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)

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

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

`a)đk:a>0,a ne 9`

`A=((sqrta+3+sqrta-3)/(a-9)).((sqrta-3)/sqrta)`

`=((2sqrtx)/(a-9)).((sqrta-3)/sqrta)`

`=2/(sqrta+3)`

`b)A>1/2`

`<=>2/(sqrta+3)>1/2`

`<=>sqrta+3<4`

`<=>sqrta<1`

`<=>a<1`

KẾt hợp đkxđ:`0<x<1`

ĐKXĐ: \(\left\{{}\begin{matrix}a>0\\a\ne9\end{matrix}\right.\)

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

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

\(=\dfrac{2\sqrt{a}}{\sqrt{a}+3}\cdot\dfrac{1}{\sqrt{a}}\)

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

b) Để \(A>\dfrac{1}{2}\) thì \(A-\dfrac{1}{2}>0\)

\(\Leftrightarrow\dfrac{2}{\sqrt{a}+3}-\dfrac{1}{2}>0\)

\(\Leftrightarrow\dfrac{4-\left(\sqrt{a}+3\right)}{2\left(\sqrt{a}+3\right)}>0\)

mà \(2\left(\sqrt{a}+3\right)>0\forall a\)

nên \(1-\sqrt{a}>0\)

\(\Leftrightarrow\sqrt{a}< 1\)

hay a<1

Kết hợp ĐKXĐ, ta được: 0<a<1

cứ nhân tung ra là ra

ĐKXĐ: \(x>0;a\ne9\)

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

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

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

b.

\(A>\dfrac{1}{2}\Rightarrow\dfrac{2}{\sqrt{a}+3}>\dfrac{1}{2}\Rightarrow\sqrt{a}+3< 4\)

\(\Rightarrow\sqrt{a}< 1\Rightarrow a< 1\)

Kết hợp ĐKXĐ \(\Rightarrow0< a< 1\)

1. ĐKXĐ: \(x\ne\pm1\)

2. \(A=\left(\dfrac{x+1}{x-1}-\dfrac{x+3}{x+1}\right)\cdot\dfrac{x+1}{2}\)

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

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

\(=\dfrac{6x-2}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{x+1}{2}\)

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

\(=\dfrac{x-3}{x-1}\)

3. Tại x = 5, A có giá trị là:

\(\dfrac{5-3}{5-1}=\dfrac{1}{2}\)

4. \(A=\dfrac{x-3}{x-1}\) \(=\dfrac{x-1-3}{x-1}=1-\dfrac{3}{x-1}\)

Để A nguyên => \(3⋮\left(x-1\right)\) hay \(\left(x-1\right)\inƯ\left(3\right)=\left\{1;-1;3;-3\right\}\)

\(\Rightarrow\left\{{}\begin{matrix}x-1=1\\x-1=-1\\x-1=3\\x-1=-3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2\left(tmđk\right)\\x=0\left(tmđk\right)\\x=4\left(tmđk\right)\\x=-2\left(tmđk\right)\end{matrix}\right.\)

Vậy: A nguyên khi \(x=\left\{2;0;4;-2\right\}\)

a. ĐKXĐ: \(x\ne\pm1\)

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

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

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

\(=\left(x-1\right)\left(x+1\right)\left[\dfrac{-x^2+3}{\left(x-1\right)\left(x+1\right)}\right]\)

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

\(=-x^2+3\)

c. Thay x = 3 vào A ta được:

\(-\left(3\right)^2+3=-6\)

Vậy: Giá trị của A tại x = 3 là -6

a) ĐKXĐ: \(x\ne1;x\ne-1.\)

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

\(=\left(x^2-1\right).\dfrac{x+1-x+1-x^2+1}{x^2-1}=-x^2+3.\)

c) Thay x = 3 (TMĐK) vào A: \(-3^2+3=-6.\)

Bài 1: 

a: \(A=\dfrac{x^2-3+x+3}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x+3}{x}=\dfrac{x\left(x+1\right)}{x\left(x-3\right)}=\dfrac{x+1}{x-3}\)

b: Để A=3 thì 3x-9=x+1

=>2x=10

hay x=5

Bài 2: 

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

\(=\dfrac{-6}{x-2}\cdot\dfrac{1}{2}=\dfrac{-3}{x-2}\)

b: Để A nguyên thì \(x-2\in\left\{1;-1;3;-3\right\}\)

hay \(x\in\left\{3;1;5;-1\right\}\)

đkxđ:\(x\ne5,x\ne-5\)

\(\dfrac{2x}{\left(x-5\right)\left(x+5\right)}-\dfrac{5}{x-5}-\dfrac{1}{x+5}\)

\(\dfrac{2x}{\left(x-5\right)\left(x+5\right)}-\dfrac{5x+25}{\left(x-5\right)\left(x+5\right)}-\dfrac{x-5}{\left(x-5\right)\left(x+5\right)}\)

\(\dfrac{2x-5x-25-x+5}{\left(x-5\right)\left(x+5\right)}=\dfrac{-4x-20}{\left(x-5\right)\left(x+5\right)}=\dfrac{-4\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}=-\dfrac{4}{x-5}\)

thay x=1 vào bt A, ta được:

\(-\dfrac{4}{1-5}=1\)

a, Với \(a\ge0;a\ne9\)

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

\(=\left(\frac{2\sqrt{a}}{a-9}\right)\left(\frac{\sqrt{a}-3}{\sqrt{a}}\right)=\frac{2}{\sqrt{a}+3}\)

b, Ta có : \(\frac{2}{\sqrt{a}+3}>\frac{1}{2}\Rightarrow\frac{2}{\sqrt{a}+3}-\frac{1}{2}>0\)

\(\Leftrightarrow\frac{1-\sqrt{a}}{2\sqrt{a}+6}>0\Rightarrow1-\sqrt{a}>0\)vì \(2\sqrt{a}+6>0\)