\(a,\frac{1}{\sqrt{2}+\sqrt{3}-\sqrt{6}}=\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}}{\left(\sqrt{2}+\sqrt{3}-\sqrt{6}\right)\left(\sqrt{2}+\sqrt{3}+\sqrt{6}\right)}=\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}}{\left(\sqrt{2}+\sqrt{3}\right)^2-\sqrt{6}^2}\)

\(=\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}}{2\sqrt{6}-1}=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{6}\right)\left(2\sqrt{6}+1\right)}{2\sqrt{6}^2-1^2}=\frac{4\sqrt{3}+6\sqrt{2}+12+\sqrt{2}+\sqrt{3}+\sqrt{6}}{11}\)\(=\frac{\sqrt{6}+5\sqrt{3}+7\sqrt{2}+12}{11}\)

\(b,\frac{1}{\sqrt{x}+\sqrt{y}+\sqrt{z}}=\frac{\sqrt{x}+\sqrt{y}-\sqrt{z}}{\left(\sqrt{z}+\sqrt{y}+\sqrt{z}\right)\left(\sqrt{x}+\sqrt{y}-\sqrt{z}\right)}=\frac{\sqrt{x}+\sqrt{y}-\sqrt{z}}{\left(\sqrt{x}+\sqrt{y}\right)^2-\sqrt{z}^2}\)

\(=\frac{\sqrt{x}+\sqrt{y}-\sqrt{z}}{x+2\sqrt{xy}+y-z}\)

\(\dfrac{5}{2\sqrt{5}}=\dfrac{5\sqrt{5}}{2.5}=\dfrac{\sqrt{5}}{2}\)

\(\dfrac{2\sqrt{2}+2}{5\sqrt{2}}=\dfrac{4+2\sqrt{2}}{5.2}=\dfrac{2+\sqrt{2}}{5}\)

\(\dfrac{y+b\sqrt{y}}{b\sqrt{y}}=\dfrac{y\sqrt{y}+by}{by}=\dfrac{\sqrt{y}+b}{b}\left(y>0;b\ne0\right)\)

\(\dfrac{5}{2\sqrt{5}}=\dfrac{\sqrt{5}\cdot\sqrt{5}}{2\sqrt{5}}=\dfrac{\sqrt{5}}{2}\)

\(\dfrac{2\sqrt{2}+2}{5\sqrt{2}}=\dfrac{2\sqrt{2}+\sqrt{2}\cdot\sqrt{2}}{5\sqrt{2}}=\dfrac{\sqrt{2}\cdot\left(2+\sqrt{2}\right)}{5\sqrt{2}}=\dfrac{2+\sqrt{2}}{5}\)

\(\dfrac{y+b\sqrt{y}}{b\sqrt{y}}=\dfrac{\sqrt{y}\cdot\sqrt{y}+b\sqrt{y}}{b\sqrt{y}}=\dfrac{\sqrt{y}\left(\sqrt{y}+b\right)}{b\sqrt{y}}=\dfrac{\sqrt{y}+b}{y}\)

Trước hết, ta cần tính giá trị của a và b trong G và H:
$$G^2 = \frac{1}{a+b} \Rightarrow a+b = \frac{1}{G^2}$$
$$H^2 = 4a - 4\sqrt{ab} + 4b = 4(\sqrt{a} - \sqrt{b})^2 \Rightarrow \sqrt{a} - \sqrt{b} = \frac{H}{2}$$
Từ đó, suy ra được:
$$\sqrt{a} + \sqrt{b} = \frac{1}{G}\sqrt{\frac{1}{G^2} + 4}$$
$$\Rightarrow 2\sqrt{a} = \frac{1}{G}\sqrt{\frac{1}{G^2} + 4} + H$$
$$\Rightarrow a = \left(\frac{1}{G}\sqrt{\frac{1}{G^2} + 4} + H\right)^2/4$$
$$\Rightarrow b = \left(\frac{1}{G}\sqrt{\frac{1}{G^2} + 4} - H\right)^2/4$$

Tiếp theo, ta tính giá trị của F:
$$F = 6\sqrt{3} + \sqrt{2} = 6\sqrt{3} + \sqrt{2}\frac{\sqrt{6}+\sqrt{2}}{2} = 6\sqrt{3} + 3\sqrt{2} + 3\sqrt{6}$$

Cuối cùng, ta tính giá trị của K:
$$K = 2xy\left(2\sqrt{x} + 3\sqrt{y}\right) = 2\sqrt{xy}(4\sqrt{x} + 6\sqrt{y})$$

Vậy, ta đã tính được giá trị của F, G, H và K.

a, căn a trừ b/ a-b^2

b, 2 căn a + 2 căn b /a-b

c, 6 căn a trừ 6 / 4a-1

a) căn a-b/a-b^2

b) 2(căn a+căn b)/a-b

c) 3(2 căn a-1)/4a-1

d)2xy(2 căn x-3 căn y)/4x-9y

a) √a -b /a-b²

b) 2√a+2b / a-b²

c) 6√a -3 / 4a-1

d) 4xy√x -6xy√y / 4x-9y

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

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

\(\dfrac{3}{\sqrt{10}+\sqrt{7}}=\dfrac{3\left(\sqrt{10}-\sqrt{7}\right)}{\left(\sqrt{10}+\sqrt{7}\right)\left(\sqrt{10}-\sqrt{7}\right)}=\dfrac{3\left(\sqrt{10}-\sqrt{7}\right)}{10-7}=\dfrac{3\left(\sqrt{10}-\sqrt{7}\right)}{3}=\sqrt{10}-\sqrt{7}\)

\(\dfrac{2}{\sqrt{6}-\sqrt{5}}=\dfrac{2\left(\sqrt{6}+\sqrt{5}\right)}{\left(\sqrt{6}-\sqrt{5}\right)\left(\sqrt{6}+\sqrt{5}\right)}=\dfrac{2\left(\sqrt{6}+\sqrt{5}\right)}{6-5}=2\left(\sqrt{6}+\sqrt{5}\right)\)

ĐS: