Question

Chứng minh đẳng thức sau

a) \(\dfrac{x-2}{x+1}=\dfrac{x^2-3x+2}{x^2-1}\)với x khác cộng trừ 1

b) \(\dfrac{4u^3-u}{5-10u}=-\dfrac{2u^3+u}{5}\), u khác \(\dfrac{1}{2}\)

Answer 1

a: \(\dfrac{x^2-3x+2}{x^2-1}=\dfrac{\left(x-2\right)\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{x-2}{x+1}\)

Answer 2

\(a,VP=\dfrac{\left(x-1\right)\left(x-2\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{x-2}{x+1}=VP\\ b,VT=\dfrac{u\left(4u^2-1\right)}{5\left(1-2u\right)}=\dfrac{-u\left(1-2u\right)\left(1+2u\right)}{5\left(1-2u\right)}=\dfrac{-u\left(1+2u\right)}{5}=-\dfrac{2u^2+u}{5}=VP\)

Answer 3

a) ta có:
(x-2)(x²-1)=x³-x-2x²+2=x³-2x²-x+2
(x+1)(x²-3x+2)=x³-3x²+2x+x²-3x+2=x³-2x²-x+2
\(\Rightarrow\)\(\left(x-2\right)\left(x^2-1\right)=\left(x+1\right)\left(x^2-3x+2\right)\Rightarrow\dfrac{x-2}{x+1}=\dfrac{x^2-3x+2}{x^2-1}\)