\(\dfrac{3}{\sqrt{5}+\sqrt{2}}+\dfrac{3}{\sqrt{5}-\sqrt{2}}\)
5 câu:
1) \(\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{6}+2}-\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{6}-2}\)
2) \(\dfrac{3}{\sqrt{5}-\sqrt{2}}-\dfrac{2}{2-\sqrt{2}}+\dfrac{1}{\sqrt{3}+\sqrt{2}}\)
3) \(\dfrac{12}{\sqrt{5}+1}-\dfrac{4}{\sqrt{5}+2}+\dfrac{20}{3+\sqrt{5}}\)
4) \(\dfrac{5}{3-\sqrt{7}}-\dfrac{2}{\sqrt{2}+\sqrt{3}}-\dfrac{1}{\sqrt{2}-1}\)
5) \(\dfrac{\sqrt{12}-6}{\sqrt{8}-\sqrt{24}}-\dfrac{3+\sqrt{3}}{\sqrt{3}}-\dfrac{4}{\sqrt{7}-1}\)
1. \(\dfrac{-2}{\sqrt{3}-1}\)
2. \(\dfrac{5}{1-\sqrt{6}}\)
3. \(\dfrac{2+\sqrt{5}}{2-\sqrt{5}}\)
4. \(\dfrac{1}{5+2\sqrt{6}}\)
5. \(\dfrac{\sqrt{5}+2}{\sqrt{5}-2}\)
6. \(\dfrac{5\sqrt{2}-2\sqrt{5}}{\sqrt{2}-\sqrt{5}}\)
7. \(\dfrac{\sqrt{20}-3\sqrt{10}}{3-\sqrt{2}}\)
8. \(\dfrac{6-2\sqrt{5}}{3+\sqrt{5}}\)
9. \(\dfrac{9+4\sqrt{5}}{\sqrt{5}+2}\)
\(\dfrac{\sqrt{6}-\sqrt{3}}{\sqrt{2}-1}+\dfrac{3+\sqrt{3}}{\sqrt{3}+1}+\dfrac{2}{\sqrt{2}+1}-\dfrac{4}{\sqrt{2}}\)
\(\dfrac{4}{\sqrt{5}+1}+\dfrac{5}{\sqrt{5}+2}+\dfrac{5}{\sqrt{5}+3}\)
\(\dfrac{\sqrt{6}-\sqrt{3}}{\sqrt{2}-1}+\dfrac{3+\sqrt{3}}{\sqrt{3}+1}+\dfrac{2}{\sqrt{2}+1}-\dfrac{4}{\sqrt{2}}\)
\(=\dfrac{\sqrt{3}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}+\dfrac{\sqrt{3}\left(\sqrt{3}+1\right)}{\sqrt{3}+1}+\dfrac{2\sqrt{2}}{2+\sqrt{2}}-\dfrac{4\sqrt{2}+4}{2+\sqrt{2}}\)
\(=\sqrt{3}+\sqrt{3}+\dfrac{-2\sqrt{2}-4}{2+\sqrt{2}}\)
\(=2\sqrt{3}+\dfrac{-2\left(2+\sqrt{2}\right)}{2+\sqrt{2}}\)
\(=2\sqrt{3}-2\)
\(------\)
\(\dfrac{4}{\sqrt{5}+1}+\dfrac{5}{\sqrt{5}+2}+\dfrac{5}{\sqrt{5}+3}\)
\(=\dfrac{4\left(\sqrt{5}-1\right)}{\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)}+\dfrac{5\left(\sqrt{5}-2\right)}{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}+\dfrac{5\left(\sqrt{5}-3\right)}{\left(\sqrt{5}+3\right)\left(\sqrt{5}-3\right)}\)
\(=\dfrac{4\sqrt{5}-4}{5-1}+\dfrac{5\sqrt{5}-10}{5-4}+\dfrac{5\sqrt{5}-15}{5-9}\)
\(=5\sqrt{5}-10+\left(\dfrac{4\sqrt{5}-4}{4}+\dfrac{5\sqrt{5}-15}{-4}\right)\)
\(=\dfrac{4\cdot\left(5\sqrt{5}-10\right)}{4}+\left(\dfrac{4\sqrt{5}-4}{4}-\dfrac{5\sqrt{5}-15}{4}\right)\)
\(=\dfrac{20\sqrt{5}-40}{4}+\dfrac{-\sqrt{5}+11}{4}\)
\(=\dfrac{19\sqrt{5}-29}{4}\)
#Ayumu
Thực hiến phép tính :
a, \(\dfrac{1}{3+\sqrt{2}}+\dfrac{1}{3-\sqrt{2}}\)
b, \(\dfrac{2}{3\sqrt{2}-4}-\dfrac{2}{3\sqrt{2}+4}\)
c, \(\dfrac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}+\dfrac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}\)
d, \(\dfrac{3}{2\sqrt{2}-3\sqrt{3}}-\dfrac{3}{2\sqrt{2}+3\sqrt{3}}\)
e, \(\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}\)
g, \(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-6\sqrt{20}}}}\)
\(a,=\dfrac{3-\sqrt{2}+3+\sqrt{2}}{\left(3+\sqrt{2}\right)\left(3-\sqrt{2}\right)}=\dfrac{6}{-1}=-6\\ b,=\dfrac{6\sqrt{2}+8-6\sqrt{2}+8}{\left(3\sqrt{2}-4\right)\left(3\sqrt{2}+4\right)}=\dfrac{16}{2}=8\\ c,=\dfrac{\left(\sqrt{5}-\sqrt{3}\right)^2+\left(\sqrt{5}+\sqrt{3}\right)^2}{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}\\ =\dfrac{8-2\sqrt{15}+8+2\sqrt{15}}{2}=\dfrac{16}{2}=8\)
\(d,=\dfrac{6\sqrt{2}+9\sqrt{3}-6\sqrt{2}+9\sqrt{3}}{\left(2\sqrt{2}-3\sqrt{3}\right)\left(2\sqrt{2}+3\sqrt{3}\right)}=\dfrac{18\sqrt{3}}{-19}=\dfrac{-18\sqrt{3}}{19}\\ e,=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}\\ =\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}\\ =\sqrt{\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}}\\ =\sqrt{\sqrt{5}-\sqrt{5}+1}=\sqrt{1}=1\)
Tính:
1) \(\dfrac{1}{1+\sqrt{5}}+\dfrac{1}{\sqrt{5}-1}\)
2) \(\dfrac{1}{\sqrt{5}+\sqrt{3}}-\dfrac{1}{\sqrt{5}-\sqrt{3}}\)
3) \(\dfrac{1}{\sqrt{2}-\sqrt{3}}-\dfrac{1}{\sqrt{3}+\sqrt{2}}\)
4) \(\dfrac{1}{3+\sqrt{5}}+\dfrac{1}{\sqrt{5}-3}\)
5) \(\dfrac{1}{\sqrt{2}-\sqrt{6}}-\dfrac{1}{\sqrt{6}+\sqrt{2}}\)
LM CHI TIẾT GIÚP MK NHÉ
4: Ta có: \(\dfrac{1}{3+\sqrt{5}}-\dfrac{1}{3-\sqrt{5}}\)
\(=\dfrac{3-\sqrt{5}-3-\sqrt{5}}{4}\)
\(=\dfrac{-\sqrt{5}}{2}\)
11) \(\dfrac{5+\sqrt{5}}{5-\sqrt{5}}\) + \(\dfrac{5-\sqrt{5}}{5+\sqrt{5}}\)
12) \(\dfrac{3+2\sqrt{3}}{\sqrt{3}}\) + \(\dfrac{2+\sqrt{2}}{\sqrt{2}+1}\) - \(\dfrac{1}{2-\sqrt{3}}\)
11.
\(\dfrac{5+\sqrt{5}}{5-\sqrt{5}}+\dfrac{5-\sqrt{5}}{5+\sqrt{5}}\)
\(=\dfrac{\left(5+\sqrt{5}\right)^2}{\left(5-\sqrt{5}\right)\left(5+\sqrt{5}\right)}+\dfrac{\left(5-\sqrt{5}\right)^2}{\left(5-\sqrt{5}\right)\left(5+\sqrt{5}\right)}\)
\(=\dfrac{25+5+10\sqrt{5}}{20}+\dfrac{25+5-10\sqrt{5}}{20}\)
\(=3\)
12.
\(\dfrac{3+2\sqrt{3}}{\sqrt{3}}+\dfrac{2+\sqrt{2}}{\sqrt{2}+1}-\dfrac{1}{2-\sqrt{3}}\)
\(=\dfrac{\sqrt{3}\left(\sqrt{3}+2\right)}{\sqrt{3}}+\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}-\dfrac{2+\sqrt{3}}{4-3}\)
\(=\sqrt{3}+2+\sqrt{2}-2-\sqrt{3}\)
\(=\sqrt{2}\)
11: Ta có: \(\dfrac{5+\sqrt{5}}{5-\sqrt{5}}+\dfrac{5-\sqrt{5}}{5+\sqrt{5}}\)
\(=\dfrac{30+10\sqrt{5}+30-10\sqrt{5}}{20}\)
=3
12: Ta có: \(\dfrac{3+2\sqrt{3}}{\sqrt{3}}+\dfrac{2+\sqrt{2}}{\sqrt{2}+1}-\dfrac{1}{2-\sqrt{3}}\)
\(=2+\sqrt{3}+\sqrt{2}-2-\sqrt{3}\)
\(=\sqrt{2}\)
1) \(\dfrac{2}{\sqrt{5}-2}+\dfrac{-2}{\sqrt{5}+2}\)
2) \(\dfrac{4}{1-\sqrt{3}}+\dfrac{\sqrt{3}-1}{\sqrt{3}+1}\)
3) \(\dfrac{\sqrt{2}-1}{\sqrt{2}+1}-\dfrac{3-\sqrt{2}}{3+\sqrt{2}}\)
4) \(\dfrac{6}{1-\sqrt{3}}-\dfrac{3\sqrt{3}-3}{\sqrt{3}+1}\)
5) \(\dfrac{\sqrt{5}+\sqrt{6}}{\sqrt{5}-\sqrt{6}}+\dfrac{\sqrt{6}-\sqrt{5}}{\sqrt{6}+\sqrt{5}}\)
4: Ta có: \(\dfrac{6}{1-\sqrt{3}}-\dfrac{3\sqrt{3}+3}{\sqrt{3}+1}\)
\(=-3-3\sqrt{3}-3\)
\(=-6-3\sqrt{3}\)
Tính:
1) \(\dfrac{3}{1-\sqrt{2}}+\dfrac{\sqrt{2}-1}{\sqrt{2}+1}\)
2) \(\dfrac{\sqrt{5}-1}{\sqrt{5}+1}+\dfrac{6}{1-\sqrt{5}}\)
3) \(\dfrac{\sqrt{2}+\sqrt{3}}{2-\sqrt{6}}+\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{6}+2}\)
4) \(\dfrac{3+\sqrt{3}}{\sqrt{3}}+\dfrac{\sqrt{6}-\sqrt{3}}{1-\sqrt{2}}\)
5) \(\dfrac{2-\sqrt{2}}{1-\sqrt{2}}+\dfrac{\sqrt{2}-\sqrt{6}}{\sqrt{3}-1}\)
5: Ta có: \(\dfrac{2-\sqrt{2}}{1-\sqrt{2}}+\dfrac{\sqrt{2}-\sqrt{6}}{\sqrt{3}-1}\)
\(=-\sqrt{2}-\sqrt{2}\)
\(=-2\sqrt{2}\)
Tính:
1) \(\dfrac{1}{3-2\sqrt{2}}+\dfrac{1}{2+\sqrt{5}}\)
2) \(\dfrac{1}{3-2\sqrt{2}}-\dfrac{1}{3+2\sqrt{2}}\)
3) \(\dfrac{1}{\sqrt{5}-\sqrt{7}}+\dfrac{2}{1-\sqrt{7}}\)
4) \(\dfrac{1}{5+2\sqrt{6}}-\dfrac{1}{5-2\sqrt{6}}\)
5) \(-\dfrac{1}{\sqrt{2}-\sqrt{3}}\)\(-\dfrac{3}{\sqrt{18}+2\sqrt{3}}\)
1: ta có: \(\dfrac{1}{3-2\sqrt{2}}+\dfrac{1}{\sqrt{5}+2}\)
\(=3+2\sqrt{2}+\sqrt{5}-2\)
\(=2\sqrt{2}+\sqrt{5}+1\)
2: Ta có: \(\dfrac{1}{3-2\sqrt{2}}-\dfrac{1}{3+2\sqrt{2}}\)
\(=3+2\sqrt{2}-3+2\sqrt{2}\)
\(=4\sqrt{2}\)
\(\)1) \(\dfrac{5+2\sqrt{5}}{\sqrt{5}+\sqrt{2}}\)
2) \(\dfrac{2\sqrt{6}-\sqrt{10}}{4\sqrt{3}-2\sqrt{5}}\)
3) \(\dfrac{1}{2\sqrt{2}-3\sqrt{3}}\)
4) \(\sqrt{\dfrac{3-\sqrt{5}}{3+\sqrt{5}}}\)