\(\frac{x-2}{x-1}=\frac{x+4}{x+7}\) ĐKXĐ: \(x\ne1;x\ne-7\)
\(\Rightarrow\left(x-2\right)\left(x+7\right)=\left(x-1\right)\left(x+4\right)\)
\(\Leftrightarrow x^2-2x+7x-14=x^2-x+4x-4\)
\(\Leftrightarrow x^2+5x-14=x^2+3x-4\)
\(\Leftrightarrow x^2-x^2+5x-3x=-4+14\)
\(\Leftrightarrow2x=10\)
\(\Leftrightarrow x=5\)( thỏa mãn điều kiện xác định)
vậy x=5
Ta có:\(\frac{x-2}{x-1}=\frac{x+4}{x+7}\)
\(\Rightarrow\left(x-2\right)\left(x-7\right)=\left(x-1\right)\left(x+4\right)\)
\(\Rightarrow x^2-9x+14=x^2+3x-4\)
\(\Rightarrow x^2-9x+14-x^2-3x+4=0\)
\(\Rightarrow18-12x=0\)
\(\Rightarrow x=\frac{18}{12}\)
