Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(x^3+5x^2+5x+1\)
\(=\left(x+1\right)\left(x^2+x+1\right)+5x\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2+6x+1\right)\)
\(3x\left(x+1\right)^2-5x^2\left(x+1\right)+7\left(x+1\right)\)
\(=\left(x+1\right)\left(3x^2+3x-5x^2+7\right)\)
\(=\left(x+1\right)\left(-2x^2+3x+7\right)\)
\(4x\left(x+1\right)^2-5x^2\left(x+1\right)-4\left(x+1\right) =\left(x+1\right)\left[4x\left(x+1\right)-5x^2-4\right]=\left(x+1\right)\left(4x^2+4x-5x^2-4\right)=\left(x+1\right)\left(-x^2+4x-4\right)=-\left(x+1\right)\left(x-2\right)^2\)
\(4x\left(x+1\right)^2-5x^2\left(x+1\right)-4\left(x+1\right)\)
\(=4x\left(x^2+2x+1\right)-5x^2\left(x+1\right)-4\left(x+1\right)\)
\(=4x^3+8x^2+4x-5x^3-5x-4x-4\)
\(=-x^3+8x^2-5x-4\)
-Đặt \(t=\left(x^2-x+1\right)\)
\(\left(x^2-x+1\right)^2-5x\left(x^2-x+1\right)+4x^2\)
\(=t^2-5xt+4x^2\)
\(=t^2-4xt-xt+4x^2\)
\(=t\left(t-4x\right)-x\left(t-4x\right)\)
\(=\left(t-4x\right)\left(t-x\right)\)
\(=\left(x^2-x+1-4x\right)\left(x^2-x+1-x\right)\)
\(=\left(x^2-5x+1\right)\left(x^2-2x +1\right)\)
\(=\left(x^2-5x+1\right)\left(x-1\right)^2\)
\(x^4-5x^2+4=\left(x^2-4\right)\left(x^2-1\right)=\left(x-2\right)\left(x+2\right)\left(x-1\right)\left(x+1\right)\)
\(\left(x^2+5x\right)^2-2\left(x^2+5x\right)-24\\ =\left[\left(x^2+5x\right)^2-6\left(x^2+5x\right)\right]+\left[4\left(x^2+5x\right)-24\right]\\ =\left(x^2+5x\right)\left(x^2+5x-6\right)+4\left(x^2+5x-6\right)\\ =\left(x^2+5x-6\right)\left(x^2+5x+4\right)\\ =\left(x^2-x+6x-6\right)\left(x^2+4x+x+4\right)\\ =\left[x\left(x-1\right)+6\left(x-1\right)\right]+\left[x\left(x+4\right)+\left(x+4\right)\right]\\ =\left(x-1\right)\left(x+1\right)\left(x+4\right)\left(x+6\right)\)
`#3107.101107`
a)
`A = 2x^2 + 5x^3 + x^2y`
`= x^2 * (2 + 5x + y)`
b)
`5x(x - 1) + 15(x - 1)`
`= (5x + 15)(x - 1)`
`= 5(x + 3)(x - 1)`
\(=x^2-5x+\dfrac{25}{4}-\dfrac{29}{4}\)
\(=\left(x-\dfrac{5}{2}\right)^2-\dfrac{29}{4}\)
\(=\left(x-\dfrac{5}{2}-\dfrac{\sqrt{29}}{2}\right)\left(x-\dfrac{5}{2}+\dfrac{\sqrt{29}}{2}\right)\)