Giải hệ phương trình : $\left\{\begin{array}{l}x(x+2y)=10\\\frac{6}{(x+y)^2}(\frac{12}{x^2}+\frac{10}{y^2})+\frac{13}{(x+y)^3}(\frac{2}{x}+\frac{2}{y})=\frac{5}{7}\end{array}\right.$
Giải hệ phương trình : $\left\{\begin{array}{l}x(x+2y)=10\\\frac{6}{(x+y)^2}(\frac{12}{x^2}+\frac{10}{y^2})+\frac{13}{(x+y)^3}(\frac{2}{x}+\frac{2}{y})=\frac{5}{7}\end{array}\right.$
Giải thích các bước giải:
Ta có : $x(x+2y)=10\to y=\dfrac{1}{2}(\dfrac{10}{x}-x)$
$\to \dfrac{6}{(x+\dfrac{1}{2}(\dfrac{10}{x}-x))^2}(\dfrac{12}{x^2}+\dfrac{10}{(\dfrac{1}{2}(\dfrac{10}{x}-x))^2})+\dfrac{13}{(x+\dfrac{1}{2}(\dfrac{10}{x}-x))^3}(\dfrac{2}{x}+\dfrac{2}{\dfrac{1}{2}(\dfrac{10}{x}-x)})=\dfrac{5}{7}$
$\to \dfrac{6}{\left(x+\dfrac{1}{2}\left(\dfrac{10}{x}-x\right)\right)^2}\left(\dfrac{12}{x^2}+\dfrac{10}{\left(\dfrac{1}{2}\left(\dfrac{10}{x}-x\right)\right)^2}\right)+\dfrac{13}{\left(x+\dfrac{1}{2}\left(\dfrac{10}{x}-x\right)\right)^3}\left(\dfrac{2}{x}+\dfrac{2}{\dfrac{1}{2}\left(\dfrac{10}{x}-x\right)}\right)=\dfrac{5}{7}$
$\to \dfrac{24\left(52x^4-240x^2+1200\right)}{\left(x^2+10\right)^2\left(x+\sqrt{10}\right)^2\left(x-\sqrt{10}\right)^2}-\dfrac{208x^2}{\left(x^2+10\right)^2\left(x^2-10\right)}=\dfrac{5}{7}$
$\to \dfrac{24\left(52x^4-240x^2+1200\right)}{\left(x^2+10\right)^2\left(x+\sqrt{10}\right)^2\left(x-\sqrt{10}\right)^2}\cdot \:7\left(x^2+10\right)^2\left(x+\sqrt{10}\right)^2\left(x-\sqrt{10}\right)^2-\dfrac{208x^2}{\left(x^2+10\right)^2\left(x^2-10\right)}\dfrac{24\left(52x^4-240x^2+1200\right)}{\left(x^2+10\right)^2\left(x+\sqrt{10}\right)^2\left(x-\sqrt{10}\right)^2}\cdot \:7\left(x^2+10\right)^2\left(x+\sqrt{10}\right)^2\left(x-\sqrt{10}\right)^2-\dfrac{208x^2}{\left(x^2+10\right)^2\left(x^2-10\right)}$
$\to 168\left(52x^4-240x^2+1200\right)-1456x^2\left(x+\sqrt{10}\right)\left(x-\sqrt{10}\right)=5\left(x^2+10\right)^2\left(x+\sqrt{10}\right)^2\left(x-\sqrt{10}\right)^2$
$\to 5x^8-8280x^4+25760x^2-151600=0$
$\to x=\sqrt{39.30061\dots },\:x=-\sqrt{39.30061\dots }$