Tính dùm tui bài 1 ik phép chia phân thức. ĐANG CẦN GẤP
a)\(\dfrac{-x^2+4x-4}{x^2-1}\\ =\dfrac{-\left(x^2-4x+4\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{-\left(x-2\right)^2}{\left(x-1\right)\left(x+1\right)}\)
b) \(\dfrac{2-x}{1-x^4}=\left(2-x\right):\left(1-x^4\right)=\dfrac{2}{\dfrac{x}{x^4}}\)
Bài 1:
\(=-\dfrac{x^2+4x-4}{x^2-1}.\dfrac{1-x^4}{2-x}=\dfrac{-\left(x^2-4x+4\right)}{\left(x+1\right)\left(x-1\right)}.\dfrac{-\left(x+1\right)\left(x-1\right)\left(1+x^2\right)}{2-x}\)\(=1+x^2\)
bài 2:2+(3+x/(x+1))/(2-3x)=2
=>(3+x/(x+1))/(2-3x)=0
=>3+x/(x+1)=0
=>(3x+3+x)/(x+1)=0
=>(4x+3)/(x+1)=0
=>4x+3=0
=>x=-4/3
