(3x-2)³+(2x-3)³-(5x-5)³=0

=> (3x-2)³ =0 => 3x-2=0 => 3x=0+2 => 3x=2 => x= 2:3 = 2/3 

=> (2x-3)³=0 => 2x-3=0 => 2x=0+3 => 2x=3 => x=3:2 = 3/2

=>(5x-5)³=0 => 5x-5=0 => 5x=0+5 => 5x=5 => x=5:5 = 1

Vậy x ={2/3; 3/2; 1}

HT và $$$

CHẮC LÀ PHẢI CHIM TO PHẢI CHIM TOOOOOOOOOOOOOOOOOOOOOOOOOOOO

\(x^2+5y^2-4xy-4x+2y+13\)

\(=x^2+4y^2+4-4xy-4x+8y+y^2-6y+9\)

\(=\left(x-2y-2\right)^2+\left(y-3\right)^2=0\)

\(\Leftrightarrow\hept{\begin{cases}x-2y-2=0\\y-3=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=8\\y=3\end{cases}}\)

Suy ra \(M=-1+1=0\)

Chọn A. 

Câu 1: Thực hiện:

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

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

Câu 2: Khai triển hằng đẳng thức:

\(\left(2x+1\right)^2=\left(2x\right)^2+2.2x.1+1^2=4x^2+4x+1\)

\(x^3-8=\left(x^2\right)^3-2^3=\left(x-2\right).\left(x^2+x.2+2^2\right)=\left(x-2\right).\left(x^2+2x+4\right)\)

Câu 3: Phân tích đa thức thành nhân tử:

\(2x^2-4x+8=2.\left(x^2-2x+4\right)=2.\left(x-2\right)^2\)

\(12x^2y+12xy+6x+3y+3\)

\(=3.\left(4x^2y+4xy+2x+y+1\right)\)

\(=3.\left(4x^2y+2xy+2xy+2x+y+1\right)\)

\(=3.[\left(4x^2y+2xy+2x\right)+\left(2xy+y+1\right)]\)

\(=3.[2x.\left(2xy+y+1\right)+\left(2xy+y+1\right)]\)

\(=3.\left(2x+1\right).\left(2xy+y+1\right)\)

x² + x +4x +5