Question

B=\([\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3\left(\sqrt{x}+3\right)}{x-9}]\div[\frac{2\sqrt{x}-2}{\sqrt{x}-3}\times1]\)\(-1]\)

a) Rút gọn 

b) Tính x đế B<-1

c) Tính x để B đạt GTNN

Answer 1

a) đk: \(x\ge0;x\ne9\)

Ta có:

\(B=\left[\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3\left(\sqrt{x}+3\right)}{x-9}\right]\div\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)

\(B=\frac{2\sqrt{x}\left(\sqrt{x}-3\right)+\left(\sqrt{x}+3\right)\sqrt{x}-3\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\div\frac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\)

\(B=\frac{2x-6\sqrt{x}+x+3\sqrt{x}-3\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\div\frac{\sqrt{x}+1}{\sqrt{x}-3}\)

\(B=\frac{3x-6\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\frac{\sqrt{x}-3}{\sqrt{x}+1}\)

\(B=\frac{3\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\frac{\sqrt{x}-3}{\sqrt{x}+1}\)

\(B=\frac{3\left(\sqrt{x}-3\right)}{\sqrt{x}+3}=\frac{3\sqrt{x}-9}{\sqrt{x}+3}\)

b) \(B< -1\Leftrightarrow\frac{3\sqrt{x}-9}{\sqrt{x}+3}+1< 0\)

\(\Leftrightarrow\frac{4\sqrt{x}-6}{\sqrt{x}+3}< 0\) , mà \(\sqrt{x}+3\ge3>0\left(\forall x\right)\)

=> \(4\sqrt{x}-6< 0\)

\(\Leftrightarrow4\sqrt{x}< 6\)

\(\Rightarrow\sqrt{x}< \frac{3}{2}\)

\(\Rightarrow x< \frac{9}{4}\)

Vậy \(0\le x< \frac{9}{4}\)

Answer 2

c) Ta có: \(B=\frac{3\sqrt{x}-9}{\sqrt{x}+3}=\frac{3\left(\sqrt{x}+3\right)-18}{\sqrt{x}+3}=3-\frac{18}{\sqrt{x}+3}\)

Vì \(\sqrt{x}+3\ge3\Rightarrow\frac{18}{\sqrt{x}+3}\le6\)

\(\Leftrightarrow3-\frac{18}{\sqrt{x}+3}\ge-3\)

\(\Rightarrow A\ge-3\)

Dấu "=" xảy ra khi: \(\sqrt{x}+3=3\Rightarrow x=0\)

Vậy \(Min_A=-3\Leftrightarrow x=0\)