Đề bài: Tìm x
Ta có : \(\frac{3}{x-5}=\frac{-4}{x+2}\)
\(\Rightarrow3\left(x+2\right)=-4\left(x-5\right)\)
\(3x+6=-4x+20\)
\(3x+4x=-6+20\)
\(7x=14\)
\(x=14\div7\)
\(x=2\)
Vậy \(x=2\).
\(\frac{3}{x-5}=\frac{-4}{x+2}\)
\(\Rightarrow3x+6=-4x+20\)
\(\Rightarrow7x=14\)
\(\Rightarrow x=2\)
3/x-5= - 4/x+2
=>3.(x+2)=-( - 4).(x-5)
=> 3x+3.2= - 4x-(-20)
=>3x+6= - 4x+20
=>4x+3x=20-6
=> 7x=14
=>x=14:7
=>x=2
\({3\over x-5}={-4\over x+2} \)
\(3x+6=-4x+20\)
\(=>7x=14=>x=2\)
