Question

Answer 1

Bài 3:

1: Thay x=16 vào A, ta được:

\(A=\frac{16-9}{\sqrt{16}}=\frac74\)

2: \(B=\frac{2}{\sqrt{x}-3}-\frac{\sqrt{x}+4}{x-9}\)

\(=\frac{2}{\sqrt{x}-3}-\frac{\sqrt{x}+4}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\frac{2\left(\sqrt{x}+3\right)-\sqrt{x}-4}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\frac{\sqrt{x}+2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\frac{\sqrt{x}+2}{x-9}\)

3: P=AB

\(=\frac{x-9}{\sqrt{x}}\cdot\frac{\sqrt{x}+2}{x-9}=\frac{\sqrt{x}+2}{\sqrt{x}}\)

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

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

=>\(\sqrt{P}-1>0\)

mà \(\sqrt{P}>0\)

nên \(\sqrt{P}\cdot\left(\sqrt{P}-1\right)>0\)

=>\(P-\sqrt{P}>0\)

=>\(P>\sqrt{P}\)

Bài 2:

1: Thay x=4 vào A, ta được:

\(A=\frac{\sqrt4-2}{\sqrt4}=\frac{2-2}{2}=0\)

2: \(B=\frac{\sqrt{x}-1}{\sqrt{x}-3}-\frac{7\sqrt{x}-9}{x-9}\)

\(=\frac{\sqrt{x}-1}{\sqrt{x}-3}-\frac{7\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)-7\sqrt{x}+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\frac{x+2\sqrt{x}-3-7\sqrt{x}+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\frac{x-5\sqrt{x}+6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\frac{\sqrt{x}-2}{\sqrt{x}+3}\)

P=A:B

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

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

3: P>=2

=>P-2>=0

=>\(\frac{\sqrt{x}+3-2\sqrt{x}}{\sqrt{x}}\ge0\)

=>\(\frac{-\sqrt{x}+3}{\sqrt{x}}\ge0\)

=>\(-\sqrt{x}+3\ge0\)

=>\(-\sqrt{x}\ge-3\)

=>\(\sqrt{x}\le3\)

=>0<x<=9

Kết hợp ĐKXĐ, ta được: 0<x<9