\(\frac{1}{\sqrt{3}+\sqrt{2}+1}\)
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TY
0
3 tháng 8 2020
a/ \(\frac{1}{2+\sqrt{3}}-\frac{1}{2-\sqrt{3}}+5\sqrt{3}\)
\(=\frac{2-\sqrt{3}}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}-\frac{2+\sqrt{3}}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}+5\sqrt{3}\)
\(=\frac{2-\sqrt{3}}{4-3}-\frac{2+\sqrt{3}}{4-3}+5\sqrt{3}\)
\(=2-\sqrt{3}-2-\sqrt{3}+5\sqrt{3}\)
\(=3\sqrt{3}\)
Vậy..
3 tháng 8 2020
b/ \(\frac{1}{\sqrt{5}+2}-\sqrt{9+4\sqrt{5}}\)
\(=\frac{1}{\sqrt{5}+2}-\sqrt{\left(\sqrt{5}+2\right)^2}\)
\(=\frac{1}{\sqrt{5}+2}-\left|\sqrt{5}+2\right|\)
\(=\frac{\sqrt{5}-2}{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}-\sqrt{5}-2\)
\(=\sqrt{5}-2-\sqrt{5}-2\)
\(=-4\)
Vậy..
NB
1
17 tháng 8 2020
Ta có : \(\frac{1-\sqrt{2}}{2\sqrt{3}-3\sqrt{2}}=\frac{\left(1-\sqrt{2}\right)\left(2\sqrt{3}+3\sqrt{2}\right)}{\left(2\sqrt{3}-3\sqrt{2}\right)\left(2\sqrt{3}+3\sqrt{2}\right)}=\frac{2\sqrt{3}+3\sqrt{2}-2\sqrt{6}-6}{12-18}\)
\(=\frac{\sqrt{12}+\sqrt{18}-\sqrt{24}-\sqrt{36}}{-6}\)\(=\frac{-\sqrt{12}-\sqrt{18}+\sqrt{24}+\sqrt{36}}{6}\)
NH
1
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
14 tháng 7 2017
\(=\frac{\sqrt{3}+\sqrt{2}-1}{\left(\sqrt{3}+\sqrt{2}+1\right)\left(\sqrt{3}+\sqrt{2}-1\right)}=\frac{\sqrt{3}+\sqrt{2}-1}{\left(\sqrt{3}+\sqrt{2}\right)^2-1}=\)
\(=\frac{\sqrt{3}+\sqrt{2}-1}{3+2\sqrt{3}\sqrt{2}+2-1}=\frac{\sqrt{3}+\sqrt{2}-1}{2\sqrt{3}\sqrt{2}+4}=\frac{\left(\sqrt{3}+\sqrt{2}-1\right)\left(2\sqrt{3}\sqrt{2}-4\right)}{4.3.2-16}\)
\(=\frac{2.3.\sqrt{2}-4.\sqrt{3}+2.2.\sqrt{3}-4.\sqrt{2}-2.\sqrt{3}.\sqrt{2}+4}{8}=\)
\(=\frac{2.\sqrt{2}-2.\sqrt{3}.\sqrt{2}+4}{8}=\frac{\sqrt{2}-\sqrt{6}+2}{4}\)
NV
Nguyễn Việt Lâm
Giáo viên
16 tháng 9 2020
\(=\frac{\sqrt[3]{9}-\sqrt[3]{6}+\sqrt[3]{4}}{\left(\sqrt[3]{3}+\sqrt[3]{2}\right)\left(\sqrt[3]{9}-\sqrt[3]{6}+\sqrt[3]{4}\right)}=\frac{\sqrt[3]{9}-\sqrt[3]{6}+\sqrt[3]{4}}{5}\)
DD
2
4 tháng 8 2018
\(\frac{1-\sqrt{2}}{2\sqrt{3}-3\sqrt{2}}\)
\(=\frac{\left(1-\sqrt{2}\right)\left(2\sqrt{3}+3\sqrt{2}\right)}{\left(2\sqrt{3}\right)^2-\left(3\sqrt{2}\right)^2}\)
\(=\frac{2\sqrt{3}+3\sqrt{2}-2\sqrt{6}-6}{12-18}\)
\(=\frac{\sqrt{6}\left(\sqrt{2}+\sqrt{3}-2-\sqrt{6}\right)}{-6}\)
\(=2+\sqrt{6}-\sqrt{3}-\sqrt{2}\)
25 tháng 7 2019
\(\frac{1-\sqrt{2}}{2\sqrt{3}-3\sqrt{2}}\)
\(=\frac{\left(1-\sqrt{2}\right)\left(2\sqrt{3}+3\sqrt{2}\right)}{\left(2\sqrt{3}\right)^2-\left(3\sqrt{2}\right)^2}\)
\(=\frac{2\sqrt{3}+3\sqrt{2}-2\sqrt{6}-6}{12-18}\)
\(=\frac{\sqrt{6}\left(\sqrt{2}+\sqrt{3}-2-\sqrt{6}\right)}{-6}\)
\(=2+\sqrt{6}-\sqrt{3}-\sqrt{2}\)
Hông chắc !!!
AT
0
NQ
5 tháng 8 2019
a) \(\frac{3}{\sqrt{5}}=\frac{3\sqrt{5}}{\sqrt{5}.\sqrt{5}}=\frac{3\sqrt{5}}{5}\)
\(\frac{2\sqrt{3}}{\sqrt{2}}=\frac{2\sqrt{3}.\sqrt{2}}{\sqrt{2}.\sqrt{2}}=\frac{2\sqrt{6}}{2}=\sqrt{6}\)
\(\frac{a}{\sqrt{b}}=\frac{a\sqrt{b}}{\sqrt{b}.\sqrt{b}}=\frac{a\sqrt{b}}{b}\)
\(\frac{x+1}{\sqrt{x^2-1}}=\frac{\left(x+1\right)\left(\sqrt{x^2-1}\right)}{\left(\sqrt{x^2-1}\right)\left(\sqrt{x^2-1}\right)}\) = \(\frac{\left(\sqrt{x^2-1}\right)\left(x+1\right)}{x^2-1}\)
\(\frac{1}{\sqrt{3}+\sqrt{2}+1}=\frac{\sqrt{3}-\sqrt{2}-1}{\left(\sqrt{3}+\sqrt{2}+1\right)\left(\sqrt{3}-\sqrt{2}-1\right)}.\)
\(\frac{\sqrt{3}-\sqrt{2}-1}{\sqrt{3}^2-\left(\sqrt{2}+1\right)^2}=\frac{\sqrt{3}-\sqrt{2}-1}{3-\left(2+2\sqrt{2}+1\right)}\)
\(=\frac{\sqrt{3}-\sqrt{2}-1}{3-2-2\sqrt{2}-1}=\frac{\sqrt{3}-\sqrt{2}-1}{-2\sqrt{2}}\)
\(=\frac{-\sqrt{2}\left(\sqrt{3}-\sqrt{2}-1\right)}{-\sqrt{2}\left(-2\sqrt{2}\right)}=\frac{2-\sqrt{6}+\sqrt{2}}{4}\)