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(x^2 - 1)[tex] \sqrt{x^2-4} [/tex][tex] \leq [/tex] 0
x^4-8x^2+16[tex] \leq [/tex] 0
(x^2 - 1)[tex] \sqrt{x^2-4} [/tex][tex] \leq [/tex] 0
x^4-8x^2+16[tex] \leq [/tex] 0
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Verified answer
ОДЗx²-4≥0
x=-2 x=2
x∈(-∞;-2] U [2;∞)
√(x²-4)≥0⇒x²-1≤0
(x-1)(x+1)≤0
x=-1 U x=1
-1≤x≤1∉ОДЗ
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