Bài 4:
1: Thay x=16 vào A, ta được:
\(A=\dfrac{6}{16-3\cdot4}=\dfrac{6}{16-12}=\dfrac{6}{4}=\dfrac{3}{2}\)
2: P=A:B
\(=\dfrac{6}{x-3\sqrt{x}}:\left(\dfrac{2\sqrt{x}}{x-9}-\dfrac{2}{\sqrt{x}+3}\right)\)
\(=\dfrac{6}{\sqrt{x}\left(\sqrt{x}-3\right)}:\left(\dfrac{2\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\dfrac{2}{\sqrt{x}+3}\right)\)
\(=\dfrac{6}{\sqrt{x}\left(\sqrt{x}-3\right)}:\dfrac{2\sqrt{x}-2\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{6}{\sqrt{x}\left(\sqrt{x}-3\right)}\cdot\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{6}=\dfrac{\sqrt{x}+3}{\sqrt{x}}\)
3: \(P-1=\dfrac{\sqrt{x}+3-\sqrt{x}}{\sqrt{x}}=\dfrac{3}{\sqrt[]{x}}>0\)
=>P>1
Bài 5:
1: Thay x=4 vào A, ta được:
\(A=\dfrac{4+2\cdot\sqrt{4}}{4}=\dfrac{4+2\cdot2}{4}=\dfrac{4+4}{4}=2\)
2: \(B=\dfrac{\sqrt{x}-1}{\sqrt{x}}+\dfrac{2\sqrt{x}+1}{x+\sqrt{x}}\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}}+\dfrac{2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)+2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x+2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\)