Question

Помогите пожалуйста. Срочно ​

Answer 1

$1.\\\begin{cases}\frac1x+\frac1y=\frac56\\2y-x=1\end{cases}\\O.D.3.:\;x\neq0,\;y\neq0\\\\\begin{cases}\frac{x+y}{xy}=\frac56\\x=2y-1\end{cases}\\\\\\\frac{2y-1+y}{(2y-1)y}=\frac56\\\\\frac{3y-1}{2y^2-y}=\frac56\\\\(3y-1)\cdot6=(2y^2-y)\cdot5\\18y-6=10y^2-5y\\10y^2-23y+6=0\\D=529-4\cdot10\cdot6=529-240=289=(\pm17)^2\\y_{1,2}=\frac{23\pm17}{20}\\y_1=\frac6{20}=\frac3{10}=0,3\\y_2=2\\\\\boxed{\begin{cases}x=-0,4\\y=0,3\end{cases}}$

$2.\\\begin{cases}x^2-y^2=9\\xy=20\end{cases}\Rightarrow\begin{cases}(x-y)(x+y)=9\\y=\frac{20}x\end{cases}\\\\\\\left(x-\frac{20}x\right)\left(x+\frac{20}x\right)=9\\\\\frac{x^2-20}x\cdot\frac{x^2+20}x=9\\\\x^4-400=9x^2\\x^4-9x^2-400=0\\x^2=t,\;x^4=t^2,\;t\geq0\\\\t^2-9t-400=0\\D=81-4\cdot1\cdot(-400)=81+1600=1681=(\pm41)^2\\t_{1,2}=\frac{9\pm41}2\\t_1=-16\;-\;He\;nogx.\\t_2=25\\x^2=25\Rightarrow x=\pm5\\\\\boxed{\begin{cases}x=5\\y=4\end{cases}\quad\quad\quad u\quad\quad\quad\begin{cases}x=-5\\y=-4\end{cases}}$

$3.\\\begin{cases}x^2-xy+y^2=14\\x-3y=10\end{cases}\Rightarrow\begin{cases}x^2-xy+y^2=14\\x=3y+10\end{cases}\\\\\\(3y+10)^2-(3y+10)y+y^2=14\\9y^2+60y+100-3y^2-10y+y^2=14\\7y^2+50y+86=0\\D=2500-4\cdot7\cdot86=2500-2408=92=(\pm\sqrt{92})^2=(\pm2\sqrt{23})^2\\y_{1,2}=\frac{-50\pm2\sqrt{23}}{14}\\y_1=-\frac{25+\sqrt{23}}7\\y_2=-\frac{25-\sqrt{23}}7$

$\boxed{\begin{cases}x_1=10-\frac{75+3\sqrt{23}}7\\y_1=-\frac{25+\sqrt{23}}7\end{cases}\quad\quad\quad u\quad\quad\quad\begin{cases}x_2=10-\frac{75-3\sqrt{23}}7\\y_2=-\frac{25-\sqrt{23}}7\end{cases}}$