Предмет: Алгебра, автор: wehehelp

help please help sos sos​

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Ответы

Автор ответа: Universalka
0

3*25^{x+0,5} +4^{2x+1,5} \leq 22*20^{x}\\\\3*25^{x}*25^{0,5}+4^{2x}*4^{1,5}-22*20^{x}\leq0\\\\3*25^{x}*5+16^{x}*8-22*20^{x}\leq0|:16^{x}>0\\\\15*(\frac{25}{16})^{x}-22*(\frac{20}{16})^{x}+8*(\frac{16}{16})^{x}\leq0\\\\15*(\frac{5}{4})^{2x }-22*(\frac{5}{4})^{x}+8\leq 0\\\\(\frac{5}{4})^{x}=m,m>0 \ \Rightarrow (\frac{5}{4})^{2x}=m^{2}\\\\15m^{2}-22m+8\leq0\\\\15m^{2}-22m+8=0\\\\\frac{D}{4}=(-11)^{2}-15*8=121-120=1

m_{1}=\frac{11+1}{15}=\frac{12}{15}=\frac{4}{5}\\\\m_{2} =\frac{11-1}{15}=\frac{10}{15}=\frac{2}{3}\\\\15m^{2}-22m+8=15(m-\frac{4}{5})(m-\frac{2}{3}) \\\\(m-\frac{4}{5})(m-\frac{2}{3}) \leq0

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______[2/3]______[4/5]_______m

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1)m\geq  \frac{2}{3}\\\\(\frac{5}{4})^{x}\geq \frac{2}{3}\\\\log_{0,8} (\frac{5}{4})^{x}\geq log_{0,8}\frac{2}{3}\\\\x\geq log_{0,8}\frac{2}{3}\\\\2)m\leq \frac{4}{5}\\\\(\frac{5}{4})^{x} \leq(\frac{5}{4})^{-1}\\\\x\leq-1\\\\\boxed{x\in [log_{0,8} \frac{2}{3};-1]}

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