Giải hệ phương trình: (Editor's note: This translates to "Solve a system of equations:") $$ \begin{cases}\sqrt[4]{x}\left(\dfrac{1}{4}+\dfrac{2\sqrt{x}+\sqrt{y}}{x+y}\right)=2 \\[8pt] \sqrt[4]{y}\left(\dfrac{1}{4}-\dfrac{2\sqrt{x}+\sqrt{y}}{x+y}\right) =1\end{cases} $$

  • 3
    $\begingroup$ differential equations? $\endgroup$
    – user52188
    Jan 18, 2013 at 0:48
  • $\begingroup$ I added the algebra-precalculus tag, but didn't remove the DEs tag, just in case this somehow relates the DEs. I don't think it does, though. $\endgroup$
    – apnorton
    Jan 18, 2013 at 0:59
  • $\begingroup$ "Pre-calculus" doesn't seem like an appropriate description. $\endgroup$ Jan 18, 2013 at 1:01

2 Answers 2


Divide the first equation by $\sqrt[4]{x}$ and the second by $\sqrt[4]{y}$. Then adding and subtracting the two resulting equations gives us the new pair of simultaneous equations: $$ \frac{2}{\sqrt[4]{x}}+\frac{1}{\sqrt[4]{y}}=\frac{1}{2} \\ \frac{2}{\sqrt[4]{x}}-\frac{1}{\sqrt[4]{y}}=2\frac{2\sqrt{x}+\sqrt{y}}{x+y} \, . $$

Multiplying these two equations together, we have: $$ \frac{4}{\sqrt{x}}-\frac{1}{\sqrt{y}}=\frac{2\sqrt{x}+\sqrt{y}}{x+y} \, . $$ Clearing denominators yields: $$ (4\sqrt{y}-\sqrt{x})(x+y)=2x\sqrt{y}+y\sqrt{x} $$ which after some algebra can be reduced to $$ (x+2y)(2\sqrt{y}-\sqrt{x})=0 \, . $$

So either $x=-2y$ or $\sqrt{x}=2\sqrt{y}$. If $x$ and $y$ are positive and real the first is clearly impossible and the second is equivalent to $x=4y$. (If they're not positive and real, we have to worry more about branch cuts than I really want to.) Now, if $x=4y$ we have $$ \frac{\sqrt{2}}{\sqrt[4]{y}}+\frac{1}{\sqrt[4]{y}}=\frac{1}{2} \, , $$ from the top equation in this answer. So $\sqrt[4]{y}=2(1+\sqrt{2})$, yielding the solution $$ x=64(1+\sqrt{2})^4 \\ y=16(1+\sqrt{2})^4 \, , $$ which upon expanding is precisely what Robert Israel got from Maple in the other answer.


According to Maple, there is one real solution, $$x = 1088+768 \sqrt{2},\ y = 272+192 \sqrt{2}$$


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