2a^4=(1-a)^2=a^2-2a+1

\(A=\frac{2a-3}{\sqrt{2\left(a^2-4a+4\right)}+2a^2}=\frac{2a-3}{\sqrt{2}!\left(a-2\right)!+2a^2}\)a> 2 không thể là nghiệm=> a<2

\(A=\frac{2a-3}{\sqrt{2}\left(2-a\right)+2a^2}=\frac{2a-3}{2a^2-\sqrt{2}a+2\sqrt{2}}=\frac{2a-3}{\sqrt{2}\left(\sqrt{2}a^2-a-1+3\right)}\)

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

bạn giải thích rõ hơn được không ?

a là nghiệm =>\(\sqrt{2}a^2+a-1=0\Rightarrow\sqrt{2}a^2=1-a\\\)\(2a^4=\left(1-a\right)^2=1^2-2a+a^2\)

Thay 2a^4=...vào ==> 

ĐKXĐ: \(a\ge2\)

Ta có: \(\sqrt{a+2\sqrt{2a-4}}-\sqrt{a-2\sqrt{2a-4}}\)

\(=\sqrt{a-2+2\cdot\sqrt{a-2}\cdot\sqrt{2}+2}-\sqrt{a-2-2\cdot\sqrt{a-2}\cdot\sqrt{2}+2}\)

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

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

\(=\sqrt{a-2}+\sqrt{2}-\left|\sqrt{a-2}-\sqrt{2}\right|\)(*)

Trường hợp 1: \(a\ge4\)

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

\(=\sqrt{a-2}+\sqrt{2}-\sqrt{a-2}+\sqrt{2}\)

\(=2\sqrt{2}\)

Trường hợp 2: a<4

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

\(=\sqrt{a-2}+\sqrt{2}-\sqrt{2}+\sqrt{a-2}\)

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

1) Để biểu thức có nghĩa thì \(a^2+2a-3\ge0\)

\(\Leftrightarrow\left(a+3\right)\left(a-1\right)\ge0\)

\(\Leftrightarrow\left[{}\begin{matrix}a-1\ge0\\a+3\le0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a\ge1\\a\le-3\end{matrix}\right.\)

2) Để biểu thức có nghĩa thì \(\left\{{}\begin{matrix}a-1\ge0\\a\ne0\end{matrix}\right.\Leftrightarrow a\ge1\)

3) Để biểu thức có nghĩa thì \(a>0\)

4) Để biểu thức có nghĩa thì \(\left\{{}\begin{matrix}a\ne-\dfrac{1}{2}\\\left[{}\begin{matrix}a-1\ge0\\2a+1< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a\ne-\dfrac{1}{2}\\\left[{}\begin{matrix}a\ge1\\a< -\dfrac{1}{2}\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a\ge1\\a< -\dfrac{1}{2}\end{matrix}\right.\)

1) Để biểu thức có nghĩa  \(\Rightarrow a^2+2a-3\ge0\Rightarrow\left(a-1\right)\left(a+3\right)\ge0\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}a-1\ge0\\a+3\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}a-1\le0\\a+3\le0\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}a\ge1\\a\le-3\end{matrix}\right.\)

2) Để biểu thức có nghĩa \(\Rightarrow\dfrac{\left(a-1\right)^3}{a^2}\ge0\Rightarrow\left\{{}\begin{matrix}\left(a-1\right)^3\ge0\\a\ne0\end{matrix}\right.\Rightarrow a\ge1\)

3) Để biểu thức có nghĩa \(\Rightarrow\dfrac{a^2+1}{2a}\ge0\Rightarrow2a>0\Rightarrow a>0\)

4) Để biểu thức có nghĩa \(\Rightarrow\dfrac{a-1}{2a+1}\ge0\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}a-1\ge0\\2a+1>0\end{matrix}\right.\\\left\{{}\begin{matrix}a-1\le0\\2a+1< 0\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}a\ge1\\a< -\dfrac{1}{2}\end{matrix}\right.\)

\(a,\dfrac{a\sqrt{a}-8+2a-4\sqrt{a}}{a-4}\left(dk:a\ne4\right)\)

\(=\dfrac{a\sqrt{a}-4\sqrt{a}-8+2a}{a-4}\)

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

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

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

\(b,\dfrac{12\sqrt{6}}{\sqrt{7+2\sqrt{6}}-\sqrt{7-2\sqrt{6}}}\\ =\dfrac{12\sqrt{6}}{\sqrt{\left(\sqrt{6}+1\right)^2}-\sqrt{\left(\sqrt{6}-1\right)^2}}\\ =\dfrac{12\sqrt{6}}{\left|\sqrt{6}+1\right|-\left|\sqrt{6}-1\right|}\\ =\dfrac{12\sqrt{6}}{\sqrt{6}+1-\sqrt{6}+1}\\ =\dfrac{12\sqrt{6}}{2}\\ =6\sqrt{6}\)