ai giúp với ạ :<

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

\(=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)

\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3}\)

\(=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)

1: Ta có: \(A=\left(\dfrac{x-5\sqrt{x}}{x-25}-1\right):\left(\dfrac{25-x}{x+2\sqrt{x}-15}-\dfrac{\sqrt{x}+3}{\sqrt{x}+5}-\dfrac{\sqrt{x}-5}{\sqrt{x}-3}\right)\)

\(=\left(\dfrac{x-5\sqrt{x}-x+25}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\right):\dfrac{25-x-x+9-x+25}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{-5}{\sqrt{x}+5}\cdot\dfrac{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}{-3x+59}\)

\(=\dfrac{-5\left(\sqrt{x}-3\right)}{-3x+59}\)

\(=\dfrac{5\sqrt{x}-15}{3x-59}\)

1: Ta có: \(A=\left(\dfrac{x-5\sqrt{x}}{x-25}-1\right):\left(\dfrac{25-x}{x+2\sqrt{x}-15}-\dfrac{\sqrt{x}+3}{\sqrt{x}+5}-\dfrac{\sqrt{x}-5}{\sqrt{x}-3}\right)\)

\(=\left(\dfrac{x-5\sqrt{x}-x+25}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\right):\dfrac{25-x-x+9-x+25}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{-5}{\sqrt{x}+5}\cdot\dfrac{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}{-3x+59}\)

\(=\dfrac{-5\left(\sqrt{x}-3\right)}{-3x+59}\)

\(=\dfrac{5\sqrt{x}-15}{3x-59}\)​

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

\(=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)

\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3}\)

\(=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)

3: Ta có: \(A=\left(\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\dfrac{\sqrt{x}-2}{x-1}\right)\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}}\)

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

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

\(=\dfrac{2}{x-1}\)

a) Ta có: \(\left(\dfrac{1}{2-\sqrt{3}}-\dfrac{3}{\sqrt{7}-2}\right):\dfrac{2}{\sqrt{7}+\sqrt{3}}\)

\(=\left(2+\sqrt{3}-\sqrt{7}-2\right):\dfrac{\left(\sqrt{7}-\sqrt{3}\right)}{2}\)

\(=\dfrac{-\left(\sqrt{7}-\sqrt{3}\right)}{1}\cdot\dfrac{2}{\sqrt{7}-\sqrt{3}}\)

=-2

b) Ta có: \(\left(\dfrac{x-\sqrt{x}}{1-\sqrt{x}}-1\right):\left(\sqrt{x}-x\right)+\dfrac{1}{x}\)

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

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

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

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

đk : \(x\ge0,x\ne1\)

\(=>P=\left[\dfrac{2\left(\sqrt{x}+2\right)-5}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\right]:\left[\dfrac{x+\sqrt{x}-2+3-x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\right]\)

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

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

b,\(x=6-2\sqrt{5}=\left(\sqrt{5}-1\right)^2\) thay vào P

\(=>P=\dfrac{2\sqrt{\left(\sqrt{5}-1\right)^2}-1}{\sqrt{\left(\sqrt{5}-1\right)^2}+1}=\dfrac{2\sqrt{5}-3}{\sqrt{5}}\)

c,\(=>\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}=\dfrac{1}{\sqrt{x}}=>2x-\sqrt{x}=\sqrt{x}+1\)

\(=>2x-2\sqrt{x}-1=0< =>2\left(x-\sqrt{x}-\dfrac{1}{2}\right)=0\)

\(=>x-\sqrt{x}-\dfrac{1}{2}=>\Delta=1-4\left(-\dfrac{1}{2}\right)=3>0=>\left[{}\begin{matrix}x1=\dfrac{1+\sqrt{3}}{2}\\x2=\dfrac{1-\sqrt{3}}{2}\end{matrix}\right.\)

đối chiếu đk loại x2 còn x1 thỏa

a: Ta có: \(A=\left(\dfrac{1}{\sqrt{x}+2}+\dfrac{1}{\sqrt{x}-2}\right)\cdot\dfrac{x-4}{3\sqrt{x}}\)

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

\(=\dfrac{2}{3}\)

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

\(x=\dfrac{9-4\sqrt{5}-9-4\sqrt{5}}{\left(2+\sqrt{5}\right)\left(2-\sqrt{5}\right)}:2\sqrt{5}=\dfrac{-8\sqrt{5}}{-2\sqrt{5}}=4\\ \Leftrightarrow\sqrt{x}=2\\ \Leftrightarrow A=\dfrac{2-1}{2+2}=\dfrac{1}{4}\)

help

  => đề sai rồi bạn

2)

ĐK: \(x\ge0;x\ne4\)

Biểu thức trở thành:

\(\dfrac{\left(\sqrt{a}+3\right)\left(\sqrt{a}+2\right)}{a-4}-\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}{a-4}-\dfrac{4\sqrt{a}-4}{a-4}\\ =\dfrac{a+2\sqrt{a}+3\sqrt{a}+6}{a-4}-\dfrac{a-2\sqrt{a}-\sqrt{a}+2}{a-4}-\dfrac{4\sqrt{a}-4}{a-4}\\ =\dfrac{a+5\sqrt{a}+6-a+3\sqrt{a}-2-4\sqrt{a}+4}{a-4}\\ =\dfrac{4\sqrt{a}+8}{a-4}\\ =\dfrac{4\left(\sqrt{a}+2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}\\ =\dfrac{4}{\sqrt{a}-2}\)

1:

\(\left(\dfrac{x+2\sqrt{x}-7}{x-9}+\dfrac{\sqrt{x}+1}{3-\sqrt{x}}\right):\left(\dfrac{1}{\sqrt{x}+3}-\dfrac{1}{\sqrt{x}-1}\right)\)

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

\(=\dfrac{x+2\sqrt{x}-8-x-4\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}{-4}\)

\(=\dfrac{-2\sqrt{x}-11}{-4}\cdot\dfrac{\sqrt{x}-1}{\sqrt{x}-3}=\dfrac{\left(2\sqrt{x}+11\right)\left(\sqrt{x}-1\right)}{4\left(\sqrt{x}-3\right)}\)

Với `x >= 0,x ne 4` có:

`M=[(\sqrt{x}+1)(\sqrt{x}+2)+2\sqrt{x}(\sqrt{x}-2)-2-5\sqrt{x}]/[(\sqrt{x}-2)(\sqrt{x}+2)]`

`M=[x+3\sqrt{x}+2+2x-4\sqrt{x}-2-5\sqrt{x}]/[(\sqrt{x}-2)(\sqrt{x}+2)]`

`M=[3x-6\sqrt{x}]/[(\sqrt{x}-2)(\sqrt{x}+2)]=[3\sqrt{x}]/[\sqrt{x}+2]`

____________

`N=(1/[\sqrt{a}-1]-1/\sqrt{a}):([\sqrt{a}+1]/[\sqrt{a}-2]-[\sqrt{a}+2]/[\sqrt{a}-1])`

      - Biểu thức `N` là như vầy?

Với `a > 0,a ne 1,a ne 4` có:

`N=[\sqrt{a}-\sqrt{a}+1]/[\sqrt{a}(\sqrt{a}-1)]:[(\sqrt{a}+1)(\sqrt{a}-1)-(\sqrt{a}+2)(\sqrt{a}-2)]/[(\sqrt{a}-2)(\sqrt{a}-1)]`

`N=1/[\sqrt{a}(\sqrt{a}-1)].[(\sqrt{a}-2)(\sqrt{a}-1)]/[a-1-a+4]`

`N=[\sqrt{a}-2]/[3\sqrt{a}]`

Với \(x\ge0;x\ne4\)

Khi đó:

\(M=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{x-4}+\dfrac{2\sqrt{x}\left(\sqrt{x}-2\right)}{x-4}-\dfrac{2+5\sqrt{x}}{x-4}\\ =\dfrac{x+2\sqrt{x}+\sqrt{x}+2}{x-4}+\dfrac{2x-4\sqrt{x}}{x-4}-\dfrac{2+5\sqrt{x}}{x-4}\\ =\dfrac{x+3\sqrt{x}+2+2x-4\sqrt{x}-2-5\sqrt{x}}{x-4}\\ =\dfrac{3x-6\sqrt{x}}{x-4}\\ =\dfrac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\\ =\dfrac{3\sqrt{x}}{\sqrt{x}+2}\)

Với \(a>0;a\ne1;a\ne4\) 

Khi đó:

\(N=(\dfrac{\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}-\dfrac{\sqrt{a}-1}{\sqrt{a}\left(\sqrt{a}-1\right)}):\left(\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}-\dfrac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\right)\\ =\left(\dfrac{\sqrt{a}}{a-\sqrt{a}}-\dfrac{\sqrt{a}-1}{a-\sqrt{a}}\right):\left(\dfrac{a-1}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}-\dfrac{a-4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\right)\\ =\dfrac{1}{a-\sqrt{a}}:\dfrac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\\ =\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}.\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3}\\ =\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{\sqrt{a}\left(\sqrt{a}-1\right).3}\\ =\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)