Question

giải hệ pt: \(\left\{{}\begin{matrix}x+y-\sqrt{xy}=3\\\sqrt{x+1}+\sqrt{y+1}=4\end{matrix}\right.\)

Answer 1

áp dụng BĐT Cauchy ngược dấu:

\(x+y=3+\sqrt{xy}\le3+\frac{x+y}{2}\)

\(=>\frac{x+y}{2}\le3=>x+y\le6\left(1\right)\)

\(4=\sqrt{x+1}+\sqrt{y+1}\)

=> \(8=\sqrt{4\left(x+1\right)}+\sqrt{4\left(y+1\right)}\le\frac{4\left(x+1\right)}{2}+\frac{4\left(y+1\right)}{2}=>x+y\ge6\left(2\right)\)

từ (1) và (2) => x+ y = 6

<=> \(\left\{{}\begin{matrix}x=y\\4=\left(x+1\right)=>x=y=3\\4=\left(y+1\right)\end{matrix}\right.\)

vậy hpt có No (x; y) = (3;3)