`a,` \(\left\{{}\begin{matrix}x-2\ne0\Leftrightarrow x\ne2\\x+2\ne0\Leftrightarrow x\ne-2\end{matrix}\right.\)

\(b,A=\dfrac{x^2}{x^2-4}-\dfrac{x}{x-2}-\dfrac{2}{x+2}\\ =\dfrac{x^2}{\left(x-2\right)\left(x+2\right)}-\dfrac{x}{x-2}-\dfrac{2}{x+2}\\ =\dfrac{x^2-x\left(x+2\right)-2\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\\ =\dfrac{x^2-x^2-2x-2x+4}{\left(x-2\right)\left(x+2\right)}\\ =\dfrac{-4x+4}{x-4}\)

`c,` Để `A=2` ta có

 \(\dfrac{-4x+4}{x-4}=2\left(x\ne4\right)\\ \Leftrightarrow\dfrac{-4x+4}{x-4}=\dfrac{2\left(x-4\right)}{x-4}\\ \Leftrightarrow-4x+5=2x-8\\ \Leftrightarrow-6x=-13\\ \Leftrightarrow x=\dfrac{13}{6}\)

`a, xy +y^2 -x-y`

`=(xy+y^2)-(x+y)`

`= y(x+y)-(x+y)`

`= (x+y)(y-1)`

`b, (x^2y^2 -8)^2 -1`

`= (x^2y^2 -8)^2 -1^2`

`=(x^2y^2 -8-1)(x^2y^2-8+1)`

`=(x^2y^2 -9)(x^2y^2-7)`

`x^8+3x^4+4`

`=x^8+4x^4+4-x^4`

`=(x^4)^2 + 2.x^4. 2+2^2-x^4`

`=(x^4+2)^2 -(x^2)^2`

`=(x^4+2-x^2)(x^4+2+x^2)`

\(A=\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\\ =\left[\left(\sqrt{4+\sqrt{15}}\right)\left(\sqrt{4+\sqrt{15}}\right)\right]\left[\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right)\right]\sqrt{4-\sqrt{15}}\\ =\left[\left(\sqrt{4+\sqrt{15}}\right)\left(\sqrt{4-\sqrt{15}}\right)\right]\cdot\sqrt{4+\sqrt{15}}\cdot\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right)\\ =1\cdot\sqrt{8+2\sqrt{15}}\cdot\left(\sqrt{5}-\sqrt{3}\right)\\ =\sqrt{3+2\sqrt{15}+5}\cdot\left(\sqrt{5}-\sqrt{3}\right)\\ =\sqrt{\left(\sqrt{3}+\sqrt{5}\right)^2}\cdot\left(\sqrt{5}-\sqrt{3}\right)\\ =\left(\sqrt{3}+\sqrt{5}\right)\left(\sqrt{5}-\sqrt{3}\right)\\ =5-3\\ =2\)

\(B=\dfrac{x+27}{x-9}+\dfrac{1}{\sqrt{x}+3}+\dfrac{6}{3-\sqrt{x}}\\ =\dfrac{x+27}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\dfrac{1}{\sqrt{x}+3}-\dfrac{6}{\sqrt{x}-3}\\ =\dfrac{x+27+\sqrt{x}-3-\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\\ =\dfrac{x-21}{x-9}\)

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

\(\dfrac{1}{5}\left(x+\dfrac{1}{5}\right)+\dfrac{2}{5}\left(x+\dfrac{5}{3}\right)=\dfrac{98}{75}\\ \Rightarrow\dfrac{1}{5}x+\dfrac{1}{25}+\dfrac{2}{5}x+\dfrac{2}{3}=\dfrac{98}{75}\\ \Rightarrow\left(\dfrac{1}{5}x+\dfrac{2}{5}x\right)+\left(\dfrac{1}{25}+\dfrac{2}{3}\right)=\dfrac{98}{75}\\ \Rightarrow\dfrac{3}{5}x+\dfrac{53}{75}=\dfrac{98}{75}\\ \Rightarrow\dfrac{3}{5}x=\dfrac{98}{75}-\dfrac{53}{75}\\ \Rightarrow\dfrac{3}{5}x=\dfrac{45}{75}=\dfrac{3}{5}\\ \Rightarrow x=\dfrac{3}{5}:\dfrac{3}{5}\\ \Rightarrow x=1\)

\(\sqrt{160}\cdot\sqrt{10}+\sqrt{8}\cdot\sqrt{0,5}\\ =\sqrt{1600}+\sqrt{8\cdot0,5}\\ =\sqrt{16\cdot100}+\sqrt{4}\\ =4\cdot10+2\\ =40+2\\ =42\\ \sqrt{\left(3-\sqrt{7}\right)^2}+\sqrt{\left(2-\sqrt{7}\right)^2}\\ =\left|3-\sqrt{7}\right|+\left|2-\sqrt{7}\right|\\ =3-\sqrt{7}+\sqrt{7}-2\\ =1\)

\(3\sqrt{50}-4\sqrt{18}+\dfrac{1}{4}\sqrt{32}\\ =3\sqrt{25\cdot2}-4\sqrt{9\cdot2}+\dfrac{1}{4}\sqrt{16\cdot2}\\ =3\cdot5\sqrt{2}-4\cdot3\sqrt{2}+\dfrac{1}{4}\cdot4\sqrt{2}\\ =15\sqrt{2}-12\sqrt{2}+\sqrt{2}\\ =\left(15-12+1\right)\sqrt{2}\\ =4\sqrt{2}\)

\(\dfrac{2+\sqrt{3}}{2-\sqrt{3}}+\dfrac{2-\sqrt{3}}{2+\sqrt{3}}\\ =\dfrac{\left(2+\sqrt{3}\right)^2}{4-3}+\dfrac{\left(2-\sqrt{3}\right)^2}{4-3}\\ =4+4\sqrt{3}+3+4-4\sqrt{3}+3\\ =14\)