a: \(\dfrac{3}{\sqrt{2}}+\sqrt{\dfrac{1}{2}}-2\sqrt{18}+\sqrt{\left(1-\sqrt{2}\right)^2}\)

\(=\dfrac{3}{2}\sqrt{2}+\dfrac{1}{2}\sqrt{2}-2\cdot3\sqrt{2}+\left|1-\sqrt{2}\right|\)

\(=2\sqrt{2}-6\sqrt{2}+\sqrt{2}-1=-3\sqrt{2}-1\)

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

\(=\dfrac{1}{\sqrt{3}}+\dfrac{1}{\sqrt{18}}+\dfrac{\sqrt{3}-\sqrt{2}}{12}\)

\(=\dfrac{4\sqrt{3}+2\sqrt{2}+\sqrt{3}-\sqrt{2}}{12}\)

\(=\dfrac{5\sqrt{3}+\sqrt{2}}{12}\)

c: \(\sqrt[3]{\dfrac{3}{4}}\cdot\sqrt[3]{\dfrac{9}{16}}=\sqrt[3]{\dfrac{3}{4}\cdot\dfrac{9}{16}}=\sqrt[3]{\dfrac{27}{64}}=\dfrac{3}{4}\)

d: \(\dfrac{\sqrt[3]{54}}{\sqrt[3]{-2}}=\sqrt[3]{\dfrac{54}{-2}}=-\sqrt[3]{27}=-3\)

e: \(\sqrt[3]{5\sqrt{2}+7}-\sqrt[3]{5\sqrt{2}+7}=0\)

\(\dfrac{6-\sqrt{6}}{\sqrt{6}-1}+\dfrac{6-\sqrt{6}}{\sqrt{6}}\)

\(=\dfrac{\sqrt{6}\cdot\sqrt{6}-\sqrt{6}}{\sqrt{6}-1}+\dfrac{\sqrt{6}\cdot\sqrt{6}-\sqrt{6}}{\sqrt{6}}\)

\(=\dfrac{\sqrt{6}\left(\sqrt{6}-1\right)}{\sqrt{6}-1}+\dfrac{\sqrt{6}\left(\sqrt{6}-1\right)}{\sqrt{6}}\)

\(=\dfrac{\sqrt{6}}{1}+\dfrac{\sqrt{6}-1}{1}\)

\(=\sqrt{6}+\sqrt{6}-1\)

\(=2\sqrt{6}-1\)

=======================

\(\dfrac{1}{\sqrt{2}-\sqrt{3}}-\dfrac{3}{\sqrt{18}+2\sqrt{3}}\)

\(=\dfrac{1}{\sqrt{2}-\sqrt{3}}-\dfrac{3}{\sqrt{6}\cdot\sqrt{3}+\sqrt{6}\cdot\sqrt{2}}\)

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

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

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

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

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

1, \(\dfrac{1}{3-2\sqrt{2}}-\dfrac{1}{3+2\sqrt{2}}=\dfrac{3+2\sqrt{2}}{9-8}-\dfrac{3-2\sqrt{2}}{9-8}\)

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

2, \(\dfrac{1}{\sqrt{2}-\sqrt{3}}-\dfrac{3}{\sqrt{18}+2\sqrt{3}}=\dfrac{1}{\sqrt{2}-\sqrt{3}}-\dfrac{3}{\sqrt{18}+\sqrt{12}}\)

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

\(=\dfrac{2\sqrt{3}+3\sqrt{2}-3\sqrt{2}+3\sqrt{3}}{-\sqrt{6}}=\dfrac{5\sqrt{3}}{-\sqrt{6}}=-5\sqrt{18}=-15\sqrt{2}\)

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

\(=2\sqrt{5}+4-2\sqrt{5}+4=8\)

tương tự 

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

Bài 1:

ĐKXĐ: \(\dfrac{5}{x^2+6}>=0\)

=>\(x^2+6>0\)

mà \(x^2+6>=6>0\forall x\)

nên \(x\in R\)

Bài 2:

a: Sửa đề: \(\dfrac{3}{\sqrt{2}}+\sqrt{\dfrac{1}{2}}-2\cdot\sqrt{18}+\sqrt{\left(1-\sqrt{2}\right)^2}\)

\(=\dfrac{3}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-2\cdot3\sqrt{2}+\left|1-\sqrt{2}\right|\)

\(=2\sqrt{2}-6\sqrt{2}+\sqrt{2}-1=-3\sqrt{2}-1\)

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

\(=\dfrac{\sqrt{6}+1}{3\sqrt{2}}+\dfrac{\sqrt{3}-\sqrt{2}}{6}\)

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

c: \(\sqrt[3]{\dfrac{3}{4}}\cdot\sqrt[3]{\dfrac{9}{16}}=\sqrt[3]{\dfrac{3}{4}\cdot\dfrac{9}{16}}=\sqrt[3]{\dfrac{27}{64}}=\dfrac{3}{4}\)

d: \(\sqrt[3]{54}=\sqrt[3]{27\cdot2}=3\sqrt[3]{2}\)

e: \(\dfrac{\sqrt[3]{54}}{\sqrt[3]{-2}}=\sqrt[3]{\dfrac{54}{-2}}=\sqrt[3]{-27}=-3\)

f: \(\sqrt[3]{5\sqrt{2}+7}-\sqrt[3]{5\sqrt{2}-7}\)

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

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

1: \(=\sqrt{5}-\dfrac{\sqrt{5}}{2}=\dfrac{\sqrt{5}}{2}\)

2: \(=\dfrac{4+2\sqrt{3}+4-2\sqrt{3}}{2}=\dfrac{8}{2}=4\)

4: \(=\dfrac{-3+5\sqrt{3}}{11}+\dfrac{3+5\sqrt{3}}{11}=\dfrac{10\sqrt{3}}{11}\)

\(\dfrac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}+\dfrac{5-2\sqrt{5}}{2\sqrt{5}-4}\)

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

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

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

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

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

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

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

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

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

\(=-1\)

\(\dfrac{3\sqrt{18}-2\sqrt{8}}{\sqrt{50}}\)

\(=\dfrac{3\cdot3\sqrt{2}-2\cdot2\sqrt{2}}{5\sqrt{2}}\)

\(=\dfrac{9\sqrt{2}-4\sqrt{2}}{5\sqrt{2}}\)

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

\(=1\)

a) Ta có: \(A=\sqrt{20}-2\sqrt{45}+3\sqrt{18}+\sqrt{72}\)

\(=2\sqrt{5}-6\sqrt{5}+9\sqrt{2}+6\sqrt{2}\)

\(=-4\sqrt{5}+15\sqrt{2}\)

b) Ta có: \(B=4\sqrt{\left(\sqrt{3}-1\right)^2}+2\sqrt{12}+4\sqrt{\dfrac{1}{2}}\)

\(=4\left(\sqrt{3}-1\right)+2\cdot2\sqrt{3}+\dfrac{4}{\sqrt{2}}\)

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

\(=8\sqrt{3}+2\sqrt{2}-4\)

c) Ta có: \(C=\left(3+\dfrac{3-\sqrt{3}}{\sqrt{3}-1}\right)\left(3-\dfrac{3+\sqrt{3}}{1+\sqrt{3}}\right)\)

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

=9-3

=6

d) Ta có: \(D=\dfrac{1}{2+\sqrt{3}}+\dfrac{1}{2-\sqrt{3}}\)

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

=4

4: Ta có: \(\dfrac{6}{1-\sqrt{3}}-\dfrac{3\sqrt{3}+3}{\sqrt{3}+1}\)

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

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

d: \(=\sqrt{5}\left(\sqrt{3}-1\right)-\dfrac{\sqrt{5}\left(\sqrt{5}-2\right)}{2\left(\sqrt{5}-2\right)}\)

=căn 5-1/2*căn 5

=1/2*căn 5

e: \(=\dfrac{2\left(\sqrt{8}-\sqrt{3}\right)}{\sqrt{6}\left(\sqrt{3}-\sqrt{8}\right)}-\dfrac{1}{\sqrt{6}}=\dfrac{2}{\sqrt{6}}-\dfrac{1}{\sqrt{6}}=\dfrac{1}{\sqrt{6}}\)

f:=2+căn 3+căn 2-2-căn 3=căn 2