Question

Tìm GTNN của biểu thức :

D = \(x+2y-\sqrt{2x-1}-5\sqrt{4y-3}+13\)  (x ≥ 1/2, y ≥ 3/4)

Helppp!!! :(

Answer 1

\(D=x+2y-\sqrt{2x-1}-5\cdot\sqrt{4y-3}+13\)

\(=\frac12\left(2x+4y-2\sqrt{2x-1}-10\sqrt{4y-3}+26\right)\)

\(=\frac12\left\lbrack2x-1-2\sqrt{2x-1}+1+4y-3-10\cdot\sqrt{4y-3}+25+4\right\rbrack\)

\(=\frac12\left\lbrack\left(\sqrt{2x-1}-1\right)^2+\left(\sqrt{4y-3}-5\right)^2+4\right\rbrack\ge\frac12\cdot\left\lbrack0+0+4\right\rbrack=2\forall x,y\) thỏa mãn ĐKXĐ

Dấu '=' xảy ra khi \(\begin{cases}2x-1=1\\ 4y-3=25\end{cases}\Rightarrow\begin{cases}2x=2\\ 4y=28\end{cases}\Rightarrow\begin{cases}x=1\\ y=7\end{cases}\)