Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(x^4+4\)
= \(\left(x^2+2\right)^2-4x^2\)
= \(\left(x^2-2x+2\right)\left(x^2+2x+2\right)\)
b, \(x^5+x+1=x^5-x^2+x^2+x+1\\ =x^2\left(x^3-1\right)+\left(x^2+x+1\right)\\ =x^2\left(x-1\right)\left(x^2+x+1\right)+\left(x^2+x+1\right)\\ =\left(x^2+x+1\right)\left[x^2\left(x+1\right)+1\right]\\ =\left(x^2+x+1\right)\left(x^3-x^2+1\right)\)
Ta có:
(x+a)(x+2a)(x+3a)(x+4a) + a4
=(x+a)(x+4a)(x+3a)(x+2a) +a4
=(x2+5ax+4a2)(x2+5ax+6a2) + a4
Đặt x2+5ax+5a2=y
=>(x2+5ax+4a2)(x2+5ax+6a2) + a4=(y-a2)(y+a2)+a4
=y2-a4+a4
=y2
=(x2+5ax+5a2)2
k mik nha
\(1,=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\\ 2,=\left(x+y\right)^3\\ 3,=\left(2y-z\right)\left(4x+7y\right)\\ 4,=\left(x+2\right)^2\\ 5,Sửa:x\left(x-2\right)-x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
\(x^4+4=x^4+4x^2+4-4x^2\)
\(=\left(x^2+2\right)^2-\left(2x\right)^2\)
\(=\left(x^2-2x+2\right)\left(x^2+2x+2\right)\)