\(3x\left(x-7\right)+2xy-14y\)

\(=3x\left(x-7\right)+2y\left(x-7\right)\)

\(=\left(x-7\right)\left(3x+2y\right)\)

#\(Toru\)

b) Ta có: \(x^3-x^2y-xy^2+y^3\)

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

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

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

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

(x-1)(3x-4x)

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

3x² - 3xy

= 3x ( x - y )

`3x-3y+a(x-y)`

`=3(x-y)+a(x-y)`

`=(x-y)(a+3)`

\(3x-3y+a\left(x-y\right)=\left(x-y\right)\left(a+3\right)\)

`3x^2 - 6x`

`= 3x.x-3x.2`

`=3x(x-2)`

Câu 1:

\(4x^2+16x-9\)

\(=4x^2+18x-2x-9\)

\(=2x\left(2x+9\right)-\left(2x+9\right)\)

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

Câu 2:

\(6x^2-11x+3=0\)

\(\Leftrightarrow6x^2-2x-9x+3=0\)

\(\Leftrightarrow2x\left(3x-1\right)-3\left(3x-1\right)=0\)

\(\Leftrightarrow\left(2x-3\right)\left(3x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=3\\3x=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)

\(=3x\cdot\left(x-5y\right)+2y\cdot\left(x-5y\right)\)

\(=\left(x-5y\right)\left(3x+2y\right)\)

\(3x\left(x-5y\right)-2y\left(5y-x\right)\)

\(=3x\left(x-5y\right)+2y\left(x-5y\right)\)

\(=\left(x-5y\right)\left(3x+2y\right)\)

\(-3x^2+4x-2020\)

\(=-3\left(x^2-\frac{4}{3}x+\frac{2020}{3}\right)\)

\(=-3\left(x^2-\frac{4}{3}x+\frac{4}{9}+\frac{6056}{9}\right)\)

\(=-3\left[\left(x-\frac{2}{3}\right)^2+\frac{6056}{9}\right]\)

\(=-3\left(x-\frac{2}{3}\right)^2-\frac{6056}{3}\ge-\frac{6056}{3}\)

(Dấu "=" \(\Leftrightarrow x-\frac{2}{3}=0\Leftrightarrow x=\frac{2}{3}\))