Question

tìm GTLN

\(M=\dfrac{\sqrt{x}+1}{x-\sqrt{x}+7}\)

Answer 1

ĐKXĐ: x>=0

Ta có: \(M=\frac{\sqrt{x}+1}{x-\sqrt{x}+7}\)

=>\(\frac{1}{M}=\frac{x-\sqrt{x}+7}{\sqrt{x}+1}=\frac{x+\sqrt{x}-2\sqrt{x}-2+9}{\sqrt{x}+1}=\sqrt{x}-2+\frac{9}{\sqrt{x}+1}\)

=>\(\frac{1}{M}=\sqrt{x}+1+\frac{9}{\sqrt{x}+1}-3\ge2\cdot\sqrt{\left(\sqrt{x}+1\right)\cdot\frac{9}{\sqrt{x}+1}}-3=2\cdot3-3=3\forall x\) thỏa mãn ĐKXĐ

=>\(M\le\frac13\forall x\) thỏa mãn ĐKXĐ

Dấu '=' xảy ra khi \(\left(\sqrt{x}+1\right)^2=9\)

=>\(\sqrt{x}+1=3\)

=>\(\sqrt{x}=2\)

=>x=4(nhận)