Question

Answer 1

Bài 1:

1: Thay \(x=4-2\sqrt3=\left(\sqrt3-1\right)^2\) vào Q, ta được:

\(Q=\frac{\sqrt{\left(\sqrt3-1\right)^2}+1}{\sqrt{\left(\sqrt3-1\right)^2}+3}=\frac{\sqrt3-1+1}{\sqrt3-1+3}=\frac{\sqrt3}{2+\sqrt3}=\sqrt3\left(2-\sqrt3\right)=2\sqrt3-3\)

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

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

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

M=P:Q

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

Bài 2:

1: \(A=\left(\frac{x+3\sqrt{x}}{x-25}+\frac{1}{\sqrt{x}+5}\right):\frac{\sqrt{x}+2}{\sqrt{x}-5}\)

\(=\left(\frac{x+3\sqrt{x}}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}+\frac{1}{\sqrt{x}+5}\right)\cdot\frac{\sqrt{x}-5}{\sqrt{x}+2}\)

\(=\frac{x+3\sqrt{x}+\sqrt{x}-5}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\cdot\frac{\sqrt{x}-5}{\sqrt{x}+2}=\frac{x+4\sqrt{x}-5}{\left(\sqrt{x}+5\right)\left(\sqrt{x}+2\right)}\)

\(=\frac{\left(\sqrt{x}+5\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}+2\right)}=\frac{\sqrt{x}-1}{\sqrt{x}+2}\)

2: Thay x=4 vào A, ta được:

\(A=\frac{2-1}{2+2}=\frac14\)

3: \(A-1=\frac{\sqrt{x}-1}{\sqrt{x}+2}-1=\frac{\sqrt{x}-1-\sqrt{x}-2}{\sqrt{x}+2}=\frac{-3}{\sqrt{x}+2}<0\)

=>A<1