Thực hiện phép tínhc/ 2/(sqrt(3) - 1) - (3 + sqrt(3))/(sqrt(3) + 1)
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1 tháng 7 2021
c) Ta có: \(\left(\dfrac{6-2\sqrt{2}}{3-\sqrt{2}}-\dfrac{5}{\sqrt{5}}\right)\cdot\dfrac{1}{2-\sqrt{5}}\)
\(=\left(\dfrac{2\left(3-\sqrt{2}\right)}{3-\sqrt{2}}-\sqrt{5}\cdot\dfrac{\sqrt{5}}{\sqrt{5}}\right)\cdot\dfrac{1}{2-\sqrt{5}}\)
\(=\left(2-\sqrt{5}\right)\cdot\dfrac{1}{2-\sqrt{5}}\)
=1
d) Ta có: \(\sqrt{\left(2-\sqrt{5}\right)^2}-\sqrt{5}\)
\(=\sqrt{5}-2-\sqrt{5}\)
=-2
AH
1
8 tháng 11 2021
\(=\dfrac{2\sqrt{3}+2+2\sqrt{3}-2}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}-\dfrac{\sqrt{3}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}\\ =\dfrac{4\sqrt{3}}{2}-\sqrt{3}=2\sqrt{3}-\sqrt{3}=\sqrt{3}\)
1 tháng 8 2021
Q=\(\dfrac{\left(3+2\sqrt{3}\right)\left(\sqrt{2}+1\right)+\sqrt{3}\left(2+\sqrt{2}\right)}{\sqrt{3}\left(\sqrt{2}+1\right)}-\left(\sqrt{2}+\sqrt{3}\right)\)
Q=\(\dfrac{3+4\sqrt{3}+3\sqrt{6}+3\sqrt{2}}{\sqrt{3}\left(\sqrt{2}+1\right)}\)-\(\left(\sqrt{2}+\sqrt{3}\right)\)
Q=\(\dfrac{3+4\sqrt{3}+3\sqrt{6}+3\sqrt{2}-2\sqrt{3}-3\sqrt{2}-\sqrt{6}-3}{\sqrt{3}\left(\sqrt{2}+1\right)}\)
Q=\(\dfrac{2\sqrt{3}-2\sqrt{6}}{\sqrt{3}\left(\sqrt{2}+1\right)}\)=\(\dfrac{\sqrt{4}-\sqrt{8}}{\sqrt{2}+1}\)
H9
HT.Phong (9A5)
CTVHS
8 tháng 9 2023
\(B=\left(\dfrac{4}{1-\sqrt{5}}+\dfrac{1}{2+\sqrt{5}}-\dfrac{4}{3-\sqrt{5}}\right)\left(\sqrt{5}-6\right)\)
\(B=\left[\dfrac{4\left(1+\sqrt{5}\right)}{\left(1-\sqrt{5}\right)\left(1+\sqrt{5}\right)}+\dfrac{2-\sqrt{5}}{\left(2+\sqrt{5}\right)\left(2-\sqrt{5}\right)}-\dfrac{4\left(3+\sqrt{5}\right)}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}\right]\left(\sqrt{5}-6\right)\)
\(B=\left[\dfrac{4\left(1+\sqrt{5}\right)}{1-5}+\dfrac{2-\sqrt{5}}{4-5}-\dfrac{4\left(3+\sqrt{5}\right)}{9-5}\right]\left(\sqrt{5}-6\right)\)
\(B=\left[-\dfrac{4\left(1+\sqrt{5}\right)}{4}-\dfrac{2-\sqrt{5}}{1}-\dfrac{4\left(3+\sqrt{5}\right)}{4}\right]\left(\sqrt{5}-6\right)\)
\(B=\left(-1-\sqrt{5}-2+\sqrt{5}-3-\sqrt{5}\right)\left(\sqrt{5}-6\right)\)
\(B=\left(-\sqrt{5}-6\right)\left(\sqrt{5}-6\right)\)
\(B=-\left(\sqrt{5}+6\right)\left(\sqrt{5}-6\right)\)
\(B=-\left(5-36\right)\)
\(B=-\left(-31\right)\)
\(B=31\)
_____________________________
\(\sqrt{48}-\dfrac{\sqrt{21}-\sqrt{15}}{\sqrt{7}-\sqrt{5}}+\dfrac{2}{\sqrt{3}+1}\)
\(=4\sqrt{3}-\dfrac{\sqrt{3}\left(\sqrt{7}-\sqrt{5}\right)}{\sqrt{7}-\sqrt{5}}+\dfrac{2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\)
\(=4\sqrt{3}-\sqrt{3}-\dfrac{2\left(\sqrt{3}-1\right)}{2}\)
\(=3\sqrt{3}-\sqrt{3}+1\)
\(=2\sqrt{3}+1\)
TA
3 tháng 8 2021
\(\dfrac{3+2\sqrt{3}}{\sqrt{3}}+\dfrac{2+\sqrt{2}}{\sqrt{2}+1}-\left(2+\sqrt{3}\right)\\ =\dfrac{\sqrt{3}\left(\sqrt{3}+2\right)}{\sqrt{3}}+\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}-2-\sqrt{3}\\ =\sqrt{3}+2+\sqrt{2}-2-\sqrt{3}\\ =\sqrt{2}\)
GH
21 tháng 7 2023
\(\sqrt{\left(\sqrt{3}-1\right)^2}+\dfrac{6}{\sqrt{3}}-15\sqrt{\dfrac{1}{3}}+1\\ =\left|\sqrt{3}-1\right|+\dfrac{6}{\sqrt{3}}-15\dfrac{\sqrt{1}}{\sqrt{3}}+1\\ =\sqrt{3}-1+\dfrac{6}{\sqrt{3}}-\dfrac{15}{\sqrt{3}}+1\\ =\sqrt{3}-\dfrac{9}{\sqrt{3}}\\ =\dfrac{3}{\sqrt{3}}-\dfrac{9}{\sqrt{3}}\\ =-\dfrac{6}{\sqrt{3}}\\ =-\dfrac{\sqrt{36}}{\sqrt{3}}\\ =-\dfrac{\sqrt{3}.\sqrt{3}.\sqrt{4}}{\sqrt{3}}\\ =-\sqrt{3}.\sqrt{4}\\ =-2\sqrt{3}\)
22 tháng 12 2023
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: \(\sqrt[3]{5\sqrt{2}+7}-\sqrt[3]{5\sqrt{2}+7}=0\)
\(\dfrac{2}{\sqrt{3}-1}-\dfrac{3\sqrt{3}}{\sqrt{3}+1}\\ =\dfrac{2\left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}-\dfrac{3\sqrt{3}\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}\\ =\dfrac{2\left(\sqrt{3}+1\right)-3\sqrt{3}\left(\sqrt{3}-1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\\ =\dfrac{2\sqrt{3}+2-9+3\sqrt{3}}{2}\\ =\dfrac{5\sqrt{3}-7}{2}\)