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

\(=\frac{\left(x+2\right)\left(x+2\right)}{2\left(x-2\right)\left(x+2\right)}-\frac{8x}{2\left(x-2\right)\left(x+2\right)}\)

\(=\frac{x^2+4x+4-8x}{2\left(x-2\right)\left(x+2\right)}==\frac{x^2-4x+4}{2\left(x-2\right)\left(x+2\right)}\)

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

+) \(\frac{2x+6}{3x^2-x}-\frac{x+3}{1-3x}=\frac{2x+6}{x\left(3x-1\right)}-\frac{-x-3}{3x-1}\)

\(=\frac{2x+6}{x\left(3x-1\right)}-\frac{x\left(-x-3\right)}{x\left(3x-1\right)}\)

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

a)\(\frac{x+2}{2x-4}-\frac{4x}{x^2-4}\)

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

\(=\frac{\left(x+2\right)^2-8x}{2\left(x+2\right)\left(x-2\right)}\)

\(=\frac{x^2+4x+4-8x}{2\left(x+2\right)\left(x-2\right)}\)

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

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

\(=\frac{x-2}{2\left(x+2\right)}\)

b)\(\frac{2x+6}{3x^2-x}-\frac{x+3}{1-3x}\)

\(=\frac{2x+6}{x\left(3x-1\right)}-\frac{-x-3}{3x-1}\)

\(=\frac{2x+6+x\left(x+3\right)}{x\left(3x-1\right)}\)

\(=\frac{2x+6+x^2+3x}{x\left(3x-1\right)}\)

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

#Hoctot

\(\frac{x^3+x^2+x+1}{3x^2+6x+3}=\frac{x^2\left(x+1\right)+\left(x+1\right)}{3x^2+3x+3x+3}\)

\(=\frac{\left(x^2+1\right)\left(x+1\right)}{3x\left(x+1\right)+3\left(x+1\right)}=\frac{\left(x^2+1\right)\left(x+1\right)}{\left(3x+3\right)\left(x+1\right)}\)

\(=\frac{\left(x^2+1\right)\left(x+1\right)}{3\left(x+1\right)^2}=\frac{x^2+1}{3\left(x+1\right)}\)

\(\frac{x^3+x^2+x+1}{3x^2+6x+3}=\frac{x^2.\left(x+1\right)+\left(x+1\right)}{3.\left(x^2+2x+1\right)}=\frac{\left(x+1\right)\left(x^2+1\right)}{3.\left(x+1\right)^2}=\frac{x^2+1}{3.\left(x+1\right)}\)

ai lm đúng mk cho 2 k nha

trang (50) nha cac ban

trang 50 sách nào hả người anh em

MB² + BC² = AM² = a²

Bài1  : 

a, \(x^2-x-4y^2-2y=\left(x^2-4y^2\right)-\left(x+2y\right)\)

\(=\left(x-2y\right)\left(x+2y\right)-\left(x+2y\right)=\left(x+2y\right)\left[x-2y-1\right]\)

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

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

Bài 2 : 

Ta có : \(x^2-5x+xy-5y=x^2+xy-5x-5y\)

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

Thay x = 2019 ; y = -2020 ta được : 

\(\left(2019-5\right)\left(2019-2020\right)=2014.\left(-1\right)=-2014\)

Suy ra  \(\left(3x^4-2x^3+x^2-2x+2\right):\left(x-2\right)=3x^3+4x^2+9x+16\)dư 34

Vậy ta chọn B