Question

Answer 1

6A:

a: \(B=\frac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\frac{\sqrt{x}+3}{\sqrt{x}-2}-\frac{2\sqrt{x}+1}{3-\sqrt{x}}\)

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

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

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

b: P=A:B

\(=\frac{x+2\sqrt{x}+5}{\sqrt{x}-3}:\frac{\sqrt{x}+1}{\sqrt{x}-3}=\frac{x+2\sqrt{x}+5}{\sqrt{x}+1}\)

\(=\sqrt{x}+1+\frac{4}{\sqrt{x}+1}\ge2\cdot\sqrt{\left(\sqrt{x}+1\right)\cdot\frac{4}{\sqrt{x}+1}}=2\cdot2=4\forall x\) thỏa mãn ĐKXĐ

Dấu '=' xảy ra khi \(\left(\sqrt{x}+1\right)^2=4\)

=>\(\sqrt{x}+1=2\)

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

=>x=1(nhận)

Bài 6B:

a: \(P=\frac{\sqrt{x}}{\sqrt{x}+3}+\frac{2\sqrt{x}}{\sqrt{x}-3}-\frac{3x+9}{x-9}\)

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

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

b: \(\sqrt{x}+3\ge3\forall x\) thỏa mãn ĐKXĐ

=>\(\frac{3}{\sqrt{x}+3}\le\frac33=1\forall x\) thỏa mãn ĐKXĐ

Dấu '=' xảy ra khi x=0