Question

\(C=\frac{x}{2x-2}+\frac{x^2+1}{2-2x^2}\)

-Rút gọn C

Answer 1

Sửa đề:

\(\frac{x}{2x-2}+\frac{x^2+1}{2-2x}\\ =\frac{-x}{2-2x}+\frac{x^2+1}{2-2x}\\ =\frac{x^2-x+1}{2-2x}\)

Answer 2

Để mình làm lại :v đọc nhầm:

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

Answer 3

\(\begin{array}{l} C = \dfrac{x}{{2x - 2}} + \dfrac{{{x^2} + 1}}{{2 - 2{x^2}}}\\ C = \dfrac{x}{{2\left( {x - 1} \right)}} - \dfrac{{{x^2} + 1}}{{2\left( {{x^2} - 1} \right)}}\\ C = \dfrac{{x\left( {x + 1} \right)}}{{2\left( {x - 1} \right)}} - \dfrac{{{x^2} + 1}}{{2\left( {x - 1} \right)\left( {x + 1} \right)}}\\ C = \dfrac{{{x^2} + x - {x^2} - 1}}{{2\left( {x - 1} \right)\left( {x + 1} \right)}}\\ C = \dfrac{{x - 1}}{{2\left( {x - 1} \right)\left( {x + 1} \right)}}\\ C = \dfrac{1}{{2x + 2}} \end{array}\)