\(\dfrac{x+7}{x+4}=\dfrac{x-1}{x-2}\)
\(\Rightarrow\left(x+7\right)\left(x-2\right)=\left(x+4\right)\left(x-1\right)\)
\(\Rightarrow x\left(x-2\right)+7\left(x-2\right)=x\left(x-1\right)+4\left(x-1\right)\)
\(\Rightarrow x^2-2x+7x-14=x^2-x+4x-4\)
\(\Rightarrow x^2+5x-14=x^2+3x-4\)
\(\Rightarrow5x-14=3x-4\)
\(\Rightarrow5x=3x+10\)
\(\Rightarrow2x=10\Rightarrow x=5\)
Ta có: \(\dfrac{x+7}{x+4}=\dfrac{x-1}{x-2}\)
\(\Rightarrow\left(x+7\right)\left(x-2\right)=\left(x-1\right)\left(x+4\right)\)
\(\Rightarrow x^2-2x+7x-14=x^2-4x-x-4\)
\(\Rightarrow x^2+5x-14=x^2-5x-4\)
\(\Rightarrow x^2-x^2+5x+5x=-4+14\)
\(\Rightarrow10x=10\)
\(\Rightarrow x=1\)
