Помогите решить неравенство! √(2^(x^2+2x-10) ) ≥ (√(33+√128) -1)^x. Created by Julia0fox. algebra-ru

Помогите решить неравенство! √(2^(x^2+2x-10) ) ≥ (√(33+√128) -1)^x...

Julia0fox @Julia0fox

August 2022 1 31 Report

Помогите решить неравенство!
√(2^(x^2+2x-10) ) ≥ (√(33+√128) -1)^x

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NNNLLL54

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$\sqrt{2^{x^2+2x-10}} \geq (\sqrt{33+\sqrt{128}}-1)^{x}\\\\33+\sqrt{128}=33+\sqrt{4\cdot 32}=33+2\sqrt{32}=32+2\sqrt32+1=(1+\sqrt{32})^2\\\\\sqrt{33+\sqrtP{128}}-1=\sqrt{(1+\sqrt{32})^2}-1=|1+\sqrt{32}|-1=\\\\=1+\sqrt{32}-1=\sqrt{32}=\sqrt{2^5}=2^\frac{5}{2}\\\\\\2^{\frac{x^2+2x-10}{2} }\geq 2^{\frac{5}{2}x}\\\\Tak\; kak\; 2\ \textgreater \ 1,\; to\; \; \frac{x^2+2x-10}{2} \geq \frac{5}{2}x\, |\cdot 2\\\\x^2+2x-10 \geq 5x\\\\x^2-3x-10 \geq 0\; \; \; (x_1=-2,\; x_2=5,\; \; teor.\; Vieta)\\\\(x-5)(x+2) \geq 0$

$Znaki:\; \; ++(-2)---(5)+++\\\\x\in (-\infty ,-2\, ]\cup [\, 5,+\infty )$

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