\(\dfrac{x-3}{x+2}=\dfrac{x-5}{x+1}\)
\(\Rightarrow\left(x-3\right)\left(x+1\right)=\left(x+2\right)\left(x-5\right)\)
\(\Rightarrow x^2+x-3x-3=x^2-5x+2x-10\)
\(\Rightarrow x^2-2x-x^2+3x=-10+3\)
\(\Rightarrow x=-7\)
Vậy...........
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\(\dfrac{x-3}{x+2}=\dfrac{x-5}{x+1}\)
\(\Rightarrow\left(x-3\right)\left(x+1\right)=\left(x-5\right)\left(x+2\right)\)
\(\Rightarrow x\left(x+1\right)-3\left(x+1\right)=x\left(x+2\right)-5\left(x+2\right)\)
\(\Rightarrow x^2+x-3x-3=x^2+2x-5x-10\)
\(\Rightarrow x^2-2x-3=x^2-3x-10\)
\(\Rightarrow x^2-2x+3x=x^2-10+3\)
\(\Rightarrow x^2+x=x^2-7\)
\(\Rightarrow x=-7\)
