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24 tháng 7 2019

\(\frac{2}{\sqrt{7}-5}-\frac{2}{\sqrt{7}+5}=\frac{2\sqrt{7}+10}{\left(\sqrt{7}-5\right)\left(\sqrt{7}+5\right)}-\frac{2\sqrt{7}-10}{\left(\sqrt{7}-5\right)\left(\sqrt{7}+5\right)}=\frac{20}{7-25}=\frac{20}{-18}=\frac{10}{-9}\)

\(\frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}-\sqrt{5}}+\frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}+\sqrt{5}}=\frac{\left(\sqrt{7}+\sqrt{5}\right)^2+\left(\sqrt{7}-\sqrt{5}\right)^2}{\left(\sqrt{7}-\sqrt{5}\right)\left(\sqrt{7}+\sqrt{5}\right)}=\frac{12+2\sqrt{35}+12-2\sqrt{35}}{2}=\frac{24}{2}=12\)

\(\left(\frac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\frac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right)\frac{1}{\sqrt{7}-\sqrt{5}}=\left(\frac{-\sqrt{7}\left(1-\sqrt{2}\right)}{1-\sqrt{2}}+\frac{-\sqrt{5}\left(1-\sqrt{3}\right)}{1-\sqrt{3}}\right)\frac{1}{\sqrt{7}-\sqrt{5}}=\frac{\left(\sqrt{7}+\sqrt{5}\right)}{\sqrt{5}-\sqrt{7}}=\frac{\left(\sqrt{7}+\sqrt{5}\right)^2}{\left(\sqrt{5}-\sqrt{7}\right)\left(\sqrt{5}+\sqrt{7}\right)}=\frac{12+2\sqrt{35}}{-2}=-6-\sqrt{35}\)

28 tháng 3 2020

\(\frac{3}{\sqrt{5}-2}+\frac{2}{\sqrt{5}+3}-\frac{1}{\sqrt{5}+4}=\frac{3\left(\sqrt{5}+2\right)}{5-4}+\frac{2\left(\sqrt{5}-3\right)}{5-9}-\frac{\sqrt{5}-4}{5-16}\)

\(=3\sqrt{5}+6+\frac{2\sqrt{5}-6}{-4}+\frac{4-\sqrt{5}}{-11}=\frac{66\sqrt{5}+132}{22}+\frac{33-11\sqrt{5}}{22}+\frac{2\sqrt{5}-8}{22}\)

\(=\frac{66\sqrt{5}-11\sqrt{5}+2\sqrt{5}+132+33-8}{22}=\frac{57\sqrt{5}+157}{22}\)

3 tháng 7 2019

\(\left(\frac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\frac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right):\frac{1}{\sqrt{7}-\sqrt{5}}=\left(\frac{\sqrt{7}\left(\sqrt{2}-1\right)}{1-\sqrt{2}}+\frac{\sqrt{5}\left(\sqrt{3}-1\right)}{1-\sqrt{3}}\right):\frac{1}{\sqrt{7}-\sqrt{5}}=\left(\frac{-\sqrt{7}\left(1-\sqrt{2}\right)}{1-\sqrt{2}}+\frac{-\sqrt{5}\left(1-\sqrt{3}\right)}{1-\sqrt{3}}\right):\frac{1}{\sqrt{7}-\sqrt{5}}=\left(-\sqrt{7}-\sqrt{5}\right):\frac{1}{\sqrt{7}-\sqrt{5}}=\frac{\sqrt{5}-\sqrt{7}}{\sqrt{7}+\sqrt{5}}=\frac{\left(\sqrt{5}-\sqrt{7}\right)\left(\sqrt{5}+\sqrt{7}\right)}{\left(\sqrt{7}+\sqrt{5}\right)^2}=\frac{2}{12+2\sqrt{35}}\)

3 tháng 7 2019

\(\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}+\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}-\frac{\sqrt{5}+1}{\sqrt{5}-1}=\frac{\left(\sqrt{5}-\sqrt{3}\right)^2}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}+\frac{\left(\sqrt{5}+\sqrt{3}\right)^2}{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+3\right)}-\frac{\sqrt{5}+1}{\sqrt{5}-1}=\frac{8-2\sqrt{15}}{2}+\frac{8+2\sqrt{15}}{2}-\frac{\left(\sqrt{5}+1\right)^2}{4}=8-\frac{6+2\sqrt{5}}{4}=\frac{26-2\sqrt{5}}{4}\)

31 tháng 8 2019

\(\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}+\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}-\frac{\sqrt{5}+1}{\sqrt{5}-1}=\frac{\left(\sqrt{5}-\sqrt{3}\right)^2}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}+\frac{\left(\sqrt{5}+\sqrt{3}\right)^2}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}-\frac{\left(\sqrt{5}+1\right)^2}{\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)}=\frac{8-2\sqrt{15}+8+2\sqrt{15}}{2}-\frac{6+2\sqrt{5}}{4}=\frac{32-6-2\sqrt{5}}{4}=\frac{26-2\sqrt{5}}{4}=\frac{14-\sqrt{5}}{2}\) \(\left(\frac{9-2\sqrt{14}}{\sqrt{7}-\sqrt{2}}\right)^2-\left(\frac{9+2\sqrt{14}}{\sqrt{7}-\sqrt{2}}\right)^2=\left(\frac{9-2\sqrt{14}-9-2\sqrt{14}}{\sqrt{7}-\sqrt{2}}\right)\left(\frac{9-2\sqrt{14}+9+2\sqrt{14}}{\sqrt{7}-\sqrt{2}}\right)=\frac{-72\sqrt{14}}{\sqrt{7}-\sqrt{2}}\)

a) Ta có: \(A=\sqrt{8-2\sqrt{15}}\cdot\left(\sqrt{3}+\sqrt{5}\right)-\left(\sqrt{45}-\sqrt{20}\right)\)

\(=\sqrt{5-2\cdot\sqrt{5}\cdot\sqrt{3}+3}\cdot\left(\sqrt{5}+\sqrt{3}\right)-\sqrt{5}\left(\sqrt{9}-\sqrt{4}\right)\)

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

\(=\left|\sqrt{5}-\sqrt{3}\right|\cdot\left(\sqrt{5}+\sqrt{3}\right)-\sqrt{5}\)

\(=\left(\sqrt{5}-\sqrt{3}\right)\cdot\left(\sqrt{5}+\sqrt{3}\right)-\sqrt{5}\)(Vì \(\sqrt{5}>\sqrt{3}\))

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

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

b) Ta có: \(B=\left(\frac{\sqrt{21}-\sqrt{3}}{\sqrt{7}-1}-\frac{\sqrt{15}-\sqrt{3}}{1-\sqrt{5}}\right)\left(\frac{1}{2}\sqrt{6}-\sqrt{\frac{3}{2}}+3\sqrt{\frac{2}{3}}\right)\)

\(=\left(\frac{\sqrt{3}\left(\sqrt{7}-1\right)}{\sqrt{7}-1}+\frac{\sqrt{3}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}\right)\left(\sqrt{\frac{3}{2}}-\sqrt{\frac{3}{2}}+\sqrt{6}\right)\)

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

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

c) Ta có: \(C=2\sqrt{3}+\sqrt{7-4\sqrt{3}}+\left(\sqrt{\frac{1}{3}}-\sqrt{\frac{4}{3}}+\sqrt{3}\right):\sqrt{3}\)

\(=2\sqrt{3}+\sqrt{4-2\cdot2\cdot\sqrt{3}+3}+\sqrt{\frac{1}{3}:3}-\sqrt{\frac{4}{3}:3}+\sqrt{3:3}\)

\(=2\sqrt{3}+\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\frac{1}{9}}-\sqrt{\frac{4}{9}}+\sqrt{1}\)

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

\(=2\sqrt{3}+2-\sqrt{3}+\frac{2}{3}\)(Vì \(2>\sqrt{3}\))

\(=\sqrt{3}+\frac{8}{3}\)

d) Ta có: \(D=\left(\frac{5+\sqrt{5}}{5-\sqrt{5}}+\frac{5-\sqrt{5}}{5+\sqrt{5}}\right):\frac{1}{\sqrt{7-4\sqrt{3}}}\)

\(=\left(\frac{\left(5+\sqrt{5}\right)^2+\left(5-\sqrt{5}\right)^2}{\left(5-\sqrt{5}\right)\left(5+\sqrt{5}\right)}\right)\cdot\sqrt{4-2\cdot2\cdot\sqrt{3}+3}\)

\(=\frac{25+10\sqrt{5}+5+25-10\sqrt{5}+5}{25-5}\cdot\sqrt{\left(2-\sqrt{3}\right)^2}\)

\(=\frac{60}{20}\cdot\left|2-\sqrt{3}\right|\)

\(=3\cdot\left(2-\sqrt{3}\right)\)(Vì \(2>\sqrt{3}\))

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

a) \(=\frac{7-4\sqrt{3}+7+4\sqrt{3}}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}=\frac{14}{49-48}=14\)

b) \(=\frac{15\left(\sqrt{6}-1\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}-\frac{5\sqrt{6}}{5}+\frac{4\sqrt{3}-12\sqrt{2}}{\sqrt{6}\left(\sqrt{3}-\sqrt{2}\right)}\)

19 tháng 10 2020

a) \(\frac{3}{2+\sqrt{3}}+\frac{13}{4-\sqrt{3}}+\frac{6}{\sqrt{3}}\)

\(=\frac{3\left(2-\sqrt{3}\right)}{2^2-3}+\frac{13\left(4+\sqrt{3}\right)}{4^2-3}+\frac{6\sqrt{3}}{3}\)

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

\(=3.2+4=6+4=10\)

b) \(=\left[\frac{\left(\sqrt{14}-\sqrt{7}\right)\left(\sqrt{2}+1\right)}{2-1}+\frac{\left(\sqrt{15}-\sqrt{5}\right)\left(\sqrt{3}+1\right)}{3-1}\right]:\frac{1}{\sqrt{7}-\sqrt{5}}\)

\(=\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)=2\) (nhân bung mấy cái trong ngoặc vuông ra, rút gọn)

c) Gợi ý: \(28-10\sqrt{3}=5^2-2.5.\sqrt{3}+\sqrt{3}=\left(5-\sqrt{3}\right)^2\)

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

e) Tự làm.

20 tháng 10 2020

Cái câu c đánh nhầm:

\(=5^2-2.5.\sqrt{3}+3=\left(5-\sqrt{3}\right)^2\) nha!