Question

Доброго времени суток решить неравенство.

за !

Answer 1

Объяснение:

$25*4^{\frac{1}{2}-\frac{2}{x} }-133 *10^{-\frac{2}{x}}+4*5^{1-\frac{4}{x} }\leq 0\ \ \ \ \ \ \ x\neq 0.\\25*2^{2*(\frac{1}{2}-\frac{2}{x})} -133 *10^{-\frac{2}{x}}+4*5^{1-\frac{4}{x} }\leq 0\\25*2^{1-\frac{4}{x} } -133 *10^{-\frac{2}{x}}+4*5^{1-\frac{4}{x} }\leq 0\\$

$25*2*2^{-\frac{4}{x}} -133*10^{-\frac{2}{x}}+4*5*5^{-\frac{4}{x}}\leq 0\\50*2^{-\frac{4}{x}} -133*(2*5)^{-\frac{2}{x}}+20*5^{-\frac{4}{x}}\leq0\ |:5^{-\frac{4}{x}}\\50*(\frac{2}{5})^{-\frac{4}{x}} -133*(\frac{2}{5} )^{-\frac{2}{x}}+20\leq 0\\50*((\frac{2}{5})^{-\frac{2}{x}})^2 -133*(\frac{2}{5} )^{-\frac{2}{x}}+20\leq 0.$

Пусть  $(\frac{2}{5} )^{-\frac{2}{x}}}=t\ \ \ \ \Rightarrow\\$

$50*t^2-133t+20\leq 0\\50*t^2-133t+20=0\\D=13689\ \ \ \ \sqrt{D} =117\\t_1=\frac{16}{100}=\frac{4}{25}=0,16 \ \ \ \\ t_2=\frac{250}{100}=\frac{5}{2} =2,5 \\(t-0,16 )*(t-2,5 )\leq 0\\$

-∞__+__0,16__-__2,5__+__+∞                ⇒

t∈[0,16;2,5].

$0,16\leq t\leq 2,5\\\frac{4}{25}\leq (\frac{2}{5})^{-\frac{2}{x}} \leq \frac{5}{2}\\(\frac{2}{5})^2\leq (\frac{2}{5})^{-\frac{2}{x}} \leq (\frac{2}{5})^{-1}\\2\leq -\frac{2}{x} \leq -1 \\ 2\leq -2*x^{-1} \leq -1 \ |:(-2)\\-1 \geq x^{-1}\geq \frac{1}{2}\\ (-1)^{-1} \geq (x^{-1})^{-1}\geq( \frac{1}{2})^{-1}\\-1\geq x\geq 2.\ \ \ \ \Rightarrow\\$

ответ: x∈(-∞;-1)U(2;+∞).