Question

Giải hệ: \(\hept{\begin{cases}x\left(2\sqrt{y-1}-x\right)+y\left(2\sqrt{x-1}-y\right)=0\\x^3+y^3=16\end{cases}}\)

Answer 1

xet pt 1

\(x\left(2\sqrt{y-1}-x\right)+y\left(2\sqrt{x-1}-y\right)=0\) dk \(x,y\ge1\)

<=> \(x\left(2\sqrt{\left(y-1\right).1}-x\right)+y\left(2\sqrt{\left(x-1\right).1}-y\right)\)

\(\le x\left(y-1+1-x\right)+y\left(x-1-y\right)\)   (bdt Cosi)

\(\le x\left(y-x\right)+y\left(x-y\right)=-\left(x^2-2xy+y^2\right)=-\left(x-y\right)^2\le0=Vp\)

dau = xay ra \(\hept{\begin{cases}x-1=1\\y-1=1\\x=y\end{cases}\Leftrightarrow x=y=2}\) tmdk

thay x=y=2 vao pt 2 ta thay thoa man

vay {x;y} ={2;2}