\(\frac{x-3}{x+2}=\frac{x+1}{x-4}\)
=> \(\left(x-3\right)\left(x-4\right)=\left(x+2\right)\left(x+1\right)\)
=> \(x^2-4x-3x+12=x^2+x+2x+2\)
=> \(\left(x^2-x^2\right)+\left(12-2\right)=\left(x+2x+4x+3x\right)\) (chuyển vế)
=> \(10=10x\Rightarrow x=10:10=1\)
Vậy x = 1
áp dụng tính chất của dãy tỉ số bằng nhau ta có;
\(\frac{x-3}{x+2}=\frac{x+1}{x-4}=\frac{x-3-\left(x+1\right)}{x+2-\left(x-4\right)}=\frac{x-3-x-1}{x+2-x+4}=\frac{-4}{6}=\frac{-2}{3}\)
suy ra:
\(\frac{x-3}{x+2}=\frac{-2}{3}\Rightarrow-2\left(x+2\right)=3.\left(x-3\right)\)
\(-2x-4=3x-9\)
\(-2x-3x=-9+4\)
\(-5x=-5\)
\(x=5\)
