câu hỏi đâu ?????

câu hỏi đâu bạn 

x\(^2\)-2xy+y\(^2\)

=x\(^2\)-2.x.y+y\(^2\)

=(x-y)\(^2\)

=(x-y)(x+y)

@ĐỗPhươngThanh

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

mới học vioeud à

\(x^3-27+\left(x-3\right).\left(x+9\right)=0\)

\(\Rightarrow\left(x-3\right).\left(x^2+3x+9\right)+\left(x-3\right).\left(x+9\right)=0\)

\(\Rightarrow\left(x-3\right).\left(x^2+3x+9+x+9\right)=0\)

\(\Rightarrow\left(x-3\right).\left(x^2+4x+18\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-3=0\\x^2+4x+18=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\\left(x+2\right)^2+14=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\\left(x+2\right)^2=-14\text{(Loại)}\end{cases}}\)

Vậy \(x=3\) 

\(1+27x^3\)

\(=\left(1+3x\right).\left(1-3x+9x^2\right)\)

\(x^3+3x^2-16x-48\)

\(=x^2.\left(x+3\right)-16.\left(x+3\right)\)

\(=\left(x^2-16\right).\left(x+3\right)\)

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

        Bài làm :

Ta có :

6x³ - 11x² - x - 2

= 6x³ - 12x² + x² - 2x + x - 2

= ( 6x³ - 12x² ) + ( x² - 2x ) + ( x - 2 )

= 6x²( x - 2 ) + x( x - 2 ) + 1( x - 2 )

= ( x - 2 )( 6x² + x + 1 ) nhớ tích