u 4 - v 4 = u 2 - v 2 u 2 + v 2

= (u − v)(u + v)(u − iv)(u + iv)

a)  u 2  +  v 2  =  u 2  − ( i v ) 2  = (u − iv)(u + iv)

b)  u 4  −  v 4 = ( u 2  −  v 2 )( u 2  +  v 2 )

= (u − v)(u + v)(u − iv)(u + iv)

Sửa: `+12y^2 -> -12y^2`

\(7x^2-5xy-12y^2\\ =7x^2+7xy-12xy-12y^2\\ =7x\left(x+y\right)-12y\left(x+y\right)\\ =\left(7x-12y\right)\left(x+y\right)\)

`a^2 + ab + 2a + 2b = a(a+2) + b(a+2) = (a+b)(a+2)`

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

\(=x^2-2xy\)

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

Lời giải:

$2x^2-5x-6=2(x^2-\frac{5}{2}x+\frac{5^2}{4^2})-\frac{73}{8}$

$=2(x-\frac{5}{4})^2-\frac{73}{8}$

$=2(x-\frac{5}{4}-\frac{\sqrt{73}}{4})(x-\frac{5}{4}+\frac{\sqrt{73}}{4})$

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

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

\(=x\left(x-\sqrt{3}y\right)+\sqrt{3}y\left(x-\sqrt{3}y\right)\)

\(=\left(x-\sqrt{3}y\right)\left(x+\sqrt{3}y\right)\)

Sai mô kkk?