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\(\Leftrightarrow\left[\left(x^2-1\right)\left(x^2-7\right)\right].\left[\left(x^2-3\right)\left(x^2-5\right)\right]\le0\)
\(\Leftrightarrow\left(x^4-8x^2+7\right)\left(x^4-8x^2+15\right)\le0\)
Đặt \(x^4-8x^2=t\Rightarrow\left(t+17\right)\left(t+15\right)\le0\Rightarrow\orbr{\begin{cases}t\le-15\\t\ge-7\end{cases}}\)
Với \(t\le-15\Rightarrow x^4-8x^2+15\le0\Rightarrow3\le x^2\le5\Rightarrow\orbr{\begin{cases}\sqrt{3}\le x\le\sqrt{5}\\-\sqrt{5}\le x\le-\sqrt{3}\end{cases}}\)
Với \(t\ge-7\Rightarrow x^4-8x^2+7\ge0\Rightarrow\orbr{\begin{cases}x^2\le1\\x^2\ge7\end{cases}\Rightarrow\orbr{\begin{cases}0\le x\le1\\x\ge\sqrt{7};x\le-\sqrt{7}\end{cases}}}\)
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a) \(-1< x< 0\Rightarrow\text{[}x\text{]}=-1\)
b) \(\frac{-7}{2}< x< -3\Rightarrow\text{[}x\text{]}=-4\)
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\(\frac{x-3}{x+7}< 0\)=> x - 3 và x + 7 khác dấu mà x - 3 < x + 7 => x - 3 < 0 ; x + 7 > 0 => x < 3 ; x > -7 => -7 < x < 3
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