Ta có: \(\left(x+7\right)\left(3x-1\right)=49-x^2\)
\(\Leftrightarrow\left(x+7\right)\left(3x-1\right)=\left(7-x\right)\left(7+x\right)\)
\(\Leftrightarrow\left(x+7\right)\left(3x-1\right)-\left(7-x\right)\left(x+7\right)=0\)
\(\Leftrightarrow\left(x+7\right)\left(3x-1-7+x\right)=0\)
\(\Leftrightarrow\left(x+7\right)\left(4x-8\right)=0\)
\(\Leftrightarrow\left(x+7\right)\cdot4\cdot\left(x-2\right)=0\)
Vì 2≠0
nên \(\left[{}\begin{matrix}x+7=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-7\\x=2\end{matrix}\right.\)
Vậy: x∈{-7;2}
\(\left(x+7\right)\left(3x-1\right)=49-x^2\)
\(\Leftrightarrow\left(x+7\right)\left(3x-1\right)=\left(x+7\right)\left(7-x\right)\)
\(\Leftrightarrow3x-1=7-x\)
\(\Leftrightarrow4x=8\Leftrightarrow x=2\)
Vậy...
(x + 7)(3x - 1) = 49 - x2
\(\Leftrightarrow\) (x + 7)(3x - 1) = (7 - x)(7 + x)
\(\Leftrightarrow\) (x + 7)(3x - 1) - (7 - x)(7 + x) = 0
\(\Leftrightarrow\) (7 + x)(3x - 1 - 7 + x) = 0
\(\Leftrightarrow\) (7 + x)(4x - 8) = 0
\(\Leftrightarrow\) 4(7 + x)(x - 2) = 0
\(\Leftrightarrow\) 7 + x = 0 hoặc x - 2 = 0
\(\Leftrightarrow\) x = -7 hoặc x = 2
Vậy S = {-7; 2}
Chúc bạn học tốt!
