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

1.A

2.C

3.B

4.C

a

c

b

c

(x-1)y^2-4(x-1)y

x²-y²-x-y=(x²-y²)-(x+y)

               =(x+y)(x-y)-(x+y)

               =(x+y)(x-y-1)

X2-y2xy=(x2-y2)-y+x

=X*y x-y -x+ý

-X+ý*x-y-1

1D  2C

Câu 1: D

Câu 2: C

#)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!

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

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

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

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

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

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

(Nhớ k cho mình với nhá!)

vậy x= 0 ,25

=>x=0,25

nhớ cho mik nha

Sửa đề:

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

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

\(\Leftrightarrow3\left(x+4\right)-x\left(x+4\right)\)

\(\Leftrightarrow\left(3-x\right)\left(x+4\right)\)

Chọn D.

x 4  + 8x = x( x 3 +8)= x( x 3 + 2 3 ) = x(x + 2)(  x 2 − 2x + 4)