Question

Хотя бы 7-12
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Answer 1

$7)\; \; \sqrt{64-3x^2}=-x\; ,\; \; \; ODZ:\; \; -x \geq 0\; ,\; \; x \leq 0\\\\64-3x^2=(-x)^2\\\\64-3x^2=x^2\; ,\; \; \; 4x^2=64\; ,\; \; x^2=16\\\\x_1=-4\; ,\; \; x_2=4\notin ODZ\\\\-4\in (-\infty ,-2\; ]\\\\8)\; \; (\frac{1}{32})^{0,5x+1}=8\\\\(2^{-5})^{0,5x+1}=2^3\\\\2^{-2,5x-5}=2^3\\\\-2,5x-5=3\\\\-2,5x=8\\\\x=-3,2\\\\-3,2\in [-4,0)$

$9)\; \; \frac{6x-2}{(x-1)(x+2)} \leq 0\; ,\; \; \; \frac{2(3x-1)}{(x-1)(x+2)} \leq 0\\\\---(-2)+++[\, \frac{1}{3}\, ]---(1)+++\\\\x\in (-\infty ,-2)\cup [\, \frac{1}{3},1)\\\\10)\; \; y=cosx-2\\\\-1 \leq cosx\leq 1\\\\-1-2\leq cosx-2 \leq 1-2\\\\-3 \leq cosx-2\leq -1$

$11)\; \; f(x)=(7-2x)^4\\\\f'(x)=4\cdot (7-2x)^3\cdot (-2)=-8\cdot (7-2x)^3\\\\12)\; \; f(x)=2x+4x^3-1\\\\F(x)=2\cdot \frac{x^2}{2}+4\cdot \frac{x^4}{4} -x+C=x^2+x^4-x+C$