\(a,\dfrac{7}{\sqrt{12}}=\dfrac{7\sqrt{3}}{\sqrt{12}\cdot\sqrt{3}}\)

\(=\dfrac{7\sqrt{3}}{\sqrt{36}}=\dfrac{7\sqrt{3}}{6}\)

\(b,\dfrac{3}{2\sqrt{3}}=\dfrac{3\sqrt{3}}{2\sqrt{3}\cdot\sqrt{3}}\)

\(=\dfrac{3\sqrt{3}}{2\cdot3}=\dfrac{3\sqrt{3}}{6}=\dfrac{\sqrt{3}}{2}\)

\(c,\dfrac{1}{5\sqrt{12}}=\dfrac{\sqrt{3}}{5\cdot2\sqrt{3}\cdot\sqrt{3}}\)

\(=\dfrac{\sqrt{3}}{10\cdot3}=\dfrac{\sqrt{3}}{30}\)

\(d,\dfrac{2\sqrt{3}+3}{4\sqrt{3}}=\dfrac{\sqrt{3}\left(2+\sqrt{3}\right)}{4\sqrt{3}}\)

\(=\dfrac{2+\sqrt{3}}{4}\)

a) \(\dfrac{7}{\sqrt[]{12}}=\dfrac{7}{2\sqrt[]{3}}=\dfrac{7\sqrt[]{3}}{2\sqrt[]{3}.\sqrt[]{3}}=\dfrac{7\sqrt[]{3}}{6}\)

b) \(\dfrac{3}{2\sqrt[]{3}}=\dfrac{\sqrt[]{3}.\sqrt[]{3}}{2\sqrt[]{3}}=\dfrac{\sqrt[]{3}}{2}\)

c) \(\dfrac{1}{5\sqrt[]{12}}=\dfrac{1}{10\sqrt[]{3}}=\dfrac{\sqrt[]{3}}{10\sqrt[]{3}.\sqrt[]{3}}=\dfrac{\sqrt[]{3}}{30}\)

d) \(\dfrac{2\sqrt[]{3}+3}{4\sqrt[]{3}}=\dfrac{\sqrt[]{3}\left(2\sqrt[]{3}+3\right)}{4\sqrt[]{3}.\sqrt[]{3}}=\dfrac{3\left(2+\sqrt[]{3}\right)}{12}=\dfrac{2+\sqrt[]{3}}{4}\)

a) Ta có: \(\dfrac{7}{\sqrt{5}-\sqrt{3}+\sqrt{7}}\)

\(=\dfrac{7\left(\sqrt{5}-\sqrt{3}-\sqrt{7}\right)}{\left(\sqrt{5}-\sqrt{3}\right)^2-7}\)

\(=\dfrac{7\left(\sqrt{5}-\sqrt{3}-\sqrt{7}\right)}{1-2\sqrt{15}}\)

\(=\dfrac{7\left(\sqrt{5}-\sqrt{3}-\sqrt{7}\right)\left(1+2\sqrt{15}\right)}{1-60}\)

\(=\dfrac{-7\left(\sqrt{5}+10\sqrt{3}-\sqrt{3}-6\sqrt{5}-\sqrt{7}-2\sqrt{105}\right)}{59}\)

\(=\dfrac{-7\left(-5\sqrt{5}+9\sqrt{3}-\sqrt{7}-2\sqrt{105}\right)}{59}\)

 

a) \(\dfrac{7}{\sqrt{5}-\sqrt{3}-\sqrt{7}}\)

\(=\dfrac{7\left(\sqrt{5}-\sqrt{3}+\sqrt{7}\right)}{\left(\sqrt{5}-\sqrt{3}\right)^2-7}\)

\(=\dfrac{7\sqrt{5}-7\sqrt{3}+7\sqrt{7}}{8-2\sqrt{15}-7}\)

\(=\dfrac{7\sqrt{5}-7\sqrt{3}+7\sqrt{7}}{1-2\sqrt{15}}\)

\(=\dfrac{\left(7\sqrt{5}-7\sqrt{3}+7\sqrt{7}\right)\left(1+2\sqrt{15}\right)}{1-60}\)

\(=\dfrac{7\sqrt{5}+70\sqrt{3}-7\sqrt{3}-42\sqrt{5}+7\sqrt{7}+14\sqrt{105}}{-59}\)

\(=\dfrac{-35\sqrt{5}+63\sqrt{3}+7\sqrt{7}+14\sqrt{105}}{-59}\)

\(=\dfrac{35\sqrt{5}-63\sqrt{3}-7\sqrt{7}-14\sqrt{105}}{59}\)

a: \(=3xy\cdot\dfrac{\sqrt{2}}{\sqrt{xy}}=3\sqrt{2}\sqrt{xy}\)

b: \(=x\cdot\dfrac{\sqrt{6}}{\sqrt{x}}+\dfrac{\sqrt{6}}{3}\sqrt{x}\)

\(=\sqrt{6}\sqrt{x}+\dfrac{\sqrt{6}}{3}\sqrt{x}=\dfrac{4\sqrt{6}}{3}\cdot\sqrt{x}\)

c: \(=\sqrt{xy}+x\cdot\dfrac{\sqrt{y}}{\sqrt{x}}-y\cdot\dfrac{\sqrt{x}}{\sqrt{y}}\)

\(=\sqrt{xy}+\sqrt{xy}-\sqrt{xy}=\sqrt{xy}\)

a) Ta có: \(\dfrac{1}{\sqrt{5}-\sqrt{3}-\sqrt{2}}\)

\(=\dfrac{\sqrt{5}+\sqrt{3}+\sqrt{2}}{5-\left(\sqrt{3}+\sqrt{2}\right)^2}\)

\(=\dfrac{\sqrt{5}+\sqrt{3}+\sqrt{2}}{5-5-2\sqrt{6}}\)

\(=\dfrac{-\sqrt{5}-\sqrt{3}-\sqrt{2}}{2\sqrt{6}}\)

\(=\dfrac{-\sqrt{6}\left(\sqrt{5}+\sqrt{3}+\sqrt{2}\right)}{12}\)

b) Ta có: \(\dfrac{2}{-1-\sqrt{2}+\sqrt{3}}\)

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

\(=\dfrac{\left(-1-\sqrt{2}-\sqrt{3}\right)}{\sqrt{2}}\)

\(=\dfrac{-\sqrt{2}-2-\sqrt{6}}{2}\)

 

có nghĩa khi và
Nếu thì
Nếu thì Tương tự như vậy ta có:
Nếu thì
Nếu thì Ta có:
Điều kiện để căn thức có nghĩa là hay Do đó:
Nếu b>0 thì
Nếu thì Điều kiện để có nghĩa là hay
Cách 1.
=
Cách 2. Biến mẫu thành một bình phương rồi áp dụng quy tắc khai phương một thương: Điều kiện để có nghĩa là hay xy>0.
Do đó