.So, now, the problem reduces to finding 2^444 mod 13. That's still a big exponent, but maybe there's a pattern or cycle in the powers of 2 modulo 13 that I can exploit.I remember that Fermat's Little Theorem says that if p is a prime number and a is not divisible by p, then a^(p-1) ≡ 1 mod p. Since 13 is prime and 2 is not divisible by 13, this theorem applies. So, 2^12 ≡ 1 mod 13. That means that every 12 exponents, the powers of 2 cycle back to 1 modulo 13.So, if I can express 444 as a multiple of 12 plus some remainder, I can simplify 2^444 mod 13. Let's divide 444 by 12. 12 times 37 is 444 because 12 times 30 is 360, and 12 times 7 is 84, so 360 + 84 is 444. So, 444 is exactly 12 times 37, which means 444 = 12*37 + 0. So, the remainder is 0.Therefore, 2^444 = (2^12)^37. Since 2^12 ≡ 1 mod 13, this becomes 1^37 mod 13, which is just 1.Wait, so does that mean the remainder is 1? Let me make sure I didn't skip any steps. So, 444 ≡ 2 mod 13, so 444^444 ≡ 2^444 mod 13. Then, using Fermat's Little Theorem, since 2^12 ≡ 1 mod 13, and 444 is a multiple of 12, 2^444 ≡ (2^12)^37 ≡ 1^37 ≡ 1 mod 13. Yeah, that seems right.But just to be thorough, maybe I should check the cycle of powers of 2 modulo 13 to see if it indeed cycles every 12. Let's compute the powers:2^1 ≡ 2 mod 132^2 ≡ 4 mod 132^3 ≡ 8 mod 132^4 ≡ 16 ≡ 3 mod 13 (since 16 - 13 = 3)2^5 ≡ 2^4 * 2 ≡ 3*2 ≡ 6 mod 132^6 ≡ 6*2 ≡ 12 mod 132^7 ≡ 12*2 ≡ 24 ≡ 11 mod 13 (24 - 13 = 11)2^8 ≡ 11*2 ≡ 22 ≡ 9 mod 13 (22 - 13 = 9)2^9 ≡ 9*2 ≡ 18 ≡ 5 mod 13 (18 - 13 = 5)2^10 ≡ 5*2 ≡ 10 mod 132^11 ≡ 10*2 ≡ 20 ≡ 7 mod 13 (20 - 13 = 7)2^12 ≡ 7*2 ≡ 14 ≡ 1 mod 13Okay, so indeed, 2^12 ≡ 1 mod 13, and the cycle length is 12. So, every 12 exponents, it cycles back. Therefore, 2^444, being 2^(12*37), is 1^37 ≡ 1 mod 13.So, putting it all together, 444^444 ≡ 2^444 ≡ 1 mod 13. Therefore, the remainder is 1.Wait, but earlier I thought 444 was congruent to 3 mod 13, but after recalculating, it's actually 2. So, I need to make sure I didn't confuse the two. Let me confirm once more.444 divided by 13: 13*34 is 442, so 444 - 442 is 2. So, yes, 444 ≡ 2 mod 13. So, that part is correct.Therefore, the remainder is indeed 1. I think that's solid.Alternatively, just to explore another method, maybe using Euler's theorem or something else, but I think Fermat's Little Theorem suffices here because 13 is prime.Another way could be to look for patterns in the exponents. Since 2^12 ≡ 1, as we saw, then 2^(12k) ≡ 1^k ≡ 1 mod 13. Since 444 is 12*37, it's a multiple of 12, so 2^444 ≡ 1 mod 13.Yeah, that seems consistent.I think I'm confident now that the remainder is 1.

answer：Okay, so I have this problem here: "When x is ______, the polynomial x² + 4x + 4 reaches its minimum value." Alright, I need to figure out what value of x makes this polynomial as small as possible. Hmm, let's see.First, I remember that polynomials can have minimum or maximum values, especially quadratic ones like this. Since the coefficient of x² is positive (which is 1 in this case), the parabola opens upwards. That means it has a minimum point and no maximum point. So, I'm looking for the vertex of this parabola because that's where the minimum occurs.Now, how do I find the vertex of a quadratic equation? Well, I think there's a formula for the x-coordinate of the vertex. Let me recall. I believe it's x = -b/(2a), where a and b are the coefficients from the quadratic equation in standard form ax² + bx + c. In this case, a is 1 and b is 4. So plugging those in, x = -4/(2*1) = -4/2 = -2. So, x = -2 should be the value where the polynomial reaches its minimum.Wait, but maybe I should double-check that. Another way to find the minimum is by completing the square. Let me try that method to confirm. Starting with the polynomial x² + 4x + 4. To complete the square, I take the coefficient of x, which is 4, divide it by 2 to get 2, and then square it to get 4. So, I can rewrite the polynomial as (x + 2)². Expanding that gives x² + 4x + 4, which matches the original polynomial. So, the minimum occurs when (x + 2)² is zero because squares are always non-negative. Therefore, (x + 2)² = 0 when x + 2 = 0, which means x = -2.Hmm, both methods gave me the same answer, x = -2. That seems consistent. Maybe I can also find the derivative if I think about calculus, but I'm not sure if that's necessary here. Let me try it anyway. The derivative of x² + 4x + 4 with respect to x is 2x + 4. Setting that equal to zero to find critical points: 2x + 4 = 0 → 2x = -4 → x = -2. So, calculus also confirms that the minimum occurs at x = -2.Alright, so all three methods—vertex formula, completing the square, and calculus—lead me to the same conclusion. It seems pretty solid that x = -2 is where the polynomial reaches its minimum value. I don't think I made any mistakes here, but just to be thorough, let me plug x = -2 back into the original polynomial to see what the value is. Plugging in, we get (-2)² + 4*(-2) + 4 = 4 - 8 + 4 = 0. So, the minimum value is 0, which makes sense because (x + 2)² is zero at x = -2.I guess another way to visualize this is by graphing the polynomial. Since it's a parabola opening upwards, the vertex at (-2, 0) is the lowest point on the graph. Any other x-value would give a higher y-value. For example, if I choose x = -1, the polynomial becomes (-1)² + 4*(-1) + 4 = 1 - 4 + 4 = 1, which is higher than 0. Similarly, at x = 0, it's 0 + 0 + 4 = 4, which is also higher. So, x = -2 indeed gives the smallest value.Wait, just to make sure I didn't mix up anything, let me think about the direction of the parabola again. Since the coefficient of x² is positive, it opens upwards, so the vertex is a minimum point. If it were negative, it would open downwards, and the vertex would be a maximum. But in this case, it's positive, so minimum at x = -2.Is there any other way to approach this problem? Maybe using symmetry? The parabola is symmetric around the vertex. So, if I can find two points with the same y-value equidistant from the vertex, their midpoint would be the x-coordinate of the vertex. Let's see. Suppose I pick x = -3 and x = -1. Plugging in x = -3: (-3)² + 4*(-3) + 4 = 9 - 12 + 4 = 1. Plugging in x = -1: (-1)² + 4*(-1) + 4 = 1 - 4 + 4 = 1. So, both x = -3 and x = -1 give y = 1. The midpoint between -3 and -1 is (-3 + (-1))/2 = -4/2 = -2. So, the axis of symmetry is x = -2, which is the x-coordinate of the vertex. That's another confirmation.I think I've checked this enough ways. All methods point to x = -2 being the point where the polynomial reaches its minimum. I feel confident about this answer now.

answer：Okay, so I need to find the minimum value of the expression ( x^2 + 2xy + y^2 ) where ( x ) and ( y ) are real numbers. Hmm, let me think about how to approach this.First, I remember that quadratic expressions can often be simplified by completing the square. Maybe I can try that here. Let me see, the expression is ( x^2 + 2xy + y^2 ). I notice that this looks similar to the expansion of ( (x + y)^2 ). Let me check that:( (x + y)^2 = x^2 + 2xy + y^2 ).Oh, that's exactly the expression I have! So, ( x^2 + 2xy + y^2 ) simplifies to ( (x + y)^2 ). Now, since squares of real numbers are always non-negative, the smallest value ( (x + y)^2 ) can take is 0. That happens when ( x + y = 0 ). So, if I set ( x + y = 0 ), then ( x = -y ). Let me substitute ( x = -y ) back into the original expression to verify:( (-y)^2 + 2(-y)y + y^2 = y^2 - 2y^2 + y^2 = 0 ).Yep, that works out. So, the minimum value of the expression is 0. I guess another way to think about this is by using calculus. If I treat the expression as a function of two variables, ( f(x, y) = x^2 + 2xy + y^2 ), I can find its critical points by taking partial derivatives and setting them equal to zero.Let's compute the partial derivatives:( frac{partial f}{partial x} = 2x + 2y ),( frac{partial f}{partial y} = 2x + 2y ).Setting both partial derivatives equal to zero:1. ( 2x + 2y = 0 ),2. ( 2x + 2y = 0 ).These equations are the same, so they don't give me any new information. From the first equation, I can solve for one variable in terms of the other:( 2x + 2y = 0 ) ⇒ ( x + y = 0 ) ⇒ ( x = -y ).This is the same result I got earlier by completing the square. So, substituting ( x = -y ) into the function gives me ( f(-y, y) = (-y)^2 + 2(-y)y + y^2 = y^2 - 2y^2 + y^2 = 0 ).To confirm that this critical point is indeed a minimum, I can use the second derivative test for functions of two variables. The second partial derivatives are:( f_{xx} = 2 ),( f_{yy} = 2 ),( f_{xy} = 2 ).The Hessian matrix is:[H = begin{bmatrix}2 & 2 2 & 2 end{bmatrix}]The determinant of the Hessian is ( (2)(2) - (2)(2) = 4 - 4 = 0 ). Hmm, the determinant is zero, which means the second derivative test is inconclusive. That's a bit of a problem. Maybe I need another approach.Wait, since I already know that ( x^2 + 2xy + y^2 = (x + y)^2 ), and squares are always non-negative, the minimum must be zero. So, even though the second derivative test didn't give me a clear answer, the algebraic approach confirms that the minimum is indeed zero.Alternatively, I could consider using the method of Lagrange multipliers, but that might be overcomplicating things since the expression simplifies so nicely. Another thought: maybe I can analyze the expression by fixing one variable and seeing how the expression behaves with respect to the other. Let's say I fix ( y ) and consider the expression as a function of ( x ):( f(x) = x^2 + 2xy + y^2 ).This is a quadratic in ( x ), and since the coefficient of ( x^2 ) is positive, it opens upwards, meaning it has a minimum. The vertex of this parabola occurs at ( x = -frac{b}{2a} ), where ( a = 1 ) and ( b = 2y ). So,( x = -frac{2y}{2(1)} = -y ).Substituting ( x = -y ) back into the expression gives:( (-y)^2 + 2(-y)y + y^2 = y^2 - 2y^2 + y^2 = 0 ).Again, the minimum value is zero. I could also approach this geometrically. The expression ( x^2 + 2xy + y^2 ) represents the square of the distance from the origin to the point ( (x + y, 0) ) in a transformed coordinate system. Since distance can't be negative, the smallest distance is zero, which occurs when ( x + y = 0 ).Wait, maybe that's a bit abstract. Let me think differently. If I consider the expression ( (x + y)^2 ), it's clear that it's a square of a real number, so it's always non-negative. The smallest square is zero, achieved when ( x + y = 0 ). Therefore, the minimum value is zero.I think I've approached this problem from multiple angles: algebraic manipulation by completing the square, calculus using partial derivatives, fixing one variable and analyzing the function, and even a bit of geometric interpretation. All these methods consistently lead me to the conclusion that the minimum value of the expression is zero, achieved when ( x = -y ).Just to make sure I haven't missed anything, let me consider specific values. If I set ( x = 1 ) and ( y = -1 ), then the expression becomes ( 1 + (-2) + 1 = 0 ). If I set ( x = 2 ) and ( y = -2 ), it's ( 4 + (-8) + 4 = 0 ). What if I choose different values where ( x ) isn't exactly the negative of ( y )? For example, ( x = 1 ) and ( y = 0 ): ( 1 + 0 + 0 = 1 ), which is greater than zero. Another example: ( x = 0 ) and ( y = 0 ): ( 0 + 0 + 0 = 0 ). Wait, that's also zero, but in this case, ( x = -y ) since both are zero.So, it seems that whenever ( x = -y ), regardless of the specific values, the expression equals zero, and in all other cases, it's positive. Therefore, zero is indeed the minimum value.I think I'm confident now that the minimum value of ( x^2 + 2xy + y^2 ) over all real numbers ( x ) and ( y ) is zero.

answer：Okay, so I have this problem about a geometric sequence. Let me try to understand it step by step. The sequence is {a_n}, and each term is positive. They tell me that a_2, a_4 + 2, and a_5 form an arithmetic sequence. Also, the first term a_1 is 2. I need to find S_{10} - S_4, where S_n is the sum of the first n terms.First, let me recall what a geometric sequence is. In a geometric sequence, each term is the previous term multiplied by a common ratio, which I'll call q. So, a_n = a_1 * q^{n-1}. Since a_1 is 2, that means a_n = 2 * q^{n-1}.Now, they mention that a_2, a_4 + 2, and a_5 form an arithmetic sequence. An arithmetic sequence has a common difference between consecutive terms. So, the difference between the second and first term should be the same as the difference between the third and second term.Let me write down the terms:- a_2 = 2 * q^{2-1} = 2q- a_4 = 2 * q^{4-1} = 2q^3- a_5 = 2 * q^{5-1} = 2q^4So, the arithmetic sequence is: a_2, a_4 + 2, a_5. That translates to: 2q, 2q^3 + 2, 2q^4.Since these form an arithmetic sequence, the difference between the second and first term should equal the difference between the third and second term. So, let me set up that equation:(2q^3 + 2) - (2q) = (2q^4) - (2q^3 + 2)Let me simplify both sides:Left side: 2q^3 + 2 - 2q = 2q^3 - 2q + 2Right side: 2q^4 - 2q^3 - 2So, the equation is:2q^3 - 2q + 2 = 2q^4 - 2q^3 - 2Let me bring all terms to one side to solve for q:2q^3 - 2q + 2 - 2q^4 + 2q^3 + 2 = 0Combine like terms:-2q^4 + (2q^3 + 2q^3) + (-2q) + (2 + 2) = 0Simplify:-2q^4 + 4q^3 - 2q + 4 = 0Hmm, this looks like a quartic equation. Maybe I can factor out a common term. Let me factor out a -2:-2(q^4 - 2q^3 + q - 2) = 0Divide both sides by -2:q^4 - 2q^3 + q - 2 = 0Now, I need to solve this quartic equation. Maybe I can factor it. Let me try to factor by grouping.Group the first two terms and the last two terms:(q^4 - 2q^3) + (q - 2) = 0Factor out q^3 from the first group:q^3(q - 2) + 1(q - 2) = 0Now, factor out (q - 2):(q - 2)(q^3 + 1) = 0So, either q - 2 = 0 or q^3 + 1 = 0.Since q is positive (given that each term is positive), q cannot be negative. So, q^3 + 1 = 0 would give q = -1, which is negative. So, we discard that.Therefore, q - 2 = 0 => q = 2.Okay, so the common ratio q is 2. Now, I can find the terms of the geometric sequence.But actually, I need to find S_{10} - S_4. Let me recall the formula for the sum of the first n terms of a geometric sequence:S_n = a_1 * (q^n - 1)/(q - 1)Since a_1 = 2 and q = 2, let's compute S_{10} and S_4.First, compute S_{10}:S_{10} = 2 * (2^{10} - 1)/(2 - 1) = 2 * (1024 - 1)/1 = 2 * 1023 = 2046Then, compute S_4:S_4 = 2 * (2^4 - 1)/(2 - 1) = 2 * (16 - 1)/1 = 2 * 15 = 30Therefore, S_{10} - S_4 = 2046 - 30 = 2016.Wait, let me double-check my calculations to make sure I didn't make a mistake.Compute S_{10}:2^{10} is 1024, so 1024 - 1 = 1023. Multiply by 2: 2046. Correct.Compute S_4:2^4 is 16, so 16 - 1 = 15. Multiply by 2: 30. Correct.Subtracting: 2046 - 30 = 2016. That seems right.So, the answer is 2016.**Final Answer**boxed{2016}

answer：Okay, so I have this function f(x) = (x + a + b)(x + a - b)(x - a + b)(x - a - b), and I need to find its minimum value without using calculus. Hmm, that's interesting. I remember that sometimes you can factor expressions or use symmetry to simplify things. Let me see.First, I notice that the function is a product of four linear terms. Maybe I can pair them up in a way that makes it easier to handle. Let me try grouping the first two and the last two terms:f(x) = [(x + a + b)(x + a - b)] * [(x - a + b)(x - a - b)]Okay, each pair looks like a difference of squares. Let me expand each pair:First pair: (x + a + b)(x + a - b) = (x + a)^2 - b^2 = x^2 + 2ax + a^2 - b^2Second pair: (x - a + b)(x - a - b) = (x - a)^2 - b^2 = x^2 - 2ax + a^2 - b^2So now, f(x) becomes:f(x) = (x^2 + 2ax + a^2 - b^2)(x^2 - 2ax + a^2 - b^2)Hmm, this looks like another difference of squares. Let me set y = x^2 + a^2 - b^2. Then the first term is y + 2ax and the second term is y - 2ax. So, multiplying them together:(y + 2ax)(y - 2ax) = y^2 - (2ax)^2 = y^2 - 4a^2x^2Substituting back y = x^2 + a^2 - b^2:f(x) = (x^2 + a^2 - b^2)^2 - 4a^2x^2Let me expand (x^2 + a^2 - b^2)^2:= x^4 + 2x^2(a^2 - b^2) + (a^2 - b^2)^2So, f(x) = x^4 + 2x^2(a^2 - b^2) + (a^2 - b^2)^2 - 4a^2x^2Combine like terms:= x^4 + [2(a^2 - b^2) - 4a^2]x^2 + (a^2 - b^2)^2Simplify the coefficient of x^2:2(a^2 - b^2) - 4a^2 = 2a^2 - 2b^2 - 4a^2 = -2a^2 - 2b^2So, f(x) = x^4 - 2(a^2 + b^2)x^2 + (a^2 - b^2)^2Hmm, this is a quartic function, but it's quadratic in terms of x^2. Maybe I can let z = x^2 and rewrite the function:f(z) = z^2 - 2(a^2 + b^2)z + (a^2 - b^2)^2Now, this is a quadratic in z. The minimum of a quadratic function az^2 + bz + c is at z = -b/(2a). So, applying that here:z = [2(a^2 + b^2)] / 2 = a^2 + b^2So, the minimum occurs at z = a^2 + b^2. But z = x^2, so x^2 = a^2 + b^2, which means x = sqrt(a^2 + b^2) or x = -sqrt(a^2 + b^2). Since we're looking for the minimum value, we can plug this back into f(z):f(z) = (a^2 + b^2)^2 - 2(a^2 + b^2)(a^2 + b^2) + (a^2 - b^2)^2Simplify term by term:First term: (a^2 + b^2)^2 = a^4 + 2a^2b^2 + b^4Second term: -2(a^2 + b^2)^2 = -2a^4 - 4a^2b^2 - 2b^4Third term: (a^2 - b^2)^2 = a^4 - 2a^2b^2 + b^4Now, add them all together:(a^4 + 2a^2b^2 + b^4) + (-2a^4 - 4a^2b^2 - 2b^4) + (a^4 - 2a^2b^2 + b^4)Combine like terms:a^4 - 2a^4 + a^4 = 02a^2b^2 - 4a^2b^2 - 2a^2b^2 = -4a^2b^2b^4 - 2b^4 + b^4 = 0So, f(z) = -4a^2b^2Therefore, the minimum value of f(x) is -4a^2b^2.