Ta có :
\(\left(x+7\right)\left(3x-1\right)=49-x^2\)
\(\Leftrightarrow\)\(\left(x+7\right)\left(3x-1\right)=7^2-x^2\)
\(\Leftrightarrow\)\(\left(x+7\right)\left(3x-1\right)=\left(x+7\right)\left(7-x\right)\)
\(\Leftrightarrow\)\(\left(x+7\right)\left(3x-1\right)-\left(x+7\right)\left(7-x\right)=0\)
\(\Leftrightarrow\)\(\left(x+7\right)\left(3x-1-7+x\right)=0\)
\(\Leftrightarrow\)\(\left(x+7\right)\left(4x-8\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x+7=0\\4x-8=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-7\\4x=8\end{cases}}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-7\\x=\frac{8}{4}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-7\\x=2\end{cases}}}\)
Vậy \(x=2\) hoặc \(x=-7\)
Chúc bạn học tốt ~
Ta có: (x+7)(3x-1)=49-x^2
\(\Rightarrow\left(x+7\right)\left(3x-1\right)-49+x^2=0\)
\(\Rightarrow4x^2+20-56=0\)
\(\Rightarrow\left(2x\right)^2+2.2x.5+5^2-81=0\)
\(\Rightarrow\left(2x+5\right)^2=81\)
\(\Rightarrow2x+5=9\)hoặc \(2x+5=-9\)
\(\Rightarrow x=2\)hoặc \(x=-7\)
