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

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

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

\(a,=-\left(x-1\right)^3\left[=\left(1-x\right)^3\right]\\ b,=\left(1-x\right)^3\)

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

b. \(=\left(1-x\right)^3\)

\(3x^2-6x+9x^2=12x^2-6x=6x\left(2x-1\right)\)

\(=-3\left(x^2-\dfrac{10}{3}x+\dfrac{5}{3}\right)\\ =-3\left(x^2-2\cdot\dfrac{5}{3}x+\dfrac{25}{9}-\dfrac{10}{9}\right)\\ =\dfrac{10}{3}-3\left(x-\dfrac{5}{3}\right)^2\\ =\left[\sqrt{\dfrac{10}{3}}-\sqrt{3}\left(x-\dfrac{5}{3}\right)\right]\left[\sqrt{\dfrac{10}{3}}+\sqrt{3}\left(x-\dfrac{5}{3}\right)\right]\\ =\left(\dfrac{\sqrt{30}}{3}+\dfrac{5\sqrt{3}}{3}-x\sqrt{3}\right)\left(\dfrac{\sqrt{30}}{3}-\dfrac{5\sqrt{3}}{3}+x\sqrt{3}\right)\)

\(=\left(\dfrac{\sqrt{30}+5\sqrt{3}}{3}-x\sqrt{3}\right)\left(\dfrac{\sqrt{30}+5\sqrt{3}}{3}-x\sqrt{3}\right)\)

\(-3x^2+10x-5\)

\(=-3\left(x^2-\dfrac{10}{3}x+\dfrac{5}{3}\right)\)

\(=-3\left(x^2-2\cdot x\cdot\dfrac{5}{3}+\dfrac{25}{9}-\dfrac{10}{9}\right)\)

\(=-3\left(x-\dfrac{5+\sqrt{10}}{3}\right)\left(x-\dfrac{5-\sqrt{10}}{3}\right)\)

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

\(=2x^2-2x+5x-5\)

\(=2x\left(x-1\right)+5\left(x-1\right)\)

\(=\left(2x+5\right)\left(x-1\right)\)

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

`

` x^2 – 3x + 2`

`= x^2 – x – 2x + 2` (Tách `–3x = – x – 2x)`

`= (x^2 – x) – (2x – 2)`

=\(x^2\) -8x+5x-40

=x (x-8)+5 (x-8)

=(x-8) (x+5)

x² - 3x - 40

= x² - 8x + 5x - 40

= x(x-8) + 5(x-8)

= (x-8)(x+5)

`a, x2 + 3x - 40`

`= x2 - 8x + 5x - 40`

`= x ( x - 8 ) + 5 ( x - 8 )`

`= ( x - 8 ) ( x + 5 )`

bằng phương thức nào bạn ?

câu trả lời:

3x.x-13.x+6²

bằng phương pháp đặt nhân tử chung

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

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

3(x³-y²)