помогите с этим5^2/x≥0,2^x-3. Created by ShiroDark. algebra-ru

x≥0,2^x-3...

NNNLLL54

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$5^{\frac{2}{x}}\geq 0,2^{x-3}\; \; ,\qquad ODZ:\; x\ne 0\\\\\star \; \; 0,2=\frac{2}{10}=\frac{1}{5}=5^{-1}\; \; \star \\\\5^{\frac{2}{x}}\geq 5^{-x+3}\; \; \Rightarrow \; \; \; \frac{2}{x}\geq -x+3\; \; ,\; \; \frac{2}{x}+x-3\geq 0\; \; ,\; \; \frac{x^2-3x+2}{x}\geq 0\; ,\\\\\frac{(x-1)(x-2)}{x}\geq 0\\\\znaki:\; \; \; ---(0)+++[\; 1\; ]---[\; 2\; ]+++\\\\\underline {\; x\in (0,1\; ]\cup [\; 2,+\infty )\; }$

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sangers1959

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Ответ: x∈(0;1]U[2;+∞).

Объяснение:

5²/ˣ≥0,2ˣ⁻³

5²/ˣ≥(1/5)ˣ⁻³

5²/ₓ≥5³⁻ˣ

2/x≥3-x

3-x-(2/x)≤0

(3x-x²-2)/x≤0 |×(-1)

(x²-3x+2)/x≥0

x²-3x+2=0 D=1

x₁=1 x₁=2

(x-1)(x-2)=0 ⇒

(x-1)(x-2)/x≥0

-∞__-__0__+__1__-__2__-__+∞

x∈(0;1]U[2;+∞).

2 votes Thanks 1

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