$\left(\frac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\frac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right):\frac{1}{\sqrt{7}-\sqrt{5}}=\left(\frac{\sqrt{2}.\sqrt{7}-\sqrt{7}}{1-\sqrt{2}}+\frac{\sqrt{5}.\sqrt{3}-\sqrt{5}}{1-\sqrt{3}}\right):\frac{1}{\sqrt{7}-\sqrt{5}}$

$=\left(\frac{\sqrt{7}\left(\sqrt{2}-1\right)}{1-\sqrt{2}}+\frac{\sqrt{5}\left(\sqrt{3}-1\right)}{1-\sqrt{3}}\right).\left(\sqrt{7}-\sqrt{5}\right)=\left(-\sqrt{7}-\sqrt{5}\right).\left(\sqrt{7}-\sqrt{5}\right)$

$=-\left(\sqrt{7}+\sqrt{5}\right).\left(\sqrt{7}-\sqrt{5}\right)=-\left(7-5\right)=-2$

Đáp án cần chọn là: B

$\left(\frac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\frac{\sqrt{30}-\sqrt{6}}{\sqrt{5}-\sqrt{1}}\right):\frac{1}{2\sqrt{5}-\sqrt{6}}=\left(\frac{\sqrt{100}-\sqrt{40}}{\sqrt{5}+\sqrt{2}}+\frac{\sqrt{5}.\sqrt{6}-\sqrt{5}}{\sqrt{5}-1}\right):\frac{1}{2\sqrt{5}-\sqrt{6}}$

$=\left(\frac{\sqrt{20}\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{5}+\sqrt{2}}+\frac{\sqrt{6}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}\right):\frac{1}{2\sqrt{5}-\sqrt{6}}=\left(2\sqrt{5}+\sqrt{6}\right)\left(2\sqrt{5}-\sqrt{6}\right)$

$={\left(2\sqrt{5}\right)}^2-{\left(\sqrt{6}\right)}^2=20-6=14$

Đáp án cần chọn là: B

$P=x\sqrt{y}+y\sqrt{x}={\left(\sqrt{x}\right)}^2\sqrt{y}+{\left(\sqrt{y}\right)}^2\sqrt{x}=\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)$

$Q=x\sqrt{x}+y\sqrt{y}={\left(\sqrt{x}\right)}^3+{\left(\sqrt{y}\right)}^3=\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)$

$R=x-y={\left(\sqrt{x}\right)}^2-{\left(\sqrt{y}\right)}^2=\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)$

Vậy $R=\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)$

Đáp án cần chọn là: C

$M={\left(\sqrt{x}+\sqrt{y}\right)}^2={\left(\sqrt{x}\right)}^2+2\sqrt{x}.\sqrt{y}+{\left(\sqrt{y}\right)}^2=x+2\sqrt{xy}+y$

$N=\frac{x\sqrt{x}-y\sqrt{y}}{\sqrt{x}-\sqrt{y}}=\frac{{\left(\sqrt{x}\right)}^3-{\left(\sqrt{y}\right)}^3}{\sqrt{x}-\sqrt{y}}=\frac{\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)}{\sqrt{x}-\sqrt{y}}=x+\sqrt{xy}+y$

$P=\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)={\left(\sqrt{x}\right)}^2-{\left(\sqrt{y}\right)}^2=x-y$

Vậy $N = x + \sqrt{xy} + y$

Đáp án cần chọn là: B

Ta có

$\frac{a}{\sqrt{5}+1}+\frac{a}{\sqrt{5}-2}-\frac{a}{3-\sqrt{5}}-\sqrt{5}a$

$=\frac{a\left(\sqrt{5}-1\right)}{\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)}+\frac{a\left(\sqrt{5}+2\right)}{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}-\frac{a\left(3+\sqrt{5}\right)}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}-\sqrt{5}a$

$=\frac{a\left(\sqrt{5}-1\right)}{4}+\frac{a\left(\sqrt{5}+2\right)}{1}+\frac{a\left(3+\sqrt{5}\right)}{4}-\sqrt{5}a$

$=\frac{a\left(\sqrt{5}-1\right)+4a\left(2+\sqrt{5}\right)-a\left(3+\sqrt{5}\right)-4\sqrt{5}a}{4}$

$=\frac{a\left(\sqrt{5}-1+8+4\sqrt{5}-3-\sqrt{5}-4\sqrt{5}\right)}{4}=\frac{4a}{4}=a$

Đáp án cần chọn là: B

Ta có: $\frac{4a}{\sqrt{7}-\sqrt{3}}-\frac{2a}{2-\sqrt{2}}-\frac{a}{\sqrt{3}+\sqrt{2}}$

$=\frac{4a\left(\sqrt{7}+\sqrt{3}\right)}{\left(\sqrt{7}-\sqrt{3}\right)\left(\sqrt{7}+\sqrt{3}\right)}-\frac{2a\left(2+\sqrt{2}\right)}{\left(2-\sqrt{2}\right)\left(2+\sqrt{2}\right)}-\frac{a\left(\sqrt{3}-\sqrt{2}\right)}{\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)}$

$$\begin{gathered} =a\left(\sqrt{7}+\sqrt{3}\right)-a\left(2+\sqrt{2}\right)-a\left(\sqrt{3}-\sqrt{2}\right) \\ =a\left(\sqrt{7}+\sqrt{3}-2-\sqrt{2}-\sqrt{3}+\sqrt{2}\right) \end{gathered}$$

$=a\left(\sqrt{7}-2\right)$

Đáp án cần chọn là: D