![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^4+4x^2-5\)
\(=\left[\left(x^2\right)^2+2.x^2.2+2^2\right]-9\)
\(=\left(x^2+2\right)^2-9\)
\(=\left(x^2+2+3\right)\left(x^2+2-3\right)\)
\(=\left(x^2+5\right)\left(x^2-1\right)\)
\(=\left(x^2+5\right)\left(x+1\right)\left(x-1\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
\(1,Sửa:x^3-2x^2+x=x\left(x^2-2x+1\right)=x\left(x-1\right)^2\\ 2,=6\left(x^2+2xy+y^2\right)=6\left(x+y\right)^2\\ 3,=2y\left(y^2+4y+4\right)=2y\left(y+2\right)^2\\ 4,=5\left(x^2-2xy+y^2\right)=5\left(x-y\right)^2\)
Bài 2:
\(1,=x\left(x^2-64\right)=x\left(x-8\right)\left(x+8\right)\\ 2,=2y\left(4x^2-9\right)=2y\left(2x-3\right)\left(2x+3\right)\\ 3,=3\left(x^3-1\right)=3\left(x-1\right)\left(x^2+x+1\right)\)
Bài 3:
\(a,=5\left(x^2+2x+1-y^2\right)=5\left[\left(x+1\right)^2-y^2\right]=5\left(x-y+1\right)\left(x+y+1\right)\\ b,=3x\left(x^2-2x+1-4y^2\right)=3x\left[\left(x-1\right)^2-4y^2\right]\\ =3x\left(x-2y-1\right)\left(x+2y-1\right)\\ c,=ab\left(a-b\right)\left(a+b\right)+\left(a+b\right)^2\\ =\left(a+b\right)\left(a^2b-ab^2+a+b\right)\\ d,=2x\left(x^2-y^2-4x+4\right)=2x\left[\left(x-2\right)^2-y^2\right]\\ =2x\left(x-y-2\right)\left(x+y-2\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^6+2x^5+x^4-2x^3-2x^2+1=\left(x^3+x^2-1\right)^2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
x4 + x3 + 2x2 + x + 1
= (x4 + 2x2 + 1) + (x3 + x)
= (x2 + 1)2 + x (x2 + 1)
= (x2 + 1) ( x2 + 1 + x)
= (x2 + 1) (x + 1)2
![](https://rs.olm.vn/images/avt/0.png?1311)
\(2x^2+2y^2-x^2z+z-y^2z-2\)
\(=\left(2x^2-x^2z\right)+\left(2y^2-y^2z\right)-\left(2-z\right)\)
\(=x^2\left(2-z\right)+y^2\left(2-z\right)-\left(2-z\right)\)
\(=\left(2-z\right)\left(x^2+y^2-1\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^3-2x^2=x^2\left(x-2\right)\)
\(y^2+2y+1-x^2=\left(y+1\right)^2-x^2=\left(y+1-x\right)\left(y+1+x\right)\)
\(x^2-x-6=x^2+2x-3x-6=x\left(x+2\right)-3\left(x+2\right)=\left(x-3\right)\left(x+2\right)\)
\(x^6-x^4+2x^3+2x\)
\(=x^5x-x^3x+2x^2x+2x\)
\(=x\left(x^5-x^3+2x^2+2\right)\)