\(a,Q=\dfrac{3x-3\sqrt{x}-3}{x+\sqrt{x}-2}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}+\dfrac{\sqrt{x}-2}{1-\sqrt{x}}\left(dk:x\ge0,x\ne1\right)\)
\(=\dfrac{3x-3\sqrt{x}-3}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}-\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\)
\(=\dfrac{3x-3\sqrt{x}-3-\left(x-1\right)-\left(x-4\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\\ =\dfrac{3x-3\sqrt{x}-3-x+1-x+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\\ =\dfrac{x-3\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\\ =\dfrac{x-\sqrt{x}-2\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\\ =\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)-2\left(\sqrt{x}-\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\\ =\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\\ =\dfrac{\sqrt{x}-2}{\sqrt{x}+2}\)
\(b,x=\dfrac{1}{9}\Leftrightarrow Q=\dfrac{\sqrt{\dfrac{1}{9}}-2}{\sqrt{\dfrac{1}{9}}+2}=\dfrac{\dfrac{1}{3}-2}{\dfrac{1}{3}+2}=-\dfrac{5}{7}\)
\(c,Q=\dfrac{1}{3}\Leftrightarrow\dfrac{\sqrt{x}-2}{\sqrt{x}+2}=\dfrac{1}{3}\\ \Rightarrow3\sqrt{x}-6-\sqrt{x}-2=0\Rightarrow2\sqrt{x}=8\Rightarrow\sqrt{x}=4\Rightarrow x=16\left(tm\right)\)
Vậy \(x=16\) thì \(Q=\dfrac{1}{3}\)
\(d,Q>\dfrac{1}{2}\Leftrightarrow\dfrac{\sqrt{x}-2}{\sqrt{x}+2}>\dfrac{1}{2}\\ \Leftrightarrow\dfrac{2\left(\sqrt{x}-2\right)-\left(\sqrt{x}+2\right)}{2\left(\sqrt{x}+2\right)}>0\Leftrightarrow2\sqrt{x}-4-\sqrt{x}-2>0\Leftrightarrow\sqrt{x}>6\Leftrightarrow x>36\left(tm\right)\)
Vậy \(x>36\) thì \(Q>\dfrac{1}{2}\)
a: \(Q=\dfrac{3x-3\sqrt{x}-3-x+1-x+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{x-3\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}-2}{\sqrt{x}+2}\)
b: Khi x=1/9 thì Q=(1/3-2):(1/3+2)=-5/3:7/3=-5/7
c: Q=1/3
=>(căn x-2)/(căn x+2)=1/3
=>3*căn x-6=căn x+2
=>2căn x=8
=>x=16
d: Q>1/2
=>Q-1/2>0
=>\(\dfrac{\sqrt{x}-2}{\sqrt{x}+2}-\dfrac{1}{2}>0\)
=>\(\dfrac{2\sqrt{x}-4-\sqrt{x}-2}{\left(\sqrt{x}+2\right)\cdot2}>0\)
=>căn x-6>0
=>x>36
