a: \(Q=\dfrac{3x+3\sqrt{x}-3}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}-\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\)
\(=\dfrac{3x+3\sqrt{x}-3-\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{3x+3\sqrt{x}-3-x+1-x+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\dfrac{x+3\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
b: Khi x=4+2căn 3 thì \(Q=\dfrac{\sqrt{3}+1+1}{\sqrt{3}+1-1}=\dfrac{2+\sqrt{3}}{\sqrt{3}}=\dfrac{2\sqrt{3}+3}{3}\)
c: Q=3
=>\(3\sqrt{x}-3=\sqrt{x}+1\)
=>\(2\sqrt{x}=4\)
=>căn x=2
=>x=4(nhận)
d: Q>1/2
=>Q-1/2>0
=>\(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{1}{2}>0\)
=>\(\dfrac{2\sqrt{x}+2-\sqrt{x}+1}{2\left(\sqrt{x}-1\right)}>0\)
=>\(\dfrac{\sqrt{x}+3}{2\left(\sqrt{x}-1\right)}>0\)
=>căn x-1>0
=>x>1
e: Q nguyên khi \(\sqrt{x}+1⋮\sqrt{x}-1\)
=>\(\sqrt{x}-1+2⋮\sqrt{x}-1\)
=>\(\sqrt{x}-1\in\left\{1;-1;2;-2\right\}\)
=>\(\sqrt{x}\in\left\{2;0;3;-1\right\}\)
=>\(\sqrt{x}\in\left\{2;0;3\right\}\)
=>\(x\in\left\{0;4;9\right\}\)
