\(1,=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\\ 2,=\left(x+y\right)^3\\ 3,=\left(2y-z\right)\left(4x+7y\right)\\ 4,=\left(x+2\right)^2\\ 5,Sửa:x\left(x-2\right)-x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

1D  2C

Câu 1: D

Câu 2: C

\(4x^4+4x^3-x^2-x\)

\(=4x^3\left(x+1\right)-x\left(x+1\right)\)

\(=\left(x+1\right)\left(4x^3-x\right)\)

\(=x\left(x+1\right)\left(4x^2-1\right)\)

\(=x\left(x+1\right)\left[\left(2x\right)^2-1\right]\)

\(=x\left(x+1\right)\left(2x+1\right)\left(2x-1\right)\)

(Nhớ k cho mình với nhá!)

vậy x= 0 ,25

=>x=0,25

nhớ cho mik nha

#)Giải :

\(x^3-2x-4\)

\(=x^3+2x^2-2x^2+2x-4x-4\)

\(=x^3+2x^2+2x-2x^2-4x-4\)

\(=x\left(x^2+2x+2\right)-2\left(x^2+2x+2\right)\)

\(=\left(x-2\right)\left(x^2+2x+2\right)\)

\(x^4+2x^3+5x^2+4x-12\)

\(=x^4+x^3+6x^2+x^3+x^2+6x-2x^2-2x-12\)

\(=x^2\left(x^2+x+6\right)+x\left(x^2+x+6\right)-2\left(x^2+x+6\right)\)

\(=\left(x^2+x+6\right)\left(x^2+x-2\right)\)

\(=\left(x^2+x+6\right)\left(x-1\right)\left(x+2\right)\)

Câu 1.

Đoán được nghiệm là 2.Ta giải như sau:

\(x^3-2x-4\)

\(=x^3-2x^2+2x^2-4x+2x-4\)

\(=x^2\left(x-2\right)+2x\left(x-2\right)+2\left(x-2\right)\)

\(=\left(x-2\right)\left(x^2+2x+2\right)\)

cảm ơn nha!

x(1-2x)(1+2x)

\(=x\left(1-4x^2\right)=x\left(1-2x\right)\left(1+2x\right)\)

\(x^3-4x=x\left(x^2-4\right)=x\left(x+2\right)\left(x-2\right)\)

\(x^3+4x^2+4x+1\)

\(=x^3+3x^2+x+x^2+3x+1\)

\(=x\left(x^2+3x+1\right)+\left(x^2+3x+1\right)\)

\(=\left(x+1\right)\left(x^2+3x+1\right)\)

\(=x^2+3x+x+3=\left(x+3\right)\left(x+1\right)\)

= x^2+x+3x+3

=x(x+1)+3(x+1)

=(x+1)(x+3)

Làm thế đúng  không nhỉ?

\(=x^3 -3x^2 +3x-1 +x-1\\=(x-1)^3 +(x-1)\\=(x-1)[(x-1)^2 +1]\)

\(x^3-3x^2+4x-2=x^2\left(x-1\right)-2x\left(x-1\right)+2\left(x-1\right)=\left(x-1\right)\left(x^2-2x+2\right)\)