Bài 3:
1: Thay x=9 vào P, ta được:
\(P=\dfrac{3-3}{3}=0\)
2: \(Q=\dfrac{\sqrt{x}-1}{\sqrt{x}+4}+\dfrac{9\sqrt{x}-4}{x-16}-\dfrac{4\sqrt{x}-x}{\sqrt{x}-4}\)
\(=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-4\right)+9\sqrt{x}-4}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+4\right)}+\dfrac{x-4\sqrt{x}}{\sqrt{x}-4}\)
\(=\dfrac{x-5\sqrt{x}+4+9\sqrt{x}-4}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+4\right)}+\dfrac{x-4\sqrt{x}}{\sqrt{x}-4}\)
\(=\dfrac{x+4\sqrt{x}}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+4\right)}+\dfrac{x-4\sqrt{x}}{\sqrt{x}-4}=\dfrac{x-4\sqrt{x}+\sqrt{x}}{\sqrt{x}-4}=\dfrac{x-3\sqrt{x}}{\sqrt{x}-4}\)
3: M=PQ
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{\sqrt{x}-4}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}}=\dfrac{\left(\sqrt{x}-3\right)^2}{\sqrt{x}-4}\)
M>=0
=>\(\dfrac{\left(\sqrt{x}-3\right)^2}{\sqrt{x}-4}>=0\)
=>\(\sqrt{x}-4>0\)
=>x>16
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: \(P=A:B\)
\(=\dfrac{6}{\sqrt{x}\left(\sqrt{x}-3\right)}:\left(\dfrac{2\sqrt{x}}{x-9}-\dfrac{2}{\sqrt{x}+3}\right)\)
\(=\dfrac{6}{\sqrt{x}\left(\sqrt{x}-3\right)}:\dfrac{2\sqrt{x}-2\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{6}{\sqrt{x}\left(\sqrt{x}-3\right)}\cdot\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{2\sqrt{x}-2\sqrt{x}+6}\)
\(=\dfrac{\sqrt{x}+3}{\sqrt{x}}\)
3: \(P-1=\dfrac{\sqrt{x}+3}{\sqrt{x}}-1=\dfrac{3}{\sqrt{x}}>0\)
=>P>1
