1) Ta có: \(3\sqrt{12}+\dfrac{1}{2}\sqrt{48}-\sqrt{27}\)

\(=3\cdot2\sqrt{3}+\dfrac{1}{2}\cdot4\sqrt{3}-3\sqrt{3}\)

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

\(=5\sqrt{3}\)

2) Ta có: \(\dfrac{2}{\sqrt{3}-5}\)

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

\(=\dfrac{2\left(\sqrt{3}+5\right)}{3-25}\)

\(=\dfrac{-2\left(\sqrt{3}+5\right)}{22}\)

\(=\dfrac{-\sqrt{3}-5}{11}\)

3) Ta có: \(\sqrt{\dfrac{2}{5}}\)

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

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

\(=\dfrac{\sqrt{10}}{5}\)

Nếu em thấy các câu hỏi do lag mà bị gửi đúp (tức là rất nhiều câu hỏi giống nhau xuất hiện cùng 1 chỗ) thì xóa giúp mình nhé cho đỡ vướng. Nhưng nhớ để lại 1 câu. Cảm ơn em.

a: Ta có: \(\left(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}-\dfrac{12}{3-\sqrt{6}}\right)\left(\sqrt{6}-11\right)\)

\(=\left(3\sqrt{6}-3+2\sqrt{6}+4-12-4\sqrt{6}\right)\left(\sqrt{6}-11\right)\)

\(=\left(\sqrt{6}-11\right)\left(\sqrt{6}-11\right)\)

\(=127-22\sqrt{6}\)

b: Ta có: \(\left(1-\dfrac{5+\sqrt{5}}{1+\sqrt{5}}\right)\left(\dfrac{5-\sqrt{5}}{1-\sqrt{5}}-1\right)\)

\(=\left(1-\sqrt{5}\right)\left(-1-\sqrt{5}\right)\)

=-1+5

=4

1

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

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

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

\(=\dfrac{\left|3-\sqrt{5}\right|}{\sqrt{9-5}}\)

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

b) \(\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}\)

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

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

\(=\dfrac{\left|2-\sqrt{3}\right|}{\sqrt{4-3}}\)

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

\(=2-\sqrt{3}\)

a: \(=\sqrt{\dfrac{\left(3-\sqrt{5}\right)\left(3-\sqrt{5}\right)}{4}}=\dfrac{3-\sqrt{5}}{2}\)

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

d: \(=\left(-3+3\sqrt{6}+4+2\sqrt{6}-12-4\sqrt{6}\right)\left(\sqrt{6}+11\right)\)

=(căn 6-11)(căn 6+11)

=6-121=-115

Bài 1:
a.

\(\frac{1}{2\sqrt{2}-3\sqrt{3}}=\frac{2\sqrt{2}+3\sqrt{3}}{(2\sqrt{2}-3\sqrt{3})(2\sqrt{2}+3\sqrt{3})}=\frac{2\sqrt{2}+3\sqrt{3}}{(2\sqrt{2})^2-(3\sqrt{3})^2}=\frac{2\sqrt{2}+3\sqrt{3}}{-19}\)

b.

\(=\sqrt{\frac{(3-\sqrt{5})^2}{(3-\sqrt{5})(3+\sqrt{5})}}=\sqrt{\frac{(3-\sqrt{5})^2}{3^2-5}}=\sqrt{\frac{(3-\sqrt{5})^2}{4}}=\sqrt{(\frac{3-\sqrt{5}}{2})^2}=|\frac{3-\sqrt{5}}{2}|=\frac{3-\sqrt{5}}{2}\)

Bài 2.

a. 

\(=\frac{\sqrt{8}(\sqrt{5}+\sqrt{3})}{(\sqrt{5}-\sqrt{3})(\sqrt{5}+\sqrt{3})}=\frac{2\sqrt{2}(\sqrt{5}+\sqrt{3})}{5-3}=\sqrt{2}(\sqrt{5}+\sqrt{3})=\sqrt{10}+\sqrt{6}\)

b.

\(=\sqrt{\frac{(2-\sqrt{3})^2}{(2-\sqrt{3})(2+\sqrt{3})}}=\sqrt{\frac{(2-\sqrt{3})^2}{2^2-3}}=\sqrt{(2-\sqrt{3})^2}=|2-\sqrt{3}|=2-\sqrt{3}\)

\(B=\left(\dfrac{4}{1-\sqrt{5}}+\dfrac{1}{2+\sqrt{5}}-\dfrac{4}{3-\sqrt{5}}\right)\left(\sqrt{5}-6\right)\)

\(B=\left[\dfrac{4\left(1+\sqrt{5}\right)}{\left(1-\sqrt{5}\right)\left(1+\sqrt{5}\right)}+\dfrac{2-\sqrt{5}}{\left(2+\sqrt{5}\right)\left(2-\sqrt{5}\right)}-\dfrac{4\left(3+\sqrt{5}\right)}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}\right]\left(\sqrt{5}-6\right)\)

\(B=\left[\dfrac{4\left(1+\sqrt{5}\right)}{1-5}+\dfrac{2-\sqrt{5}}{4-5}-\dfrac{4\left(3+\sqrt{5}\right)}{9-5}\right]\left(\sqrt{5}-6\right)\)

\(B=\left[-\dfrac{4\left(1+\sqrt{5}\right)}{4}-\dfrac{2-\sqrt{5}}{1}-\dfrac{4\left(3+\sqrt{5}\right)}{4}\right]\left(\sqrt{5}-6\right)\)

\(B=\left(-1-\sqrt{5}-2+\sqrt{5}-3-\sqrt{5}\right)\left(\sqrt{5}-6\right)\)

\(B=\left(-\sqrt{5}-6\right)\left(\sqrt{5}-6\right)\)

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

\(B=-\left(5-36\right)\)

\(B=-\left(-31\right)\)

\(B=31\)

_____________________________

\(\sqrt{48}-\dfrac{\sqrt{21}-\sqrt{15}}{\sqrt{7}-\sqrt{5}}+\dfrac{2}{\sqrt{3}+1}\)

\(=4\sqrt{3}-\dfrac{\sqrt{3}\left(\sqrt{7}-\sqrt{5}\right)}{\sqrt{7}-\sqrt{5}}+\dfrac{2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\)

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

\(=3\sqrt{3}-\sqrt{3}+1\)

\(=2\sqrt{3}+1\)

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

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

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

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

ĐS:

\(a,=\dfrac{\sqrt{7}-5}{2}-\dfrac{3-\sqrt{7}}{2}+\dfrac{6\left(\sqrt{7}+2\right)}{3}-\dfrac{5\left(4-\sqrt{7}\right)}{9}\\ =\dfrac{\sqrt{7}-5-3+\sqrt{7}}{2}+2\sqrt{7}+4-\dfrac{20-5\sqrt{7}}{9}\\ =\dfrac{2\sqrt{7}-8}{2}+2\sqrt{7}+4-\dfrac{20-5\sqrt{7}}{9}\\ =\sqrt{7}-4+2\sqrt{7}+4-\dfrac{20-5\sqrt{7}}{9}\\ =\dfrac{27\sqrt{7}-20+5\sqrt{7}}{9}=\dfrac{32\sqrt{7}-20}{9}\)

\(b,=\dfrac{2\left(\sqrt{6}+2\right)}{2}+\dfrac{2\left(\sqrt{6}-2\right)}{2}+\dfrac{5\sqrt{6}}{6}\\ =\sqrt{6}+2+\sqrt{6}-2+\dfrac{5\sqrt{6}}{6}\\ =\dfrac{12\sqrt{6}+5\sqrt{6}}{6}=\dfrac{17\sqrt{6}}{6}\)

\(c,=\dfrac{\sqrt{3}+\sqrt{2}+\sqrt{5}-\sqrt{3}-\sqrt{2}+\sqrt{5}}{\left(\sqrt{3}+\sqrt{2}\right)^2-5}\\ =\dfrac{2\sqrt{5}}{5+2\sqrt{6}-5}=\dfrac{2\sqrt{5}}{2\sqrt{6}}=\dfrac{\sqrt{30}}{6}\)

\(\left(1-\dfrac{5+\sqrt{5}}{1+\sqrt{5}}\right)\left(\dfrac{5-\sqrt{5}}{1-\sqrt{5}}-1\right)=\left(1-\dfrac{\sqrt{5}\left(1+\sqrt{5}\right)}{1+\sqrt{5}}\right)\left(-\dfrac{\sqrt{5}\left(1-\sqrt{5}\right)}{1-\sqrt{5}}-1\right)=-\left(1-\sqrt{5}\right)\left(1+\sqrt{5}\right)=-\left(1-5\right)=4\)

\(\left(1-\dfrac{5+\sqrt{5}}{1+\sqrt{5}}\right)\left(\dfrac{5-\sqrt{5}}{1-\sqrt{5}}-1\right)\)

\(=\left(1-\sqrt{5}\right)\left(-1-\sqrt{5}\right)\)

\(=-1+5=4\)