Câu hỏi của ๖ۣۜTina Ss

Question

Giải phương trình

a) $\frac{2}{x+\frac{1}{1+\frac{x+1}{x-2}}}=\frac{6}{3x-1}$

b) $\frac{3x-1}{x-1}-\frac{2x+2}{x+2}+\frac{4}{x^2+2x-3}=1$

Dennis · Accepted Answer

a) Có : $\frac{2}{x+\frac{1}{1+\frac{x+1}{x-2}}}=\frac{6}{3x-1}$

$\Leftrightarrow$ $\frac{2}{x+\frac{1}{\frac{x-2+x+1}{x-2}}}=\frac{6}{3x-1}$

$\Leftrightarrow$ $\frac{2}{x+\frac{1}{\frac{2x-1}{x-2}}}=\frac{6}{3x-1}$

$\Leftrightarrow$ $\frac{2}{x+\frac{x-2}{2x-1}}=\frac{6}{3x-1}$

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

$\Leftrightarrow$ $\frac{2}{\frac{2x^2-x+x-2}{2x-1}}=\frac{6}{3x-1}$

$\Leftrightarrow$ $\frac{2}{\frac{2x^2-2}{2x-1}}=\frac{6}{3x-1}$

$\Leftrightarrow$ $\frac{2.\left(2x-1\right)}{2x^2-2}=\frac{6}{3x-1}$ $\Leftrightarrow$ $\frac{4x-2}{2x^2-2}=\frac{6}{3x-1}$

$\Leftrightarrow$ $\left(4x-2\right)\left(3x-1\right)=6\left(2x^2-2\right)$

$\Leftrightarrow$ $12x^2-4x-6x+2=12x^2-12$

$\Leftrightarrow$ $12x^2-10x+2-12x^2+12=0$

$\Leftrightarrow$ $-10x+14=0$

$\Leftrightarrow$ $-10x=-14$

$\Leftrightarrow$ $x=\frac{-14}{-10}=\frac{14}{10}=\frac{7}{5}$

Vậy $x=\frac{7}{5}$Câu trả lời: a) Có : $\frac{2}{x+\frac{1}{1+\frac{x+1}{x-2}}}=\frac{6}{3x-1}$