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17 tháng 10 2023

a) \(\dfrac{3}{2}\sqrt{6}+2\sqrt{\dfrac{2}{3}}-4\sqrt{\dfrac{3}{2}}-\dfrac{6}{\sqrt{6}}=\dfrac{3}{2}\sqrt{6}+\dfrac{2}{3}\sqrt{6}-2\sqrt{6}-\dfrac{6}{\sqrt{6}}\)

\(=\dfrac{1}{6\sqrt{6}}-\dfrac{6}{\sqrt{6}}=-\dfrac{35}{6\sqrt{6}}\)

b)\(\left(\sqrt{6}+\sqrt{5}\right)^2+\left(\sqrt{6}-\sqrt{5}\right)^2\)

\(=6+2\sqrt{30}+5+6-2\sqrt{30}+5=22\)

a: \(=\sqrt{5}+2+\sqrt{3}+1-\sqrt{5}-\sqrt{3}=3\)

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

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

\(=-\sqrt{3}\left(7+4\sqrt{3}\right)=-7\sqrt{3}-12\)

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

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

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

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

\(=-11\)

3) Ta có: \(\left(\dfrac{3}{\sqrt{5}-\sqrt{2}}+\dfrac{4}{\sqrt{6}+\sqrt{2}}\right)\left(\sqrt{3}-1\right)^2\)

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

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

\(=4\sqrt{6}-6\sqrt{2}+4\sqrt{5}-2\sqrt{15}\)

12 tháng 7 2021

còn câu 1

10 tháng 7 2017

\(A=\sqrt{8}-\sqrt{7}+5\sqrt{7}+2\sqrt{2}\\ =2\sqrt{2}-\sqrt{7}+5\sqrt{7}+2\sqrt{2}\\ =4\sqrt{2}+4\sqrt{7}\)

10 tháng 7 2017

\(B=\left(3+2\sqrt{6}+2\right)\left(25-20\sqrt{6}+24\right)\sqrt{3-2\sqrt{6}+2}\\ =\left(\sqrt{3}+\sqrt{2}\right)^2\left(5-2\sqrt{6}\right)^2\left(\sqrt{3}-\sqrt{2}\right)\\ =\left(\sqrt{3}+\sqrt{2}\right)\left(3-2\sqrt{6}+2\right)^2\\ =\left(\sqrt{3}-\sqrt{2}\right)^3\\ =9\sqrt{3}-11\sqrt{2}\)

16 tháng 10 2022

b: \(=\left(\sqrt{ab}+\dfrac{2\sqrt{ab}}{a}-\sqrt{\dfrac{a^2+1}{ab}}\right)\cdot\sqrt{ab}\)

\(=ab+\dfrac{2ab}{a}-\sqrt{a^2+1}=ab+2b-\sqrt{a^2+1}\)

c: \(=2\sqrt{6b}-6\sqrt{18}+10\sqrt{12}-\sqrt{48}\)

\(=2\sqrt{6b}-18\sqrt{2}+20\sqrt{3}-4\sqrt{3}\)

\(=2\sqrt{6n}-18\sqrt{2}+16\sqrt{3}\)

d: \(=\dfrac{\sqrt{3}\left(\sqrt{5}-\sqrt{2}\right)}{\sqrt{7}\left(\sqrt{5}-\sqrt{2}\right)}=\dfrac{\sqrt{21}}{7}\)

b: \(=\sqrt{5}-1-\sqrt{5}-1=-2\)

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

d: \(=\dfrac{\sqrt{6-2\sqrt{5}}-\sqrt{6+2\sqrt{5}}}{\sqrt{2}}\)

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

16 tháng 10 2018

ta có : \(A=\left(\dfrac{3}{2}\sqrt{6}+2\sqrt{\dfrac{2}{3}}-4\sqrt{\dfrac{3}{2}}\right)\left(3\sqrt{\dfrac{2}{3}}-\sqrt{12}-\sqrt{6}\right)\)

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

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

ta có : \(B=\dfrac{\sqrt{3+\sqrt{5}}}{\sqrt{2}}-\dfrac{\sqrt{5}-1}{2}=\dfrac{\sqrt{6+2\sqrt{5}}}{2}-\dfrac{\sqrt{5}-1}{2}\)

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

ta có : \(C=\dfrac{1}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\dfrac{1}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)

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

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

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

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

14 tháng 11 2018

Đề không khó, mỗi tội dài

14 tháng 11 2018

vậy thì bn làm hộ mik vs , mik cần gấp

19 tháng 6 2017

a) \(\dfrac{\left(5+2\sqrt{6}\right)\left(49-20\sqrt{6}\right)\sqrt{5-2\sqrt{6}}}{9\sqrt{3}-11\sqrt{2}}\)

\(=\dfrac{\left(245-100\sqrt{6}+98\sqrt{6}-240\right)\sqrt{\left(\sqrt{2}-\sqrt{3}\right)^2}}{9\sqrt{3}-11\sqrt{2}}\)

\(=\dfrac{\left(5-2\sqrt{6}\right)\left(\sqrt{3}-\sqrt{2}\right)}{9\sqrt{3}-11\sqrt{2}}\)

\(=\dfrac{5\sqrt{3}-5\sqrt{2}-2\sqrt{18}+2\sqrt{12}}{9\sqrt{3}-11\sqrt{2}}\)

\(=\dfrac{5\sqrt{3}-5\sqrt{2}-6\sqrt{2}+4\sqrt{3}}{9\sqrt{3}-11\sqrt{2}}\)

\(=\dfrac{9\sqrt{3}-11\sqrt{2}}{9\sqrt{3}-11\sqrt{2}}\)

\(=1\)

19 tháng 6 2017

b)

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

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

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

\(=\dfrac{\dfrac{\sqrt{2+\sqrt{3}}}{2}}{\dfrac{3\sqrt{3\left(2+\sqrt{3}\right)}-2\sqrt{18}+3\sqrt{2+\sqrt{3}}}{6\sqrt{3}}}\)

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

\(=\dfrac{3\sqrt{\left(2+\sqrt{3}\right)\cdot3}}{3\sqrt{6+3\sqrt{3}}-6\sqrt{2}+3\sqrt{2+\sqrt{3}}}\)

\(=\dfrac{3\sqrt{\left(2+\sqrt{3}\right)\cdot3}}{3\left(\sqrt{6+3\sqrt{3}}-2\sqrt{2}+\sqrt{2+\sqrt{3}}\right)}\)

\(=\dfrac{\sqrt{\left(2+\sqrt{3}\right)\cdot3}}{\sqrt{6+3\sqrt{3}}-2\sqrt{2}+\sqrt{2+\sqrt{3}}}\)

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

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

\(=\dfrac{\sqrt{\left(6+3\sqrt{3}\right)\cdot2}}{-2\sqrt{3}}\)

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

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

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

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

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

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

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