\(\left(\sqrt{2}+1\right)^3-\left(\sqrt{2}-1\right)^3\\ =\sqrt{2}^3+3.\sqrt{2}^2.1+3.\sqrt{2}.1^2+1^3-\left(\sqrt{2}^3-3.\sqrt{2}^2.1+3.\sqrt{2}.1^2-1^3\right)\\ =\left(\sqrt{2}^3-\sqrt{2}^3\right)+\left(3.\sqrt{2}^2.1+3.\sqrt{2}^2.1\right)+\left(3.\sqrt{2}.1^2-3.\sqrt{2}.1^2\right)+\left(1^3+1^3\right)\\ =3.2+3.2+1+1=14\)

\(\dfrac{\sqrt{7}+\sqrt{5}}{\sqrt{7}-\sqrt{5}}+\dfrac{\sqrt{7}-\sqrt{5}}{\sqrt{7}+\sqrt{5}}\\ =\dfrac{\left(\sqrt{7}+\sqrt{5}\right)^2+\left(\sqrt{7}-\sqrt{5}\right)^2}{\left(\sqrt{7}-\sqrt{5}\right)\left(\sqrt{7}+\sqrt{5}\right)}\\ =\dfrac{7+2\sqrt{35}+5+7-2\sqrt{35}+5}{\sqrt{7}^2-\sqrt{5}^2}\\ =\dfrac{24}{2}=12\)

\(\left(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\right)^2=3-\sqrt{5}+3+\sqrt{5}+2\sqrt{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}\\ =6+2\sqrt{3^2-\sqrt{5}^2}\\ =6+2\sqrt{9-5}=6+2\sqrt{4}\\ =6+2.2=10\)

\(8\dfrac{1}{6}:2\dfrac{1}{2}=\dfrac{8\times6+1}{6}:\dfrac{2\times2+1}{2}=\dfrac{49}{6}:\dfrac{5}{2}\\ =\dfrac{49}{6}\times\dfrac{2}{5}=\dfrac{49\times2}{2\times3\times5}=\dfrac{49}{3\times5}=\dfrac{49}{15}\)

\(x^2y^2+xy+\dfrac{1}{4}=\left(xy\right)^2+2.xy.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2\\ =\left(xy+\dfrac{1}{2}\right)^2\)

\(\sqrt{50}-3\sqrt{98}+2\sqrt{8}+3\sqrt{32}-5\sqrt{18}\\ =\sqrt{25}.\sqrt{2}-3.\sqrt{49}.\sqrt{2}+2.\sqrt{4}.\sqrt{2}+3.\sqrt{16}.\sqrt{2}-5.\sqrt{9}.\sqrt{2}\\ =\left(5-3.7+2.2+3.4-5.3\right).\sqrt{2}=-15\sqrt{2}\\ 14-6\sqrt{5}+\sqrt{\left(2-\sqrt{5}\right)^2}=14-6\sqrt{5}+\left|2-\sqrt{5}\right|\\ =14-6\sqrt{5}+\sqrt{5}-2=12-5\sqrt{5}\)

\(\dfrac{\sqrt{88}}{\sqrt{22}}=\sqrt{\dfrac{88}{22}}=\sqrt{4}=2\\ \dfrac{1}{3}.\sqrt{72}-3.\sqrt{50}-\dfrac{\sqrt{66}}{\sqrt{33}}\\ =\dfrac{1}{3}.\sqrt{36}.\sqrt{2}-3.\sqrt{25}.\sqrt{2}-\sqrt{\dfrac{66}{33}}\\ =\dfrac{1}{3}.6.\sqrt{2}-3.5.\sqrt{2}-\sqrt{2}\\ =\left(\dfrac{1}{3}.6-3.5-1\right).\sqrt{2}=-14\sqrt{2}\\ \dfrac{5-\sqrt{3}}{\sqrt{3}-1}=\dfrac{\left(5-\sqrt{3}\right)\left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}=\dfrac{5\sqrt{3}-3+5-\sqrt{3}}{\sqrt{3}^2-1^2}\\ =\dfrac{4\sqrt{3}+2}{2}=2\sqrt{3}+1\)

\(\left(1+\sqrt{3}-\sqrt{2}\right)\left(1+\sqrt{3}+\sqrt{2}\right)\\ =\left(1+\sqrt{3}\right)^2-\sqrt{2}^2\\ =4+2\sqrt{3}-2=2+2\sqrt{3}\)

\(\dfrac{A}{B}=\dfrac{2\sqrt{x}-2}{3}\left(ĐK:x>0\right)\\ < =>\dfrac{\sqrt{x}+2}{\sqrt{x}}:\dfrac{\sqrt{x}+2}{\sqrt{x}+1}=\dfrac{2\sqrt{x}-2}{3}\\ < =>\dfrac{\sqrt{x}+2}{\sqrt{x}}.\dfrac{\sqrt{x}+1}{\sqrt{x}+2}=\dfrac{2\sqrt{x}-2}{3}\\ < =>\dfrac{\sqrt{x}+1}{\sqrt{x}}=\dfrac{2\sqrt{x}-2}{3}\)

\(< =>\sqrt{x}\left(2\sqrt{x}-2\right)=3\left(\sqrt{x}+1\right)\\ < =>2x-2\sqrt{x}=3\sqrt{x}+3\\ < =>2x-5\sqrt{x}-3=0\\ < =>\left(\sqrt{x}-3\right)\left(2\sqrt{x}+1\right)=0\\ =>\left[{}\begin{matrix}\sqrt{x}-3=0\\2\sqrt{x}+1=0\end{matrix}\right.< =>\left[{}\begin{matrix}x=9\left(TMDK\right)\\\sqrt{x}=-\dfrac{1}{2}< 0\left(KTMDK\right)\end{matrix}\right.< =>x=9\)

\(C=-4y^2+8y-12\\ =-\left(4y^2-8y+12\right)\\ =-\left(4y^2-8y+4+8\right)\\ =-\left[\left(2y\right)^2-2.2y.2+2^2+8\right]\\ =-\left(2y-2\right)^2-8\le-8\forall y\)

\(Dấu''=''\ xảy\ ra\ khi :2y-2=0<=>y=1\)

\(Vậy\ GTLN\ của\ C\ là :-8<=>y=1\)