Explicație pas cu pas:
a)
notăm: x² - 3x = t
t² - 2t - 8 = t² + 2t - 4t - 8 = (t + 2)(t - 4)
[tex]( {x}^{2} - 3x + 2)( {x}^{2} - 3x - 4) \leqslant 0 \\ (x - 1)(x - 2)(x + 1)(x - 4) \leqslant 0[/tex]
[tex]x - 1 = 0 \implies x = 1[/tex]
[tex]x - 2 = 0 \implies x = 2[/tex]
[tex]x + 1 = 0 \implies x = - 1[/tex]
[tex]x - 4 = 0 \implies x = 4[/tex]
[tex]\implies x\in \left[- 1;1\right]\cup\left[2;4\right][/tex]
b)
[tex](x + 3)(x + 4)(x - 3)(x - 4) \geqslant 120[/tex]
[tex]( {x}^{2} - 9)( {x}^{2} - 16) - 120 \geqslant 0 \\ [/tex]
[tex]{x}^{4} - 25 {x}^{2} + 24 \geqslant 0[/tex]
[tex]({x}^{2} - 1)( {x}^{2} - 24) \geqslant 0[/tex]
[tex](x + 1)(x - 1)(x + 2 \sqrt{6})(x - 2 \sqrt{6}) \geqslant 0 \\ [/tex]
[tex]x + 1 = 0 \implies x = - 1[/tex]
[tex]x - 1 = 0 \implies x = 1[/tex]
[tex]x + 2 \sqrt{6} = 0 \implies x = - 2 \sqrt{6} [/tex]
[tex]x - 2 \sqrt{6} = 0 \implies x = 2 \sqrt{6} [/tex]
[tex]\implies x\in \left(- \infty ; - 2 \sqrt{6} \right]\cup\left[ - 1;1\right]\cup\left[2 \sqrt{6} ; + \infty \right) \\ [/tex]
