\(\Leftrightarrow11x^2+18x+22=7\sqrt{x^4+4x^2+4-4x^2}\)
\(\Leftrightarrow11x^2+18x+22=7\sqrt{\left(x^2+2\right)^2-4x^2}\)
\(\Leftrightarrow11x^2+18x+22=7\sqrt{\left(x^2+2x+2\right)\left(x^2-2x+2\right)}\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x^2+2x+2}=a>0\\\sqrt{x^2-2x+2}=b>0\end{matrix}\right.\) \(\Rightarrow10a^2+b^2=11x^2+18x+22\)
Pt trở thành:
\(10a^2+b^2=7ab\Leftrightarrow10a^2-7ab+b^2=0\)
\(\Leftrightarrow\left(2a-b\right)\left(5a-b\right)=0\Rightarrow\left[{}\begin{matrix}b=2a\\b=5a\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}b^2=4a^2\\b^2=25a^2\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2-2x+2=4\left(x^2+2x+2\right)\\x^2-2x+2=25\left(x^2+2x+2\right)\end{matrix}\right.\)
\(\Leftrightarrow...\) (casio)
Ta có: \(11x^2+18x+22=7\sqrt{x^4+4}\)
\(\Leftrightarrow11x^2+18x+22=7\left(x^2+2\right)\)
\(\Leftrightarrow11x^2+18x+22=7x^2+14\)
\(\Leftrightarrow4x^2+18x+8=0\)
\(\Delta=b^2-4ac=18^2-4\cdot4\cdot8=196\)
Vì Δ>0 nên phương trình có hai nghiệm là: \(\left\{{}\begin{matrix}x_1=\frac{-b-\sqrt{\Delta}}{2a}\\x_2=\frac{-b+\sqrt{\Delta}}{2a}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x_1=\frac{-18-\sqrt{196}}{2\cdot4}=-4\\x_2=\frac{-18+\sqrt{196}}{2\cdot4}=\frac{-1}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{-4;\frac{-1}{2}\right\}\)
