Question

Tính dùm tui bài 1 ik phép chia phân thức. ĐANG CẦN GẤP

Answer 1

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) \(\dfrac{2-x}{1-x^4}=\left(2-x\right):\left(1-x^4\right)=\dfrac{2}{\dfrac{x}{x^4}}\)

Answer 2

Bài 1:

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

Answer 3

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