Ta có:\(\Delta=a^2-4\)
Để pt có 2 nghiệm thì \(\Delta\ge0\)\(\Rightarrow a^2-4\ge0\Rightarrow\left[{}\begin{matrix}a\ge2\\a\le-2\end{matrix}\right.\)
Theo hệ thức vi-ét ta có:\(\left\{{}\begin{matrix}x_1+x_2=-a\\x_1x_2=1\end{matrix}\right.\)
\(\left(\frac{x_1}{x_2}\right)^2+\left(\frac{x_2}{x_1}\right)^2>7\)
\(\Leftrightarrow\frac{x_1^4+x_2^4-7x_1^2x_2^2}{x_1^2x_2^2}\ge0\)
\(\Leftrightarrow x_1^4+x_2^4\ge7\)
\(\Leftrightarrow x_1^4+2x_1^2x_2^2+x_2^4\ge9\)
\(\Leftrightarrow\left(x_1^2+x_2^2\right)^2\ge9\)
\(\Leftrightarrow x_1^2+x_2^2\ge3\)
\(\Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2\ge3\)
\(\Leftrightarrow a^2\ge5\)
\(\Leftrightarrow\left[{}\begin{matrix}a\ge\sqrt{5}\\a\le-\sqrt{5}\end{matrix}\right.\)(tm)
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