Question

Mọi người giúp em giải cầu đánh dấu với

Answer 1

\(2^{\sqrt{3x+2y-1}}+3^{\sqrt{2x-y-2}}=2\)

Ta có: \(\left\{{}\begin{matrix}\sqrt{3x+2y-1}\ge0\\\sqrt{2x-y-2}\ge0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}2^{\sqrt{3x+2y-1}}\ge1\\3^{\sqrt{2x-y-2}}\ge1\end{matrix}\right.\)

\(\Rightarrow2^{\sqrt{3x+2y-1}}+3^{\sqrt{2x-y-2}}\ge2\)

Dấu = xảy ra khi

\(\left\{{}\begin{matrix}3x+2y-1=0\\2x-y-2=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{7}\\y=-\dfrac{4}{7}\end{matrix}\right.\)