|7x - 5| = |3x + 3|
7x - 5 = 3x + 3
7x - 3x = -5 + 3
4x = 2
x = 2
Ta có: |7x - 5| = |3x + 3|
\(\Rightarrow\)\(\orbr{\begin{cases}7x-5=3x+3\\7x-5=-\left(3x+3\right)\end{cases}}\)
Xét 7x - 5 = 3x + 3
7x - 3x= 3 + 5
4x = 8
x = 2
Xét 7x - 5 = -(3x + 3)
7x - 5 = -3x - 3
7x + 3x = -3 + 5
10x = 2
x=\(\frac{1}{5}\)
Vậy x\(\in\){ 2; \(\frac{1}{5}\)}
\(\left|7x-5\right|=\left|3x+3\right|\)
\(TH1:7x-5=3x-3\) \(TH2:7x-5=-3x+3\)
\(7x-3x=-3+5\) \(7x+3x=3+5\)
\(4x=2\) \(10x=8\)
\(x=\frac{1}{2}\) \(x=\frac{4}{5}\)
\(\Rightarrow x\in\left\{\frac{1}{2};\frac{4}{5}\right\}\)