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31 tháng 7 2023

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

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

\(=\dfrac{30-3\sqrt{5}}{\left(2\sqrt{5}\right)^2-1}\)

\(=\dfrac{30-\sqrt{5}}{20-1}\)

\(=\dfrac{30-\sqrt{5}}{19}\)

AH
Akai Haruma
Giáo viên
19 tháng 7 2021

Bài 1:
a.

\(\frac{1}{2\sqrt{2}-3\sqrt{3}}=\frac{2\sqrt{2}+3\sqrt{3}}{(2\sqrt{2}-3\sqrt{3})(2\sqrt{2}+3\sqrt{3})}=\frac{2\sqrt{2}+3\sqrt{3}}{(2\sqrt{2})^2-(3\sqrt{3})^2}=\frac{2\sqrt{2}+3\sqrt{3}}{-19}\)

b.

\(=\sqrt{\frac{(3-\sqrt{5})^2}{(3-\sqrt{5})(3+\sqrt{5})}}=\sqrt{\frac{(3-\sqrt{5})^2}{3^2-5}}=\sqrt{\frac{(3-\sqrt{5})^2}{4}}=\sqrt{(\frac{3-\sqrt{5}}{2})^2}=|\frac{3-\sqrt{5}}{2}|=\frac{3-\sqrt{5}}{2}\)

 

AH
Akai Haruma
Giáo viên
19 tháng 7 2021

Bài 2.

a. 

\(=\frac{\sqrt{8}(\sqrt{5}+\sqrt{3})}{(\sqrt{5}-\sqrt{3})(\sqrt{5}+\sqrt{3})}=\frac{2\sqrt{2}(\sqrt{5}+\sqrt{3})}{5-3}=\sqrt{2}(\sqrt{5}+\sqrt{3})=\sqrt{10}+\sqrt{6}\)

b.

\(=\sqrt{\frac{(2-\sqrt{3})^2}{(2-\sqrt{3})(2+\sqrt{3})}}=\sqrt{\frac{(2-\sqrt{3})^2}{2^2-3}}=\sqrt{(2-\sqrt{3})^2}=|2-\sqrt{3}|=2-\sqrt{3}\)

17 tháng 12 2020

1) Ta có: \(3\sqrt{12}+\dfrac{1}{2}\sqrt{48}-\sqrt{27}\)

\(=3\cdot2\sqrt{3}+\dfrac{1}{2}\cdot4\sqrt{3}-3\sqrt{3}\)

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

\(=5\sqrt{3}\)

2) Ta có: \(\dfrac{2}{\sqrt{3}-5}\)

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

\(=\dfrac{2\left(\sqrt{3}+5\right)}{3-25}\)

\(=\dfrac{-2\left(\sqrt{3}+5\right)}{22}\)

\(=\dfrac{-\sqrt{3}-5}{11}\)

3) Ta có: \(\sqrt{\dfrac{2}{5}}\)

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

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

\(=\dfrac{\sqrt{10}}{5}\)

NV
17 tháng 12 2020

Nếu em thấy các câu hỏi do lag mà bị gửi đúp (tức là rất nhiều câu hỏi giống nhau xuất hiện cùng 1 chỗ) thì xóa giúp mình nhé cho đỡ vướng. Nhưng nhớ để lại 1 câu. Cảm ơn em.

16 tháng 7 2021

\(\dfrac{\sqrt{5}-\sqrt{3}}{\sqrt{2}}=\dfrac{\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right)}{\sqrt{2}.\sqrt{2}}=\dfrac{\sqrt{10}-\sqrt{6}}{2}\)

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

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

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

\(\dfrac{\sqrt{5}-\sqrt{3}}{\sqrt{2}}=\dfrac{\sqrt{10}-\sqrt{6}}{2}\)

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

Ta có: \(\dfrac{2}{5-\sqrt{2}-\sqrt{3}}\)

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

\(=\dfrac{2\left(5+\sqrt{2}+\sqrt{3}\right)}{25-5-2\sqrt{6}}\)

\(=\dfrac{2\left(5+\sqrt{2}+\sqrt{3}\right)}{20-2\sqrt{6}}\)

\(=\dfrac{5+\sqrt{2}+\sqrt{3}}{10-\sqrt{6}}\)

\(=\dfrac{\left(5+\sqrt{2}+\sqrt{3}\right)\left(10+\sqrt{6}\right)}{94}\)

\(=\dfrac{50+5\sqrt{6}+10\sqrt{2}+2\sqrt{3}+10\sqrt{3}+3\sqrt{2}}{94}\)

\(=\dfrac{50+5\sqrt{6}+13\sqrt{2}+12\sqrt{3}}{94}\)

13 tháng 7 2023

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

\(=\dfrac{\sqrt{10}+\sqrt{6}}{\left(\sqrt{5}\right)^2-\left(\sqrt{3}\right)^2}\)

\(=\dfrac{\sqrt{10}+\sqrt{6}}{5-3}\)

\(=\dfrac{\sqrt{10}+\sqrt{6}}{2}\)

18 tháng 12 2020

1) \(5\sqrt{8}-\dfrac{7}{2}\sqrt{72}+6\sqrt{\dfrac{1}{2}}\\ =5.\sqrt{4^2.\dfrac{1}{2}}-\dfrac{7}{2}.\sqrt{12^2.\dfrac{1}{2}}+6.\sqrt{\dfrac{1}{2}}=\left(5.4+\dfrac{7}{2}.12+6\right)\sqrt{\dfrac{1}{2}}\\ =68\sqrt{\dfrac{1}{2}}\)

2) \(\dfrac{6}{\sqrt{5}-1}=\dfrac{6.\left(\sqrt{5}+1\right)}{\left(\sqrt{5}-1\right).\left(\sqrt{5}+1\right)}=\dfrac{6\left(\sqrt{5}+1\right)}{5-1}\\ =\dfrac{6\left(\sqrt{5}+1\right)}{4}=\dfrac{3.\left(\sqrt{5+1}\right)}{2}\)

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

13 tháng 10 2023

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

\(\dfrac{2\sqrt{2}+2}{5\sqrt{2}}=\dfrac{2\sqrt{2}+\sqrt{2}\cdot\sqrt{2}}{5\sqrt{2}}=\dfrac{\sqrt{2}\cdot\left(2+\sqrt{2}\right)}{5\sqrt{2}}=\dfrac{2+\sqrt{2}}{5}\)

\(\dfrac{y+b\sqrt{y}}{b\sqrt{y}}=\dfrac{\sqrt{y}\cdot\sqrt{y}+b\sqrt{y}}{b\sqrt{y}}=\dfrac{\sqrt{y}\left(\sqrt{y}+b\right)}{b\sqrt{y}}=\dfrac{\sqrt{y}+b}{y}\)

13 tháng 10 2023

\(\dfrac{5}{2\sqrt{5}}=\dfrac{5\sqrt{5}}{2.5}=\dfrac{\sqrt{5}}{2}\)

\(\dfrac{2\sqrt{2}+2}{5\sqrt{2}}=\dfrac{4+2\sqrt{2}}{5.2}=\dfrac{2+\sqrt{2}}{5}\)

\(\dfrac{y+b\sqrt{y}}{b\sqrt{y}}=\dfrac{y\sqrt{y}+by}{by}=\dfrac{\sqrt{y}+b}{b}\left(y>0;b\ne0\right)\)

17 tháng 2 2019

Theo mình thì làm như vầy nha:

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

=\(\dfrac{1+\sqrt{5}}{\left(\sqrt{15}-\sqrt{5}\right)+\left(\sqrt{3}-1\right)}\)

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

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

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

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

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

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