a)A\(=\dfrac{1}{2+2\sqrt{a}}+\dfrac{1}{2-2\sqrt{a}}-\dfrac{a^2+1}{1-a^2}\)
A\(=\dfrac{1}{2\left(1+\sqrt{a}\right)}+\dfrac{1}{2\left(1-\sqrt{a}\right)}-\dfrac{a^2+1}{\left(1-a\right)\left(1+a\right)}\)
A\(=\dfrac{1}{2\left(1+\sqrt{a}\right)}+\dfrac{1}{2\left(1-\sqrt{a}\right)}-\dfrac{a^2+1}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)\left(1+a\right)}\)
A\(=\dfrac{\left(1-\sqrt{a}\right)\left(1+a\right)+\left(1+\sqrt{a}\right)\left(1+a\right)-\left(a^2+1\right)2}{2\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)\left(1+a\right)}\)
A\(=\dfrac{1+a-\sqrt{a}-a\sqrt{a}+1+a+\sqrt{a}+a\sqrt{a}-2a^2+2}{2\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)\left(1+a\right)}\)
A\(=\dfrac{2a-2a^2}{2\left(1-a\right)\left(1+a\right)}\)
A\(=\dfrac{2a\left(1-a\right)}{2\left(1-a\right)\left(1+a\right)}\)
A\(=\dfrac{a}{1+a}\)
b) Thay A<\(\dfrac{1}{3}\) vào bt ta có:
\(\dfrac{a}{1+a}< \dfrac{1}{3}\)
ĐKXĐ: \(1+a\ne0\Rightarrow a\ne-1\)
\(\Leftrightarrow\dfrac{a}{1+a}-\dfrac{1}{3}< 0\)
\(\Leftrightarrow\dfrac{3a-1-a}{3\left(1+a\right)}< 0\)
\(\Leftrightarrow\dfrac{2a-1}{3+3a}< 0\)
\(\Leftrightarrow2a-1< 0\)
\(\Leftrightarrow2a< 1\)
\(\Leftrightarrow a< \dfrac{1}{2}\)
vậy...
a)
\(a\ge0;a\ne1;-1\)
\(A=\dfrac{1}{2+2\sqrt{a}}+\dfrac{1}{2-2\sqrt{a}}-\dfrac{a^2+1}{1-a^2}\)
\(\Leftrightarrow A=\dfrac{4}{4-4a}-\dfrac{a^2+1}{1-a^2}\)
\(\Leftrightarrow A=\dfrac{1+a-a^2+1}{1-a^2}=-\dfrac{a^2-a-2}{1-a^2}\)
\(A=-\dfrac{\left(a-2\right)\left(a+1\right)}{\left(a+1\right)\left(1-a\right)}=-\dfrac{a-2}{1-a}=\dfrac{a-2}{a-1}\)(\(a\ge0;a\ne1;-1\))
b)
\(A< \dfrac{1}{3}\Rightarrow\dfrac{a-2}{a-1}< \dfrac{1}{3}\)
\(\Rightarrow\dfrac{a-2}{a-1}-\dfrac{1}{3}< 0\Leftrightarrow\dfrac{3a-6-a+1}{3\left(a-1\right)}< 0\)
\(\Leftrightarrow\dfrac{2a-5}{3a-3}< 0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2a-5< 0\\3a-3>0\end{matrix}\right.\\\left\{{}\begin{matrix}2a-5>0\\3a-3< 0\end{matrix}\right.\end{matrix}\right.\)
\(\left[{}\begin{matrix}\left\{{}\begin{matrix}a< \dfrac{5}{2}\\a>1\end{matrix}\right.\\\left\{{}\begin{matrix}a>\dfrac{5}{2}\\a< 1\end{matrix}\right.\end{matrix}\right.\)
\(\left\{{}\begin{matrix}a>\dfrac{5}{2}\\a< 1\end{matrix}\right.\)(l)
\(\Rightarrow1< a< \dfrac{5}{2}\) thi \(A< \dfrac{1}{3}\)
