\(\dfrac{3\left(x+\sqrt{x}-3\right)+\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{3x+3\sqrt{x}-9+x-\sqrt{x}+3\sqrt{x}-3-x+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{3x+5\sqrt{x}-8}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{\sqrt{x}+\dfrac{8}{3}}{\sqrt{x}+2}\)
b) ta có: P < \(\dfrac{15}{4}\)
=> \(\dfrac{\sqrt{x}+\dfrac{8}{2}}{\sqrt{x}+2}< \dfrac{15}{2}\)
<=> \(\dfrac{2\sqrt{x}+8-15\sqrt{x}-30}{2\left(\sqrt{x}+2\right)}< 0\)
<=> \(\dfrac{-13\sqrt{x}-22}{2\left(\sqrt{x}+2\right)}< 0\)
Vì \(2\left(\sqrt{x}+2\right)>0\) với mọi \(x\ge0;x\ne1\)
Nên -\(13\sqrt{x}-22< 0\)
<=> \(\sqrt{x}>\dfrac{-22}{13}\)
<=> \(x>\dfrac{484}{169}\) với mọi
c) có P =\(\dfrac{\sqrt{x}+\dfrac{8}{3}}{\sqrt{x}+2}=1+\dfrac{\dfrac{2}{3}}{\sqrt{x}+2}=1+\dfrac{2}{3\left(\sqrt{x}+2\right)}\)
ta có : \(\sqrt{x}\ge0\) với mọi x \(\ge\) 0 ; \(x\ne1\)
=> \(\sqrt{x}+2\ge2\) với mọi x \(\ge\) 0 ; \(x\ne1\)
=> 3(\(\sqrt{x}+2\) ) \(\ge\) 6 với mọi x.................
=> \(\dfrac{1}{3\left(\sqrt{x}+2\right)}\le6\) với mọi x ................
=> \(\dfrac{2}{3\left(\sqrt{x}+2\right)}\le\dfrac{1}{3}\) với mọi x................
=> 1 + \(\dfrac{2}{3\left(\sqrt{x}+2\right)}\le\dfrac{4}{3}\) với mọi x...............
=> P \(\le\dfrac{4}{3}\)
=> MaxP = \(\dfrac{4}{3}\)
Dấu = xảy ra <=>\(\sqrt{x}\) = 0
<=> x = 0
