`a)` A có nghĩa khi \(\left\{{}\begin{matrix}2ab\ge0\\\sqrt{a^2+b^2}\ne0\end{matrix}\right.\)

                               \(\Leftrightarrow\left\{{}\begin{matrix}a,b\ge0;a,b\le0\\a;b\ne0\end{matrix}\right.\)

                                 \(\rightarrow\left[{}\begin{matrix}a;b>0\\a;b< 0\end{matrix}\right.\)

`b)`\(a=b\)

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

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

\(A=1-\dfrac{2ab}{a^2+b^2}\)

\(A=1-\dfrac{2a^2}{2a^2}\)

\(A=1-1=0\)

\(\dfrac{2x+4}{4}-\dfrac{2x-7}{5}=\dfrac{2x-13}{6}\)

\(15\left(2x+4\right)-12\left(2x-7\right)=10\left(2x-13\right)\)
\(30x+60-24x+84=20x-130\)

\(30x-24x-20x=-130-60-84\)

\(-14x=-274\)

\(x=\dfrac{137}{7}\approx19,\left(571428\right)\)

\(\dfrac{\sqrt{x}}{\sqrt{x}-1}\)+\(\dfrac{4}{\sqrt{x}\left(\sqrt{x}-1\right)}\)x\(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)

=\(\dfrac{\sqrt{x}}{\sqrt{x}-1}\)+​​\(\dfrac{4}{\sqrt{x}\left(\sqrt{x}+1\right)}\)​​

=\(\dfrac{\sqrt{x}\cdot\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)+\(\dfrac{4\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

=\(\dfrac{x}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)+\(\dfrac{4\sqrt{x}-4}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

=\(\dfrac{x+4\sqrt{x}-4}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

Còn lại tự làm đê :))) công nhận ai ra đề mà ác vc