\(=\dfrac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{2}{\sqrt{x}-1}\)

\(=\dfrac{2}{x+\sqrt{x}+1}\)

a) ĐK: \(x\ge0,x\ne1,x\ne\frac{1}{4}\)

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

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

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

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

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

Để \(A=\frac{6-\sqrt{6}}{5}\Rightarrow\frac{x+1}{x+\sqrt{x}+1}=\frac{6-\sqrt{6}}{5}\)

\(\Rightarrow5x+5=\left(6-\sqrt{6}\right)x+\left(6-\sqrt{6}\right)\sqrt{x}+6-\sqrt{6}\)

\(\Rightarrow\left(1-\sqrt{6}\right)x+\left(6-\sqrt{6}\right)\sqrt{x}+1-\sqrt{6}=0\)

\(\Rightarrow x-\sqrt{6}.\sqrt{x}+1=0\)

\(\Rightarrow\orbr{\begin{cases}\sqrt{x}=\frac{\sqrt{2}+\sqrt{6}}{2}\\\sqrt{x}=\frac{-\sqrt{2}+\sqrt{6}}{2}\end{cases}}\Rightarrow\orbr{\begin{cases}x=2+\sqrt{3}\\x=2-\sqrt{3}\end{cases}}\left(tmđk\right)\)

b) Xét \(A-\frac{2}{3}=\frac{x+1}{x+\sqrt{x}+1}-\frac{2}{3}=\frac{3x+3-2x-2\sqrt{x}-2}{3\left(x+\sqrt{x}+1\right)}\)

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

Do \(x\ge0,x\ne1,x\ne\frac{1}{4}\Rightarrow\left(\sqrt{x}-1\right)^2>0\)

Lại có \(x+\sqrt{x}+1=\left(\sqrt{x}+\frac{1}{2}\right)+\frac{3}{4}>0\)

Nên \(A-\frac{2}{3}>0\Rightarrow A>\frac{2}{3}\).

a,\(P=\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{2}\)

\(P=\left[\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right].\dfrac{2}{\sqrt{x}-1}\)

\(P=\dfrac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{2}{\sqrt{x}-1}\)

\(P=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{2}{\sqrt{x}-1}\)

\(P=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{2}{\sqrt{x}-1}=\dfrac{2}{x+\sqrt{x}+1}\)

Vậy \(P=\dfrac{2}{x+\sqrt{x}+1}\)

b, Ta có \(x+\sqrt{x}+1=\left(x+2\sqrt{x}.\dfrac{1}{2}+\dfrac{1}{4}\right)+\dfrac{3}{4}=\left(\sqrt{x}+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\forall x\)Suy ra \(\dfrac{2}{x+\sqrt{x}+1}>0\forall x>0,x\ne1\)

hay \(P>0\forall x>0,x\ne1\)(đpcm)

Câu b bạn sửa lại đề

\(a,VT=\left[1+\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\right]\left[1-\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\right]\\ =\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)=1-x=VP\\ b,VT=\dfrac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{ab}}+\dfrac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}\\ =\sqrt{a}-\sqrt{b}+\sqrt{a}+\sqrt{b}=2\sqrt{a}=VP\)

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

1)))))))

\(\frac{2}{\sqrt{ab}}:\left(\frac{1}{\sqrt{a}}-\frac{1}{\sqrt{b}}\right)^2-\frac{a+b}{\left(\sqrt{a}-\sqrt{b}\right)^2}\)

\(=\frac{2}{\sqrt{ab}}:\frac{\left(\sqrt{b}-\sqrt{a}\right)^2}{\left(\sqrt{ab}\right)^2}-\frac{a+b}{\left(\sqrt{a}-\sqrt{b}\right)^2}\)

\(=\frac{2}{\sqrt{ab}}.\frac{\left(\sqrt{ab}\right)^2}{\left(\sqrt{a}-\sqrt{b}\right)^2}-\frac{a+b}{\left(\sqrt{a}-\sqrt{b}\right)^2}\)

\(=\frac{2\sqrt{ab}}{\left(\sqrt{a}-\sqrt{b}\right)^2}-\frac{a+b}{\left(\sqrt{a}-\sqrt{b}\right)^2}\)

\(=\frac{2\sqrt{ab}-a-b}{\left(\sqrt{a}-\sqrt{b}\right)^2}\)

\(=\frac{-\left(\sqrt{a}-\sqrt{b}\right)^2}{\left(\sqrt{a}-\sqrt{b}\right)^2}=-1\)

\(\text{VT}=\left(1+\frac{x+\sqrt{x}}{\sqrt{x}+1}\right)\left(1-\frac{x-\sqrt{x}}{\sqrt{x}-1}\right)=\left(1+\frac{\sqrt{x}.\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\right)\left(1-\frac{\sqrt{x}.\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\right)\)

\(=\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)=1-x=\text{VP(điều phải chứng minh)}\)

a/ \(Q=\left[\frac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\frac{\sqrt{x}}{x+\sqrt{x}+1}-\frac{1}{\sqrt{x}-1}\right].\frac{2}{\sqrt{x}-1}\)

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

       \(=\frac{\left(x-2\sqrt{x}+1\right).2}{\left(\sqrt{x}-1\right)^2\left(x+\sqrt{x}+1\right)}=\frac{\left(\sqrt{x}-1\right)^2.2}{\left(\sqrt{x}-1\right)^2\left(x+\sqrt{x}+1\right)}\) 

       \(=\frac{2}{x+\sqrt{x}+1}\)

b/ Ta có: \(x+\sqrt{x}+1=x+2.\frac{1}{2}.\sqrt{x}+\left(\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2+1=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\)

                      \(\Rightarrow Q=\frac{2}{x+\sqrt{x}+1}>0\). 

                                                                                  Vậy Q > 0