\(A=\left(\frac{\sqrt{x}+1}{\sqrt{x}-1}+\frac{\sqrt{x}}{\sqrt{x}+1}+\frac{\sqrt{x}}{1-x}\right):\left(\frac{\sqrt{x}+1}{\sqrt{x}-1}-\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)\)

\(A=\left(\frac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)\(\div\left(\frac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\)

\(A=\left(\frac{x+2\sqrt{x}+1+x-\sqrt{x}-\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right):\frac{x+2\sqrt{x}+1-x+2\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(A=\frac{2x+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{4\sqrt{x}}\)

\(A=\frac{2x+1}{4\sqrt{x}}\)

c, \(A=\frac{2x+1}{4\sqrt{x}}=\frac{\sqrt{x}}{2}+\frac{1}{4\sqrt{x}}\)

ap dụng cô si ta có \(\frac{\sqrt{x}}{2}+\frac{1}{4\sqrt{x}}\ge2\sqrt{\frac{\sqrt{x}}{2}\cdot\frac{1}{4\sqrt{x}}}=\frac{\sqrt{2}}{2}\)

dấu = xảy ra khi \(\frac{\sqrt{x}}{2}=\frac{1}{4\sqrt{x}}\Leftrightarrow x=\frac{1}{2}\) (tm)

Trả lời:

\(P=\frac{2\sqrt{x}-1}{\sqrt{x}+1}\left(ĐK:x\ge0;x\ne1\right)\)

+) P > 0 

\(\frac{2\sqrt{x}-1}{\sqrt{x}+1}>0\)

\(\Leftrightarrow2\sqrt{x}-1>0\) ( vì \(\sqrt{x}+1>0\) )

\(\Leftrightarrow2\sqrt{x}>1\)

\(\Leftrightarrow\sqrt{x}>\frac{1}{2}\)

\(\Leftrightarrow x>\frac{1}{4}\) 

Vậy để P > 0 thì \(x>\frac{1}{4}\) và \(x\ne1\)

+) P < 1 

\(\frac{2\sqrt{x}-1}{\sqrt{x}+1}< 1\)

\(\Leftrightarrow\frac{2\sqrt{x}-1}{\sqrt{x}+1}-1< 0\)

\(\Leftrightarrow\frac{2\sqrt{x}-1-\sqrt{x}-1}{\sqrt{x}+1}< 0\)

\(\Leftrightarrow\frac{\sqrt{x}-2}{\sqrt{x}+1}< 0\)

\(\Rightarrow\sqrt{x}-2< 0\)

\(\Leftrightarrow\sqrt{x}< 2\)

\(\Leftrightarrow x< 4\)  

Vậy để P < 1 thì \(0\le x< 4\) và \(x\ne1\)

\(D=x+1-\sqrt{x}=x-\sqrt{x}+\frac{1}{4}+\frac{3}{4}=\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{3}{4}>0\)

\(ĐKXĐ:x\ge0\)

\(\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)-2x=-4\)

\(2x+2\sqrt{x}-3\sqrt{x}-3+4-2x=0\)

\(1-\sqrt{x}=0\)

\(\sqrt{x}=1\)

\(x=1\)

\(2\ge2a+3b\ge2\sqrt{2.3.ab}\Rightarrow ab\le\frac{1}{6}\)

\(A=\frac{4}{4a^2+9b^2}+\frac{9}{ab}=\frac{4}{4a^2+9b^2}+\frac{4}{12ab}+\frac{26}{3ab}\)

\(\ge\frac{\left(2+2\right)^2}{4a^2+9b^2+12ab}+\frac{26}{3.\frac{1}{6}}\)

\(=\frac{4^2}{2^2}+52=56\)

Dấu \(=\)khi \(\hept{\begin{cases}2a=3b\\2a+3b=2\end{cases}}\Leftrightarrow\hept{\begin{cases}a=\frac{1}{2}\\b=\frac{1}{3}\end{cases}}\).