Question

Решите систему неравенств.

Answer 1

$\displaystyle\bf\\5)\\\\4^{2x} -5\cdot4^{x} +4\leq 0\\\\4^{x} =m \ , \ m>0\\\\m^{2}-5m+4\leq 0\\\\(m-1)(m-4)\leq 0$

+ + + + + [1] - - - - - [4] + + + + +   m

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m ∈ [1 , 4]

$\displaystyle\bf\\\left \{ {{m\geq 1} \atop {m\leq 4}} \right. \\\\\\\left \{ {{4^{x} \geq1 } \atop {4^{x} }\leq 4} \right. \\\\\\\left \{ {{4^{x} }\geq 4^{0} \atop {4^{x} }\leq 4} \right.\\\\\\\left \{ {{x\geq 0} \atop {x\leq 1}} \right. \\\\\\Otvet:x\in\Big[0 \ ; \ 1\Big]$

$\displaystyle\bf\\6)\\\\5^{x} \cdot 2^{1-x} +5^{x+1} \cdot 2^{-x} >7\cdot \Big(0,4\Big)^{-\frac{3}{x} } \\\\\\5^{x} \cdot 2\cdot 2^{-x} +5^{x} \cdot 5\cdot 2^{-x} >7\cdot \Big(\frac{2}{5}\Big)^{-\frac{3}{x} } \\\\\\2\cdot\Big(\frac{5}{2} \Big)^{x} +5\cdot\Big(\frac{5}{2} \Big)^{x} >7\cdot\Big(\frac{5}{2} \Big)^{\frac{3}{x} } \\\\\\7\cdot\Big(\frac{5}{2} \Big)^{x} >7\cdot\Big(\frac{5}{2} \Big)^{\frac{3}{x} }$

$\displaystyle\bf\\\Big(\frac{5}{2} \Big)^{x} >\Big(\frac{5}{2} \Big)^{\frac{3}{x} }\\\\\\x>\frac{3}{x} \\\\\\x-\frac{3}{x} >0\\\\\\\frac{x^{2} -3}{x} >0\\\\\\\frac{(x-\sqrt{3})(x+\sqrt{3} ) }{x} >0$

- - - - -  (-√3) + + + + + (0) - - - - - (√3) + + + + +

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Ответ  : x ∈ (- √3 ; 0) ∪ (√3 ; +∞)