a) A có nghĩa khi: \(\left\{{}\begin{matrix}a>0\\a\ne1\end{matrix}\right.\)

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

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

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

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

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

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

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

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

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

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

\(A=\dfrac{a}{1+a}\cdot\dfrac{a+1}{a}\)

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

\(A=1\)

Vậy giá trị của A không phụ thuộc và biến

a: ĐKXĐ: a>0; a<>1

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

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

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

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

\(\left(\sqrt{\dfrac{1+sin\alpha}{1-sin\alpha}}+\sqrt{\dfrac{1-sin\alpha}{1+sin\alpha}}\right).\dfrac{1}{\sqrt{1+tan^2\alpha}}\)

\(=\left(\sqrt{\dfrac{\left(1+sin\alpha\right)^2}{\left(1-sin\alpha\right)\left(1+sin\alpha\right)}}+\sqrt{\dfrac{\left(1-sin\alpha\right)^2}{\left(1+sin\alpha\right)\left(1-sin\alpha\right)}}\right).\dfrac{1}{\sqrt{1+\left(\dfrac{sin\alpha}{cos\alpha}\right)^2}}\)

\(=\left(\sqrt{\dfrac{\left(1+sin\alpha\right)^2}{1-sin^2\alpha}}+\sqrt{\dfrac{\left(1-sin\alpha\right)^2}{1-sin^2\alpha}}\right).\dfrac{1}{\sqrt{\dfrac{cos^2\alpha+sin^2\alpha}{cos^2\alpha}}}\)

\(=\left(\sqrt{\dfrac{\left(1+sin\alpha\right)^2}{cos^2\alpha}}+\sqrt{\dfrac{\left(1-sin\alpha\right)^2}{cos^2\alpha}}\right).\dfrac{1}{\sqrt{\dfrac{1}{cos^2\alpha}}}\)

\(=\left(\dfrac{1+sin\alpha}{cos\alpha}+\dfrac{1-sin\alpha}{cos\alpha}\right).\dfrac{1}{\dfrac{1}{cos\alpha}}=\dfrac{2}{cos\alpha}.cos\alpha=2\)

a,Ta có  \(x=4-2\sqrt{3}=\sqrt{3}^2-2\sqrt{3}+1=\left(\sqrt{3}-1\right)^2\)

\(\Rightarrow\sqrt{x}=\sqrt{\left(\sqrt{3}-1\right)^2}=\left|\sqrt{3}-1\right|=\sqrt{3}-1\)do \(\sqrt{3}-1>0\)

\(\Rightarrow A=\frac{1}{\sqrt{3}-1-1}=\frac{1}{\sqrt{3}-2}\)

b, Với \(x\ge0;x\ne1\)

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

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

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

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

Vậy biểu thức ko phụ thuộc biến x 

c, Ta có : \(\frac{2A}{B}\)hay \(\frac{2}{\sqrt{x}-1}\)để biểu thức nhận giá trị nguyên 

thì \(\sqrt{x}-1\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)

a. Ta có $x=4-2\sqrt{3}=1-\mathrm{2.1.}\sqrt{3}+3=(1-\sqrt{3}{)}^2$.

$\Rightarrow \sqrt{x}=|1-\sqrt{3}|=\sqrt{3}-1.$

1x−1=13−1−1=13−2=−2−3.

b. Với $x\ge 0$ và $x\ne 1$ ta có:

−3xxx−1−11−x):(1−x+21+x+x)

−3x(x−1)(x+x+1)+1x−1]:(1+x+x)−(x+2)1+x+x

−3x+x+x+1(x−1)(x+x+1).x+x+1x−1

x−2x+1(x−1)(x+x+1).x+x+1x−1 

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

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

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

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

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

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

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

ko biet

Bài 2: 

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

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

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

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

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

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

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

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

\(M=\left(\dfrac{2}{2\left(1-a\right)}-\dfrac{a^2+1}{1-a^2}\right)\left(\dfrac{a+1}{a}\right)\)

\(M=\left(\dfrac{1}{1-a}-\dfrac{a^2+1}{1-a^2}\right)\left(\dfrac{a+1}{a}\right)\)

\(M=\left(\dfrac{1+a-a^2-1}{\left(1-a\right)\left(1+a\right)}\right)\left(\dfrac{a+1}{a}\right)\)

\(M=\dfrac{a-a^2}{a\left(1-a\right)}\)

\(M=\dfrac{a\left(1-a\right)}{a\left(1-a\right)}=1\)

--> giá trị của M ko phụ thuộc vào A

cái này mà là toán lớp 6 hả

(? . ?)