\(\dfrac{1}{x-1}+\dfrac{-2}{3}\left(\dfrac{3}{4}-\dfrac{6}{5}\right)=\dfrac{5}{2-2x}\)
\(\dfrac{1}{x-1}+\dfrac{-2}{3}\times\dfrac{-9}{20}=\dfrac{5}{2-2x}\)
\(\dfrac{1}{x-1}+\dfrac{3}{10}=\dfrac{5}{2-2x}\)
\(\dfrac{10+3x-3}{10x-10}=\dfrac{5}{2-2x}\)
\(\dfrac{7+3x}{10x-10}=\dfrac{5}{2-2x}\)
\(\left(7+3x\right)\left(2-2x\right)=5\left(10x-10\right)\)
\(7\left(2-2x\right)+3x\left(2-2x\right)=50x-50\)
\(14-14x+6x-6x^2=50x-50\)
\(14-8x-6x^2=50x-50\)
\(14-8x-6x^2+50=50x\)
\(36+8x-6x^2=50x\)
\(36=50x+6x^2-8x\)
\(36=42x+6x^2\)
\(\dfrac{36}{6}=\dfrac{42x+6x^2}{6}\)
\(6=7x+x^2\)
\(6=x\left(7+x\right)\)
còn lại mình chịu
\(\dfrac{1}{x-1}-\dfrac{2}{3}.\left(\dfrac{3}{4}-\dfrac{6}{5}\right)=\dfrac{5}{2-2x}\) (*)
ĐKXĐ: \(x\ne1\)
(*) \(\Leftrightarrow\dfrac{1}{x-1}+\dfrac{3}{10}=-\dfrac{5}{2x-2}\)
\(\Leftrightarrow\dfrac{1}{x-1}+\dfrac{5}{2x-2}=-\dfrac{3}{10}\)
\(\Leftrightarrow\dfrac{1.2}{\left(x-1\right).2}+\dfrac{5}{2x-2}=-\dfrac{3}{10}\)
\(\Leftrightarrow\dfrac{7}{2x-2}=-\dfrac{3}{10}\)
\(\Leftrightarrow2x-2=-\dfrac{70}{3}\)
\(\Leftrightarrow2x=-\dfrac{64}{3}\)
\(\Leftrightarrow x=-\dfrac{32}{3}\)
Vậy \(x=-\dfrac{32}{3}\).
