Question

решите пожалуйста систему уравнений ​

Answer 1

Ответ:

$(1;1),\ \left(\frac{\sqrt[3]{6}}{3}; \frac{2\sqrt[3]{6}}{3}\right)$

Объяснение:

ОДЗ:

$y>0,\ x+y>0$

$\begin{cases}2-log_2y=2log_2(x+y)\\ log_2(x+y)+log_2(x^2-xy+y^2)=1 \end{cases}$

$\begin{cases}2log_22-log_2y=log_2(x+y)^2\\ log_2 \left((x+y)(x^2-xy+y^2) \right) =log_22 \end{cases}$

$\begin{cases}log_22^2-log_2y=log_2(x+y)^2\\ (x+y)(x^2-xy+y^2) =2 \end{cases}$

$\begin{cases}log_24-log_2y=log_2(x+y)^2\\(x+y)(x^2-xy+y^2) =2 \end{cases}$

$\begin{cases}log_2\frac{4}{y}=log_2(x+y)^2\\(x+y)(x^2-xy+y^2) =2 \end{cases}$

$\begin{cases}\frac{4}{y}=(x+y)^2\\(x+y)(x^2-xy+y^2) =2 \end{cases}$

$\begin{cases}y(x+y)^2=4\\(x+y)(x^2-xy+y^2) =2 \end{cases}$

1)

$y(x+y)^2=4\ \ \ |:y$

$(x+y)^2=\frac{4}{y}$

$x+y=\sqrt{\frac{4}{y}}$

$x+y=\frac{2}{\sqrt{y}}$

$x=\frac{2}{\sqrt{y}}-y$

2)

$(x+y)(x^2-xy+y^2) =2$

$\left(\frac{2}{\sqrt{y}} \right) \left( \left(\frac{2}{\sqrt{y}}-y\right)^2- \left(\frac{2}{\sqrt{y}}-y\right)y+y^2 \right) =2$

$\frac{2}{\sqrt{y}} \left( \frac{4}{y}-2\cdot \frac{2}{\sqrt y}\cdot y+y^2- \frac{2y}{\sqrt y}+y^2+y^2 \right) =2$

$\frac{2}{\sqrt{y}} \left( \frac{4}{y}-\frac{4y}{\sqrt y}- \frac{2y}{\sqrt y}+3y^2 \right)=2$

$\frac{2}{\sqrt{y}} \left( \frac{4}{y}-\frac{6y}{\sqrt y}+3y^2 \right)=2\ \ \ |:\frac{2}{\sqrt{y}}$

$\frac{4}{y}-\frac{6y}{\sqrt y}+3y^2 =\sqrt y\ \ \ |\cdot y\sqrt y$

$4\sqrt y-6y^2+3y^3\sqrt y =y^2$

$4\sqrt y+3y^3\sqrt y =y^2+6y^2$

$4\sqrt y+3y^3\sqrt y =7y^2\ \ \ |()^2$

$16y+24y^4+9y^7 =49y^4$

$16y+24y^4+9y^7-49y^4=0$

$9y^7-25y^4+16y=0\ \ \ |:y$

$9y^6-25y^3+16=0$

$9(y^3)^2-25y^3+16=0$

заменяем

$y^3=t,\ t>0$

$9t^2-25t+16=0$

$D=(-25)^2-4\cdot 9\cdot 16=625-576=49$

$\sqrt{D}=\sqrt{49}=7$

$t_1=\frac{25-7}{2\cdot 9}=\frac{18}{18}=1$

$t_2=\frac{25+7}{2\cdot 9}=\frac{32}{18}=\frac{16}{9}$

$y_1=\sqrt[3]{1}=1$

$y_2=\sqrt[3]{\frac{16}{9}}=\sqrt[3]{\frac{16\cdot 3}{9\cdot 3}}=\sqrt[3]{\frac{48}{27}}=\frac{\sqrt[3]{48}}{3}=\frac{2\sqrt[3]{6}}{3}$

$x=\frac{2}{\sqrt{y}}-y$

$x_1=\frac{2}{\sqrt{1}}-1=\frac{2}{1}-1=2-1=1$

$x_2=\frac{2}{\sqrt{\sqrt[3]{\frac{16}{9}}}}-\frac{2\sqrt[3]{6}}{3}=\frac{2}{\sqrt[6]{\frac{16}{9}}}-\frac{2\sqrt[3]{6}}{3}=\sqrt[6]\frac{2^6}{\frac{16}{9}}-\frac{2\sqrt[3]{6}}{3}=$

$\sqrt[6]{\frac{64\cdot 9}{16}}-\frac{2\sqrt[3]{6}}{3}=\sqrt[6]{36}-\frac{2\sqrt[3]{6}}{3}=\sqrt[6]{6^2}-\frac{2\sqrt[3]{6}}{3}=\sqrt[3]{6}-\frac{2\sqrt[3]{6}}{3}=\frac{\sqrt[3]{6}}{3}$

$(1;1),\ \left(\frac{\sqrt[3]{6}}{3}; \frac{2\sqrt[3]{6}}{3}\right)$