Question

[Hungkhuat I'm]

giúp với mn

Answer 1

[Edogawa Conan]

a) x = 25 => B = \(\frac{\sqrt{25}+3}{\sqrt{25}+1}=\frac{5+3}{5+1}=\frac{8}{6}=\frac{4}{3}\)

\(A=\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{1}{\sqrt{x}+2}-\frac{3\sqrt{x}}{x+\sqrt{x}-2}\)

\(A=\frac{\sqrt{x}\left(\sqrt{x}+2\right)+\sqrt{x}-1-3\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

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

3. \(S=A.B=\frac{\sqrt{x}+1}{\sqrt{x}+2}\cdot\frac{\sqrt{x}+3}{\sqrt{x}+1}=\frac{\sqrt{x}+3}{\sqrt{x}+2}=1+\frac{1}{\sqrt{x}+2}\)

Do \(\sqrt{x}+2\ge2\forall x\in R\) => \(\frac{1}{\sqrt{x}+2}\le\frac{1}{2}\) => \(S=1+\frac{1}{\sqrt{x}+2}\le1+\frac{1}{2}=\frac{3}{2}\)

Dấu "=" xảy ra<=> x=  0

Vậy MaxS = 3/2 <=> x=  0