\(a.P=1:\left(\dfrac{x+2\sqrt{x}-2}{x\sqrt{x}+1}-\dfrac{\sqrt{x}-1}{x-\sqrt{x}+1}+\dfrac{1}{\sqrt{x}+1}\right)=1:\dfrac{x+2\sqrt{x}-2-x+1+x-\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}=\dfrac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}=\dfrac{x-\sqrt{x}+1}{\sqrt{x}}\) \(b.x>0\)
\(x=7-4\sqrt{3}=4-2.2\sqrt{3}+3=\left(2-\sqrt{3}\right)^2\left(TMĐKXĐ\right)\)
⇒ \(\sqrt{x}=2-\sqrt{3}\)
Khi đó : \(\dfrac{7-4\sqrt{3}+\sqrt{3}-2+1}{2-\sqrt{3}}=\dfrac{6-3\sqrt{3}}{2-\sqrt{3}}=\dfrac{3\left(2-\sqrt{3}\right)}{2-\sqrt{3}}=3\)
\(c.P=\dfrac{x-\sqrt{x}+1}{\sqrt{x}}=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}}+\dfrac{1}{\sqrt{x}}=\sqrt{x}+\dfrac{1}{\sqrt{x}}-1\)
Áp dụng BĐT Cauchy cho các số dương , ta có :
\(\sqrt{x}+\dfrac{1}{\sqrt{x}}\) ≥ \(2\sqrt{\sqrt{x}.\dfrac{1}{\sqrt{x}}}=2\)
⇔ \(\sqrt{x}+\dfrac{1}{\sqrt{x}}-1\text{≥}2-1=1\)
⇒ \(P_{MIN}=1."="\text{⇔}x=1\left(TM\right)\)
\(d.P=2\sqrt{x}-1\text{⇔}\dfrac{x+1}{\sqrt{x}}=2\sqrt{x}\)
\(\text{⇔}2x=x+1\) \(\text{⇔}x=1\left(TM\right)\)
