\(\frac{2}{x+\frac{1}{1+\frac{x+1}{x+2}}}=\frac{6}{3x-1}\)
\(\frac{2}{x+\frac{1}{\frac{x+2+x+1}{x+2}}}=\frac{6}{3x-1}\)
\(\frac{2}{x+\frac{1}{\frac{2x+3}{x+2}}}=\frac{6}{3x-1}\)
\(\frac{2}{x+\frac{x+2}{2x+3}}=\frac{6}{3x-1}\)
\(\frac{2}{\frac{2x+3+x+2}{2x+3}}=\frac{6}{3x-1}\)
\(\frac{2}{\frac{3x+5}{2x+3}}=\frac{6}{3x-1}\)
\(\frac{4x+6}{3x+5}=\frac{6}{3x-1}\)
\(\Rightarrow\left(4x+6\right)\left(3x-1\right)=6\left(3x+5\right)\)
\(\Rightarrow12x^2-4x+18x-6=18x+30\)
\(\Rightarrow12x^2-4x+18x-18x=30+6\)
\(\Rightarrow12x^2-4x-36=0\)
\(\Rightarrow3x^2-x-9=0\)
\(\Rightarrow x^2-\frac{1}{3}x-3=0\)
\(\Rightarrow x^2-2.\frac{1}{6}x+\frac{1}{36}-\frac{1}{36}-3=0\)
\(\Rightarrow\left(x-\frac{1}{6}\right)^2-\frac{109}{36}=0\)
\(\Rightarrow\left(x-\frac{1}{6}-\frac{\sqrt{109}}{6}\right)\left(x-\frac{1}{6}+\frac{\sqrt{109}}{6}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{1+\sqrt{109}}{6}\\x=\frac{1-\sqrt{109}}{6}\end{cases}}\)
làm lại nhé, chỗ kia quy đồng sai
lần này làm theo cách khác
\(\frac{2}{x+\frac{1}{1+\frac{x+1}{x+2}}}=\frac{6}{3x-1}\)
\(\frac{2}{x+\frac{1}{\frac{x+2+x+1}{x+2}}}=\frac{2}{x-\frac{1}{3}}\)
\(\Rightarrow x+\frac{1}{\frac{2x+3}{x+2}}=x-\frac{1}{3}\)
\(\Rightarrow\frac{x+2}{2x+3}=\frac{-1}{3}\)
\(\Rightarrow\left(x+2\right).3=-1.\left(2x+3\right)\)
\(\Rightarrow3x+6=-2x-3\)
\(\Rightarrow3x+2x=-3-6\)
\(\Rightarrow5x=-9\)
\(\Rightarrow x=\frac{-9}{5}\)
vậy \(x=\frac{-9}{5}\)
