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

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

\(=3x^2-3y^2-2x^2+4xy-2y^2\)

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

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

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

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

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

\(2x^2+5x-3=\left(2x^2-x\right)+\left(6x-3\right)\)\(=x\left(2x-1\right)+3\left(2x-1\right)=\left(x+3\right)\left(2x-1\right)\)

2x²+5x-3

=2x²-x+6x-3

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

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

dễ lém nha:

2x²+5x-3=2x²-x+6x-3

             =x(2x-1)+3(2x-1)

             =(x+3)(2x-1)

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

\(a^2-4b^2\)

= \(\left(a-2b\right)\left(a+2b\right)\)

\(a^2-4b^2=a^2-\left(2b\right)^2=\left(a-2b\right)\left(a+2b\right)\)

a) \(a^2-4b^2\)

= \(\left(a-4b\right).\left(a+4b\right)\)

= \(a^2\)

Xóa

\(5x\left(x-2\right)-x+2=0\)

\(\Rightarrow5x\left(x-2\right)-\left(x-2\right)=0\)

\(\Rightarrow\left(5x-1\right)\left(x-2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}5x-1=0\\x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=2\end{matrix}\right.\)

Vậy x=1/5 hoặc x=2

a: Ta có: \(5x\left(x-2\right)-x+2=0\)

\(\Leftrightarrow\left(x-2\right)\left(5x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{5}\end{matrix}\right.\)

Lần sau bạn cần chú ý viết đầy đủ yêu cầu đề bài.

Coi đây là bài toán rút gọn/ khai triển.

Ta có:

$(1+2x)(1-2x)-x(x+2)(x-2)$

$=1^2-(2x)^2-x(x^2-2^2)$

$=1-4x^2-x(x^2-4)=-x^3-4x^2+4x+1$

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

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

\(=-x^3+x^2-5x^2+5x-x+1\)

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

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

\(=\left(1-x\right)\left(x^2+\dfrac{2x.5}{2}+\dfrac{25}{4}-\dfrac{21}{4}\right)\)

\(=\left(1-x\right)\left(\left(x+\dfrac{5}{2}\right)^2-\dfrac{21}{4}\right)\)

\(=\left(1-x\right)\left(x+\dfrac{5}{2}-\dfrac{\sqrt{21}}{2}\right)\left(x+\dfrac{5}{2}+\dfrac{\sqrt{21}}{2}\right)\)

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

(x² - x + 1)(x² + 2x -1)