a: Ta có: \(P=\left(\dfrac{1}{1-\sqrt{a}}-\dfrac{1}{1+\sqrt{a}}\right)\left(\dfrac{1}{\sqrt{a}}+1\right)\)

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

\(=\dfrac{2}{1-\sqrt{a}}\)

a: Ta có: \(P=\left(\dfrac{4a}{\sqrt{a}-1}-\dfrac{\sqrt{a}}{a-\sqrt{a}}\right)\cdot\dfrac{\sqrt{a}-1}{a^2}\)

\(=\dfrac{4a-1}{\sqrt{a}-1}\cdot\dfrac{\sqrt{a}-1}{a^2}\)

\(=\dfrac{4a-1}{a^2}\)

b: Để P=3 thì \(4a-1=3a^2\)

\(\Leftrightarrow3a^2-4a+1=0\)

\(\Leftrightarrow\left(3a-1\right)\left(a-1\right)=0\)

hay \(a=\dfrac{1}{9}\)

a) ĐK: a>0; a≠1 

Ta có: \(P=\left(\dfrac{4a}{\sqrt{a}-1}-\dfrac{\sqrt{a}}{a-\sqrt{a}}\right).\dfrac{\sqrt{a}-1}{a^2}\)

                  \(=\left(\dfrac{4a}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}-1}\right).\dfrac{\sqrt{a}-1}{a^2}\)

                  \(=\dfrac{4a-1}{\sqrt{a}-1}.\dfrac{\sqrt{a}-1}{a^2}=\dfrac{4a-1}{a^2}\)

b) Ta có: \(P=3\Leftrightarrow\dfrac{4a-1}{a^2}=3\Leftrightarrow3a^2=4a-1\Leftrightarrow3a^2-4a+1=0\)

               \(\Leftrightarrow\left(a-1\right)\left(3a-1\right)=0\)

               \(\Leftrightarrow\left[{}\begin{matrix}a=1\left(loại\right)\\a=\dfrac{1}{3}\left(tm\right)\end{matrix}\right.\)

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

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

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

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

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

b: Khi x=9/4 thì \(P=\dfrac{3}{2}:\left(\dfrac{3}{2}-1\right)=\dfrac{3}{2}:\dfrac{1}{2}=3\)

c: P<0

=>\(\dfrac{\sqrt{x}}{\sqrt{x}-1}< 0\)

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

=>\(\sqrt{x}< 1\)

=>0<=x<1

a) \(M=3\sqrt{3}-\sqrt{12}-\sqrt{\left(\sqrt{3}-1\right)^2}\)

\(M=3\sqrt{3}-2\sqrt{3}-\left|\sqrt{3}-1\right|\)

\(M=\sqrt{3}-\sqrt{3}+1\)

\(M=1\)

b) Ta có:

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

\(N=\left(\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}+\dfrac{\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\dfrac{\sqrt{a}+1}{\left(\sqrt{a}-1\right)^2}\)

\(N=\left(\dfrac{1+\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}\right)\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}\)

\(N=\dfrac{\left(\sqrt{a}+1\right)\cdot\left(\sqrt{a}-1\right)^2}{\sqrt{a}\left(\sqrt{a}-1\right)\cdot\left(\sqrt{a}+1\right)}\)

\(N=\dfrac{\sqrt{a}-1}{\sqrt{a}}\)

Theo đề ta có: \(M=2N\)

Khi: \(1=2\cdot\left(\dfrac{\sqrt{a}-1}{\sqrt{a}}\right)\)

\(\Leftrightarrow1=\dfrac{2\sqrt{a}-2}{\sqrt{a}}\)

\(\Leftrightarrow\sqrt{a}=2\sqrt{a}-2\)

\(\Leftrightarrow2\sqrt{a}-\sqrt{a}=2\)

\(\Leftrightarrow\sqrt{a}=2\)

\(\Leftrightarrow a=4\left(tm\right)\)

a) \(A=\dfrac{3\left(\sqrt{a}-1\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}-\dfrac{\sqrt{a}+1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}-\dfrac{\sqrt{a}-3}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\)

\(=\dfrac{3\sqrt{a}-3-\sqrt{a}-1-\sqrt{a}+3}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}=\dfrac{\sqrt{a}-1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}=\dfrac{1}{\sqrt{a}+1}\)

b) Với \(a=3-2\sqrt{2}\)(tmđk)

\(A=\dfrac{1}{\sqrt{a}+1}=\dfrac{\sqrt{a}-1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}=\dfrac{\sqrt{a}-1}{a-1}\)

\(=\dfrac{\sqrt{3-2\sqrt{2}}-1}{3-2\sqrt{2}-1}=\dfrac{\sqrt{\left(\sqrt{2}-1\right)^2}-1}{2-2\sqrt{2}}=\dfrac{\left|\sqrt{2}-1\right|-1}{2-2\sqrt{2}}=\dfrac{\sqrt{2}-1-1}{2-2\sqrt{2}}=\dfrac{\sqrt{2}-2}{2-2\sqrt{2}}=\dfrac{\sqrt{2}\left(1-\sqrt{2}\right)}{2\left(1-\sqrt{2}\right)}=\dfrac{\sqrt{2}}{2}\)

Bài 2: 

Ta có: \(P=\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}-\dfrac{3\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)

\(=\dfrac{15\sqrt{x}-11-3x-9\sqrt{x}+2\sqrt{x}+6-2x+2\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{-5\sqrt{x}+1}{\sqrt{x}+3}\)