11.4. Solve the equation $2^{\sqrt{x y}}+5^{\sqrt{x+y}}=3^{-x}+4^{-y}$.
Solution: The domain of the equation is defined by two conditions: $x y \geqslant 0$ and $x+y \geqslant 0$. From the first condition, it follows that either one of the numbers $x$ and $y$ is 0, or these two numbers have the same sign. From the second condition, it is clear that the case where both are negative is impossible; also, the case where one number is negative and the other is zero is impossible. Therefore, $x \geqslant 0$ and $y \geqslant 0$. Then $3^{-x} \leqslant 3^{0}=1, 4^{-y} \leqslant 4^{0}=1$, and $3^{-x}+4^{-y} \leqslant 2$, with equality being achieved only when $x=y=0$. On the other hand, $x y \geqslant 0$, so $2^{\sqrt{x y}} \geqslant 2^{0}=1$. Also, $5^{\sqrt{x+y}} \geqslant 5^{0}=1$, hence $2^{\sqrt{x y}}+5^{\sqrt{x+y}} \geqslant 2$. Thus, the left side of the given equation is not less than two, and the right side is not more than two. Therefore, they are both equal to 2, which is achieved only when $x=y=0$.

Answer: $(0 ; 0)$.

Recommendations for checking:

| is in the work | points |
| :--- | :--- |
| Correct justified answer | 7 points |
| The solution process is correct, but some of the inequalities, either a), or b), or the domain, are incorrectly written as strict; based on this, the conclusion is made that the inequality has no solutions | 6 points |
| Two inequalities a) $2^{\sqrt{x y}}+5^{\sqrt{x+y}} \geqslant 2$ and b) $3^{-x}+4^{-y} \leqslant 2$ are proven, but either an incorrect conclusion is made that the inequality has no solutions, or it is not justified that equality is achieved in the unique case $x=y=0$ | 5 points |
| One of the two inequalities a) or b) from the previous point is proven | 3 points |
| The domain of the equation is correctly found only | 1 point |
| Correct answer without justification (with incorrect justification) | 0 points |
| Any calculations not leading to a solution | not evaluated |