1. Вычислите: 2. Решите уравнение: 3. Решите неравенство: 4. Решите систему уравнений:

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arsenlevadniy_zn · Answer 1 · 2018-03-27T12:28:33+0000

$1) 2^{\frac{1}{\log_{5} 2} } + \log_{3}\log _{3} \sqrt[3]{\sqrt[3]{3}} =2^{\log_{2} 5} + \log_{3}\log _{3} 3^{ \frac{1}{9}} = 5 + \log_{3}( \frac{1}{9}\log _{3} 3) =\\= 5 + \log_{3}3^{-2} = 5 -2\log_{3}3 = 5-2=3;$

$3\cdot16^{x}+2\cdot81^{x}=5\cdot36^{x}, \\ 3\cdot2^{4x}+2\cdot3^{4x}-5\cdot(2\cdot3)^{2x}=0, \\ 3\cdot (\frac{2}{3})^{4x}-5(\frac{2}{3})^{2x}+2=0, \\ (\frac{2}{3})^{2x}=t, \\ 3t^2-5t+2=0, \\ D=1, \\ t_1= \frac{2}{3}, x_2=1, \\ (\frac{2}{3})^{2x}=\frac{2}{3}, \\ 2x=1, \\ x_1=\frac{1}{2} , \\ (\frac{2}{3})^{2x}=1, \\ 2x=0, \\ x_2=0;$

0,} \atop { x^{2} -2x-2 \leq 1;}} \right. \left \{ {{ x^{2} -2x-2>0,} \atop { x^{2} -2x-3 \leq 0;}} \right. \\ x^{2} -2x-2=0, \\ D=1^2+2=3, \\ x_1=1-\sqrt{3},x_2=1+ \sqrt{3}, \\ x^{2} -2x-3=0, \\ x_1=-1,x_2=3; \\ \left \{ {{(x-1+\sqrt{3})(x-1-\sqrt{3})>0,} \atop { (x+1)(x-3) \leq 0;}} \right. \left \{ {{ \left [ {{x<1-\sqrt{3},} \atop {x>1+\sqrt{3},}} \right. } \atop { -1 \leq x \leq 3;}} \right. " alt="3) log_{3}( x^{2} -2x-2) \leq 0, \\ \left \{ {{ x^{2} -2x-2>0,} \atop { x^{2} -2x-2 \leq 1;}} \right. \left \{ {{ x^{2} -2x-2>0,} \atop { x^{2} -2x-3 \leq 0;}} \right. \\ x^{2} -2x-2=0, \\ D=1^2+2=3, \\ x_1=1-\sqrt{3},x_2=1+ \sqrt{3}, \\ x^{2} -2x-3=0, \\ x_1=-1,x_2=3; \\ \left \{ {{(x-1+\sqrt{3})(x-1-\sqrt{3})>0,} \atop { (x+1)(x-3) \leq 0;}} \right. \left \{ {{ \left [ {{x<1-\sqrt{3},} \atop {x>1+\sqrt{3},}} \right. } \atop { -1 \leq x \leq 3;}} \right. " align="absmiddle" class="latex-formula"> $\left [ {{-1 \leq x<1-\sqrt{3},} \atop {1+\sqrt{3}<x \leq 3;}} \right. \\ x\in[-1;1-\sqrt{3})\cup(1-\sqrt{3};3];$

$\left \{ {{ 3^{x}\cdot 2^{y} =972, } \atop { \log_{ \sqrt{3} } (x-y)=2;}} \right. \left \{ {{3^{x}\cdot 2^{y} =972, } \atop {x-y=(\sqrt{3})^2;}} \right. \left \{ {{3^{y+3}\cdot 2^{y} =972, } \atop {x=y+3;}} \right. \left \{ {{3^3\cdot3^{y}\cdot 2^{y} =972, } \atop {x=y+3;}} \right.\left \{ {{6^{y} =36, } \atop {x=y+3;}} \right. \\ \left \{ {{6^{y} =6^2, } \atop {x=y+3;}} \right.\left \{ {{x=5, } \atop {y=2;}} \right.\\ (5;2).$