\(\sqrt{7x+7}+\sqrt{7x-6}+2\sqrt{49x^2+7x-42}=181-14x\) ( ĐK : \(\frac{6}{7}\le x\le\frac{181}{14}\))
\(\Leftrightarrow\sqrt{7x+7}+\sqrt{7x-6}+2\sqrt{\left(7x+7\right)\left(7x-6\right)}=-\left(7x+7\right)-\left(7x-6\right)+182\)
Đặt \(\left\{{}\begin{matrix}\sqrt{7x+7}=a\left(a\ge0\right)\\\sqrt{7x-6}=b\left(b\ge0\right)\end{matrix}\right.\)
\(\Leftrightarrow a+b+2ab=-a^2-b^2+182\)
\(\Leftrightarrow\left(a+b\right)^2+\left(a+b\right)-182=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a+b=13\left(N\right)\\a+b=-14\left(L\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{7x+7}+\sqrt{7x-6}=13\)
\(\Leftrightarrow\sqrt{49x^2+7x-42}=84-7x\)
\(\Leftrightarrow49x^2+7x-42=49x^2-1176x+7056\)
\(\Leftrightarrow1183x=7098\)
\(\Leftrightarrow x=6\left(TM\right)\)
Vậy S={6}
