Question

[Hungkhuat I'm]

mọi người giúp mình nah

Answer 1

[Edogawa Conan]

1. x = 4 => B = \(\frac{3}{\sqrt{4}-1}=\frac{3}{2-1}=3\)

2. \(P=A-B=\frac{6}{x-1}+\frac{\sqrt{x}}{\sqrt{x}+1}-\frac{3}{\sqrt{x}-1}\)

\(P=\frac{6+\sqrt{x}\left(\sqrt{x}-1\right)-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(P=\frac{6+x-\sqrt{x}-3\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(P=\frac{x-4\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{x-3\sqrt{x}-\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(P=\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}-3}{\sqrt{x}+1}=\frac{\sqrt{x}+1-4}{\sqrt{x}+1}=1-\frac{4}{\sqrt{x}+1}\)

Do \(\sqrt{x}+1\ge1\forall x\in R\) => \(\frac{4}{\sqrt{x}+1}\le4\)=> \(1-\frac{4}{\sqrt{x}+1}\ge1-4=-3\)

=> \(P\ge-3\) => \(\frac{1}{P}\le-\frac{1}{3}\)

Dấu "=" xảy ra<=> x = 0

Vậy Max1/P = -1/3 <=> x = 6