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.

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

Bài 3:

a.

\(M=\left[\frac{15(\sqrt{6}-1)}{(\sqrt{6}+1)(\sqrt{6}-1)}+\frac{4(\sqrt{6}+2)}{(\sqrt{6}-2)(\sqrt{6}+2)}-\frac{12(3+\sqrt{6})}{(3-\sqrt{6})(3+\sqrt{6})}\right](\sqrt{6}+11)\)

\(=\left[\frac{15(\sqrt{6}-1)}{6-1}+\frac{4(\sqrt{6}+2)}{6-2^2}-\frac{12(3+\sqrt{6})}{3^2-6}\right](\sqrt{6}+11)\)

\(=[3(\sqrt{6}-1)+2(\sqrt{6}+2)-4(3+\sqrt{6})](\sqrt{6}+11)=(\sqrt{6}-11)(\sqrt{6}+11)=6-11^2=-115\)

b.

\(N=\left[1-\frac{\sqrt{5}(\sqrt{5}+1)}{\sqrt{5}+1}\right].\left[\frac{\sqrt{5}(\sqrt{5}-1)}{1-\sqrt{5}}-1\right]\)

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

Em thử nhá, ko chắc đâu ạ. Em chỉ làm đc một cái thôi

Gọi biểu thức trên là A

*Chứng minh A > 1/6

Đặt \(x=\sqrt{6+\sqrt{6+\sqrt{6+...+\sqrt{6}}}}\left(\text{n dấu căn}\right)\)

Thì \(x=\sqrt{6+\sqrt{6+\sqrt{6+...+\sqrt{6}}}}< \sqrt{6+\sqrt{6+\sqrt{6+...+\sqrt{9}}}}=\sqrt{6+3}=3\) (1)

Và \(x^2-6=\sqrt{6+\sqrt{6+...+\sqrt{6}}}\left(\text{n -1 dấu căn}\right)\)

Biểu thức trở thành \(A=\frac{3-x}{9-x^2}=\frac{1}{3+x}\). Từ (1) suy ra \(A>\frac{1}{3+3}=\frac{1}{6}\)(*)

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

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

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: \(\sqrt{\dfrac{2}{5}}\)

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

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

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

\(A=\dfrac{2}{2.\sqrt[3]{2}+2+\sqrt[3]{2^2}}=\dfrac{2}{\left(\sqrt[3]{2}\right)^2+2.\left(\sqrt[3]{2}\right)+\left(\sqrt{2}\right)^2}\)

\(A=\dfrac{2.\left(\sqrt[3]{2}\right)-\left(\sqrt{2}\right)}{\left(\sqrt[3]{2}\right)-\left(\sqrt{2}\right)\left[\left(\sqrt[3]{2}\right)^2+2.\left(\sqrt[3]{2}\right)+\left(\sqrt{2}\right)^2\right]}=\dfrac{2.\left(\sqrt[3]{2}\right)-\left(\sqrt{2}\right)}{\left(\sqrt[3]{2}\right)^3-\left(\sqrt{2}\right)^3}=\dfrac{2.\left(\sqrt[3]{2}\right)-\left(\sqrt{2}\right)}{2-2\sqrt{2}}\)

\(A=\dfrac{2\left[.\left(\sqrt[3]{2}\right)-\left(\sqrt{2}\right)\right].\left(1+\sqrt{2}\right)}{2\left(1-\sqrt{2}\right)\left(1+\sqrt{2}\right)}=\left(\sqrt{2}+1\right)\left(\sqrt{2}-\sqrt[3]{2}\right)\)

ở phân thức A nhân cả tử và mẫu cho: (2\(\sqrt[3]{2}\))²-2.\(\sqrt[3]{2}\left(2+\sqrt[3]{4}\right)+\left(2-\sqrt[3]{4}\right)^2\)

ở phân thức B nhân cả tử và mẫu cho :(2\(\sqrt[3]{2}\))²+\(2.\sqrt[3]{2}\left(2-\sqrt[3]{4}\right)+\left(2-\sqrt[3]{4}\right)^2\)

\(a,\frac{2\sqrt{10}-5}{4-\sqrt{10}}=\frac{\left(2\sqrt{10}-5\right)\left(4+\sqrt{10}\right)}{\left(4-\sqrt{10}\right)\left(4+\sqrt{10}\right)}=\frac{20+6\sqrt{10}-5\sqrt{10}-9}{16-10}.\)

\(=\frac{11-\sqrt{10}}{6}\)

\(b,=\frac{\left(9-2\sqrt{2}\right)\left(3\sqrt{6}+2\sqrt{2}\right)}{\left(3\sqrt{6}-2\sqrt{2}\right)\left(3\sqrt{6}+2\sqrt{2}\right)}=\frac{\left(9-2\sqrt{2}\right)\left(3\sqrt{6}+2\sqrt{2}\right)}{54-8}\)

\(=\frac{\left(9-2\sqrt{2}\right)\left(3\sqrt{6}+2\sqrt{2}\right)}{46}\)

Ta có: \(\frac{1}{\sqrt[3]{9}-\sqrt[3]{6}+\sqrt[3]{4}}=\)\(\frac{\sqrt[3]{3}+\sqrt[3]{2}}{\left(\sqrt[3]{2}+\sqrt[3]{3}\right)\left(\sqrt[3]{9}-\sqrt[3]{6}+\sqrt[3]{4}\right)}=\frac{\sqrt[3]{2}+\sqrt[3]{3}}{\left(\sqrt[3]{2}\right)^3+\left(\sqrt[3]{3}\right)^3}=\frac{\sqrt[3]{2}+\sqrt[3]{3}}{5}\)