`x^2 -4x+4-y^2`

`=(x^2 -4x+4)-y^2`

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

`=(x-2-y)(x-2+y)`

`x^2+2xy+y^2-x-y`

`=(x^2+2xy+y^2) -(x+y)`

`=(x+y)^2 -(x+y)`

`=(x+y)(x+y-1)`

`x^2-2xy+y^2-9`

`=(x^2-2xy+y^2)-3^2`

`=(x-y)^2-3^3`

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

Tách ra đi cậu.

chia dọc đi bé oi=)))

Bên cạnh đó, một số tính năng mới như tạo Hội/nhóm để cùng thi đua

`->` Trời ước mơ của em;-;;;; em cũng mong đổi tên nữa để giống với nhóm mong ra sớm ;-;;;.

\(\dfrac{1}{x^2-3x+2}-\dfrac{1}{x-2}=2\left(đkxđ:x\ne1;x\ne2\right)\\ \Leftrightarrow\dfrac{1}{\left(x-2\right)\left(x-1\right)}-\dfrac{1}{x-2}=2\\ \Leftrightarrow\dfrac{1}{\left(x-2\right)\left(x-1\right)}-\dfrac{x-1}{\left(x-2\right)\left(x-1\right)}=\dfrac{2\left(x-2\right)\left(x-1\right)}{\left(x-2\right)\left(x-1\right)}\\ \Rightarrow1-x+1=\left(2x-4\right)\left(x-1\right)\\ \Leftrightarrow2-x=2x^2-2x-4x+4\\ \Leftrightarrow-x-2x^2+2x+4x=4-2\\ \Leftrightarrow-2x^2+5x-2=0\\ \Leftrightarrow-\left(2x^2-x-4x+2\right)=0\\ \Leftrightarrow-\left[\left(2x^2-x\right)-\left(4x-2\right)\right]=0\)

\(\Leftrightarrow-\left[x\left(2x-1\right)-2\left(2x-1\right)\right]=0\\ \Leftrightarrow-\left(2x-1\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}-2x+1=0\\x-2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}-2x=-1\\x=2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=2\end{matrix}\right.\)

\(\dfrac{1}{x-2}+\dfrac{1}{x^2+x-6}=\dfrac{1}{x+3}-1\left(đkxđ:x\ne2;x\ne-3\right)\\ \Leftrightarrow\dfrac{1}{x-2}+\dfrac{1}{\left(x-2\right)\left(x+3\right)}=\dfrac{1}{x+3}-1\\ \Leftrightarrow\dfrac{x+3}{\left(x-2\right)\left(x+3\right)}+\dfrac{1}{\left(x-2\right)\left(x+3\right)}=\dfrac{x-2}{\left(x-2\right)\left(x+3\right)}-\dfrac{\left(x-2\right)\left(x+3\right)}{\left(x-2\right)\left(x+3\right)}\\ \Rightarrow x+3+1=x-2-\left(x^2+3x-2x-6\right)\\ \Leftrightarrow x+4=x-2-x^2-3x+2x+6\\ \Leftrightarrow x-x+x^2+3x-2x=-2+6-4\\ \Leftrightarrow x^2+x=0\)

\(\Leftrightarrow x\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)

`1/(x-2) -x/(x^2-4)`

`= 1/(x-2) - x/((x-2)(x+2))`

`= (x+2)/((x-2)(x+2))  - x/((x-2)(x+2))`

`= (x+2-x)/((x-2)(x+2)) `

`= 2/(x-4)`

Bài `3`

\(a,đkxđ:\left\{{}\begin{matrix}x-3\ne0\\x-2\ne0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x\ne3\\x\ne2\end{matrix}\right.\)

\(b,A=\dfrac{x^2-4}{\left(x-3\right)\left(x-2\right)}\\ =\dfrac{\left(x-2\right)\left(x+2\right)}{\left(x-3\right)\left(x-2\right)}\\ =\dfrac{x+2}{x-3}\)

Bài `2`

`a,` không rõ .

`b,9x^2-6x+1`

`=(3x)^2 - 2.3x.1+1^2`

`=(3x-1)^2`

`c,x^3-9x+2x^2y+xy^2`

`= x(x^2 -9+ 2xy +y^2)`

`=x[(x^2 +2xy+y^2)-9]`

`=x[(x+y)^2 -3^2]`

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