Question

Tìm GTLN

a) A=\(\sqrt{x}\)- x

b) B = x + \(\sqrt{3-x}\)

Answer 1

a/ ĐKXĐ: \(x\ge0\)

\(A=\frac{1}{4}-\frac{1}{4}+\sqrt{x}-x=\frac{1}{4}-\left(x-\sqrt{x}+\frac{1}{4}\right)=\frac{1}{4}-\left(\sqrt{x}-\frac{1}{2}\right)^2\le\frac{1}{4}\)

\(A_{max}=\frac{1}{4}\) khi \(\sqrt{x}-\frac{1}{2}=0\Rightarrow x=\frac{1}{4}\)

b/ ĐKXĐ: \(x\le3\)

\(B=\frac{13}{4}-\left(3-x-\sqrt{3-x}+\frac{1}{4}\right)=\frac{13}{4}-\left(\sqrt{3-x}-\frac{1}{2}\right)^2\le\frac{13}{4}\)

\(B_{max}=\frac{13}{4}\) khi \(3-x=\frac{1}{4}\Rightarrow x=\frac{11}{4}\)