\(30\cdot\left(x+2\right)-6\cdot\left(x-5\right)-24\cdot x=100\)
\(\Leftrightarrow30x+60-6x+30-24x=100\)
\(\Leftrightarrow0x+90=100\)
=> PT vô nghiệm
b) \(|7x+3|=66\)
\(\Leftrightarrow\orbr{\begin{cases}7x+3=66\\7x+3=-66\end{cases}\Leftrightarrow\orbr{\begin{cases}7x=63\\7x=-69\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=9\\x=\frac{-69}{7}\end{cases}}}\)
\(\text{30.(x+2)-6.(x-5)-24.x=100}\)
\(\Leftrightarrow30x+60-6x+30-24x=100\)
\(\Leftrightarrow90=100\)(vô lí)
Vậy phương trình vô nghiệm
\(|7x+3|=66\)
TH1 \(7x+3=66\Leftrightarrow x=\frac{66-3}{7}=9\)
TH2 \(7x+3=-66\Leftrightarrow x=\frac{-66-3}{7}=-\frac{69}{7}\)
