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