a) ta có: \(\frac{x-3}{6}=\frac{3}{2x-6}\)
\(\Rightarrow\left(x-3\right).\left(2x-6\right)=6.3\)
\(\Rightarrow x.\left(2x-6\right)-3.\left(2x-6\right)=18\)
\(2x^2-6x-6x+18=18\)
\(2x^2-12x+18=18\)
\(2x^2-12x=0\)
\(2x.\left(x-6\right)=0\)
\(\Rightarrow2x=0\Rightarrow x=0\)
\(x-6=0\Rightarrow x=6\)
KL: x =0 hoặc x = 6
b) ta có: \(\frac{x+6}{x+2}=\frac{2x+3}{2x-1}\)
\(\Rightarrow\left(x+6\right).\left(2x-1\right)=\left(x+2\right).\left(2x+3\right)\)
\(\Rightarrow x.\left(2x-1\right)+6.\left(2x-1\right)=x.\left(2x+3\right)+2.\left(2x+3\right)\)
\(2x^2-x+12x-6=2x^2+3x+4x+6\)
\(2x^2+11x-6=2x^2+7x+6\)
\(\Rightarrow2x^2+11x-2x^2-7x=6+6\)
\(3x=12\)
\(x=12:3\)
\(x=4\)
\(\frac{x-3}{6}=\frac{3}{2x-6}\Leftrightarrow\left(x-3\right).\left(2x-6\right)=18\Leftrightarrow2x^2-12x+18=18\Leftrightarrow2x^2-12x=0\)
\(\Leftrightarrow x^2-6x=0\Leftrightarrow x.\left(x-6\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=6\end{cases}}\)
\(\frac{x+6}{x+2}=\frac{2x+3}{2x-1}\Leftrightarrow\left(x+6\right).\left(2x-1\right)=\left(2x+3\right).\left(x+2\right)\)
\(\Leftrightarrow2x^2+11x-6=2x^2+7x+6\Leftrightarrow4x-12=0\Leftrightarrow x=3\)
\(\frac{x-3}{6}=\frac{3}{2x-6}\)
\(\Rightarrow\left(x-3\right)\left(2x-6\right)=18\)
\(\Rightarrow2x^2-6x-6x+18=18\)
\(\Rightarrow2x^2-12x=0\)
\(\Rightarrow\left(2x-12\right).x=0\)
\(\Rightarrow\hept{\begin{cases}x=0\\2x-12=0\end{cases}\Rightarrow\hept{\begin{cases}x=0\\x=6\end{cases}}}\)
