Question

пожалуйста хоть одну люди добрые​

Answer 1

$1)\; \; (5^{1/4}:2^{3/4}-2^{1/4}:5^{3/4})\cdot \sqrt[4]{1000}=\Big (\frac{5^{\frac{1}{4}}}{2^{\frac{3}{4}}}-\frac{2^{\frac{1}{4}}}{5^{\frac{3}{4}}}\Big )\cdot \sqrt[4]{1000}=\\\\=\frac{5^{\frac{1}{4}}\cdot 5^{\frac{3}{4}}-2^{\frac{1}{4}}\cdot 2^{\frac{3}{4}}}{2^{\frac{3}{4}}\, \cdot \, 5^{\frac{3}{4}}}\cdot \sqrt[4]{1000}=\frac{5-2}{\sqrt[4]{10^3}}\cdot \sqrt[4]{10^3}=3$

$\frac{m^{\sqrt3}\, \cdot \, n^{\sqrt3}}{(mn)^{2+\sqrt3}}=\frac{(mn)^{\sqrt3}}{(mn)^{2+\sqrt3}}=\frac{1}{(mn)^2}=\frac{1}{m^2n^2}$

$2)\; \; 2^{x+1}+2^{x-1}-3^{x-1}=3^{x-2}-2^{x-3}+2\cdot 3^{x-3}\\\\2\cdot 2^{x}+\frac{1}{2}\cdot 2^{x}-\frac{1}{3}\cdot 3^{x}=\frac{1}{9}\cdot 3^{x}-\frac{1}{8}\cdot 2^{x}+\frac{1}{27}\cdot 2\cdot 3^{x}\\\\\frac{21}{8}\cdot 2^{x}=\frac{14}{27}\cdot 3^{x}\\\\\frac{21}{8}\cdot \frac{27}{14}\cdot (\frac{2}{3})^{x}=1\; \; ,\; \; \; \frac{3\cdot 27}{8\cdot 2}\cdot (\frac{2}{3})^{x} =1\; \; ,\; \; (\frac{2}{3})^{x}=(\frac{2}{3})^4\; \; \Rightarrow \; \; \; \underline {x=4}$

$3)\; \; \left \{ {{x^2+xy-y^2=11} \atop {x-2y=1}} \right.\; \; \left \{ {{4y^2+4y+1+2y^2+y-y^2=11} \atop {x=2y+1}} \right.\; \; \left \{ {{5y^2+5y-10=0} \atop {x=2y+1}} \right.\\\\\left \{ {{y_1=1\; ,\; y_2=-2} \atop {x_1=3\; ,\; x_2=-3}} \right. \; \; \; Otvet:\; \; (3,1)\; ,\; (-3,-2)$

$4)\; \; \left \{ {{x+2y=4} \atop {x^2+xy=y-5}} \right.\; \; \left \{ {{x=4-2y\qquad \qquad \qquad } \atop {16-16y+4y^2+4y-2y^2-y+5=0}} \right.\; \; \left \{ {{x=4-2y\qquad } \atop {2y^2-13y+21=0}} \right.\\\\\left \{ {{x_1=0\; ,\; x_2=-3} \atop {y_1=2\; ,\; y_2=3,5}} \right.\; \; \; Otvet:\; \; (\, 0\, ;2)\; ,\; (-3\, ;\, 3,5)$