Помогите решить алгебру.Спасибо!

Question 1

Question 2

y,\; x>-y\\\\\left \{ {{x+y=\frac{1}{9}} \atop {x-y=25}} \right. \; \left \{ {{2x=\frac{226}{9}} \atop {2y=-\frac{224}{9}}} \right. \; \left \{ {{x=\frac{113}{9}} \atop {y=-\frac{112}{9}}} \right. " alt="1)\quad \left \{ {{log_{1/3}(x+y)=2} \atop {log_5(x-y)=2}} \right. \; ,\; \; ODZ:\; x>y,\; x>-y\\\\\left \{ {{x+y=\frac{1}{9}} \atop {x-y=25}} \right. \; \left \{ {{2x=\frac{226}{9}} \atop {2y=-\frac{224}{9}}} \right. \; \left \{ {{x=\frac{113}{9}} \atop {y=-\frac{112}{9}}} \right. " align="absmiddle" class="latex-formula">

$2)\; \; \left \{ {{log_4x+log_4y=1+log_49} \atop {x+y-20=0}} \right. \; ,\; \; ODZ:\; \; x\ \textgreater \ 0,\; y\ \textgreater \ 0\\\\\left \{ {{log_4(xy)=log_4(4\cdot 9)} \atop {x+y=20}} \right. \; \left \{ {{xy=36} \atop {y=20-x}} \right. \\\\x\, (20-x)=36\\\\-x^2+20x=36\\\\x^2-20x+36=0\\\\x_1=2,\; \; x_2=18$

NNNLLL54_zn · Answer 1 · 2018-05-02T15:17:43+0000

y,\; x>-y\\\\\left \{ {{x+y=\frac{1}{9}} \atop {x-y=25}} \right. \; \left \{ {{2x=\frac{226}{9}} \atop {2y=-\frac{224}{9}}} \right. \; \left \{ {{x=\frac{113}{9}} \atop {y=-\frac{112}{9}}} \right. " alt="1)\quad \left \{ {{log_{1/3}(x+y)=2} \atop {log_5(x-y)=2}} \right. \; ,\; \; ODZ:\; x>y,\; x>-y\\\\\left \{ {{x+y=\frac{1}{9}} \atop {x-y=25}} \right. \; \left \{ {{2x=\frac{226}{9}} \atop {2y=-\frac{224}{9}}} \right. \; \left \{ {{x=\frac{113}{9}} \atop {y=-\frac{112}{9}}} \right. " align="absmiddle" class="latex-formula">

$2)\; \; \left \{ {{log_4x+log_4y=1+log_49} \atop {x+y-20=0}} \right. \; ,\; \; ODZ:\; \; x\ \textgreater \ 0,\; y\ \textgreater \ 0\\\\\left \{ {{log_4(xy)=log_4(4\cdot 9)} \atop {x+y=20}} \right. \; \left \{ {{xy=36} \atop {y=20-x}} \right. \\\\x\, (20-x)=36\\\\-x^2+20x=36\\\\x^2-20x+36=0\\\\x_1=2,\; \; x_2=18$