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DT
0
VV
0
6 tháng 6 2017
#Giải:
a)\(\sqrt{27}\)+\(\sqrt{75}\)-\(\sqrt{\dfrac{1}{3}}\)=8\(\sqrt{3}\)-\(\sqrt{\dfrac{1}{3}}\)=\(\dfrac{23\sqrt{3}}{3}\).
b)\(\sqrt{4+2\sqrt{3}}\)-\(\sqrt{4-2\sqrt{3}}\)=2.
c)\(\dfrac{3}{\sqrt{7}+\sqrt{2}}\)+\(\dfrac{2}{3+\sqrt{7}}\)+\(\dfrac{2-\sqrt{2}}{\sqrt{2}-1}\)=1,093+\(\dfrac{2-\sqrt{2}}{\sqrt{2}-1}\)=2,507.
HT
7 tháng 6 2017
a) = \(3\sqrt{3}+5\sqrt{3}-\dfrac{1}{\sqrt{3}}\)
= \(3\sqrt{3}+5\sqrt{3}-\dfrac{3}{\sqrt{3}}\)
= \(\dfrac{23\sqrt{3}}{3}\)
b) = \(\sqrt{\left(1+\sqrt{3}\right)^2}-\sqrt{\left(1-\sqrt{3}\right)^2}\)
= \(1+\sqrt{3}-\left(\sqrt{3}-1\right)\)
= \(1+\sqrt{3}-\sqrt{3}+1\)
= 2
c) = \(\dfrac{3\left(\sqrt{7}-\sqrt{2}\right)}{5}+\dfrac{2\left(3-\sqrt{7}\right)}{2}+\left(2-\sqrt{2}\right)\left(\sqrt{2}+1\right)\)
= \(3\sqrt{7}-3\sqrt{2}+3-\sqrt{7}+2\sqrt{2}+2-2-\sqrt{2}\)
= \(\dfrac{3\sqrt{7}-3\sqrt{2}}{5}+3-\sqrt{7}+\sqrt{2}\)
= \(\dfrac{3\sqrt{7}-3\sqrt{2}-5\sqrt{7}+5\sqrt{2}}{5}+3\)
= \(\dfrac{-2\sqrt{7}+2\sqrt{2}}{5}+3\)
\(\approx2,5\)
G
17 tháng 11 2014
Bạn viết sai phân số cuối cùng.
\(\frac{1}{2\sqrt{1}+1\sqrt{2}}=\frac{2\sqrt{1}-1\sqrt{2}}{\left(2\sqrt{1}+1\sqrt{2}\right)\left(2\sqrt{1}-1\sqrt{2}\right)}=\frac{2\sqrt{1}-1\sqrt{2}}{\left(2\sqrt{1}\right)^2-\left(1\sqrt{2}\right)^2}=\frac{2\sqrt{1}-1\sqrt{2}}{2^21-1^22}=\frac{2\sqrt{1}-1\sqrt{2}}{1.2}=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}\)
Tương tự:
\(\frac{1}{3\sqrt{2}+2\sqrt{3}}=\frac{3\sqrt{2}-2\sqrt{3}}{3^22-2^23}=\frac{3\sqrt{2}-2\sqrt{3}}{2.3}=\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}\)
....
\(\frac{1}{25\sqrt{24}+24\sqrt{25}}=\frac{25\sqrt{24}-24\sqrt{25}}{25^224-24^225}=\frac{25\sqrt{24}-24\sqrt{25}}{25.24}=\frac{1}{\sqrt{24}}-\frac{1}{\sqrt{25}}\)
Vậy \(P=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{25}}=\frac{1}{1}-\frac{1}{5}=\frac{4}{5}\)
SL
9
KN
20 tháng 11 2019
\(\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{24}+\sqrt{25}}\)
\(=\frac{\left(\sqrt{2}-\sqrt{1}\right)\left(\sqrt{2}+\sqrt{1}\right)}{\sqrt{1}+\sqrt{2}}+\frac{\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)}{\sqrt{2}+\sqrt{3}}\)
\(+...+\frac{\left(\sqrt{25}-\sqrt{24}\right)\left(\sqrt{25}+\sqrt{24}\right)}{\sqrt{24}+\sqrt{25}}\)
\(=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{25}-\sqrt{24}\)
\(=\sqrt{25}-1=5-1=4\)
30 tháng 10 2016
\(\frac{1}{\sqrt{1}\sqrt{2}}+\frac{1}{\sqrt{2}\sqrt{3}}+...+\frac{1}{\sqrt{24}\sqrt{25}}\)
\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{24}}-\frac{1}{\sqrt{25}}\)
\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{25}}\)
25 tháng 10 2022
a: \(=2\cdot\dfrac{5}{4}-3\cdot\dfrac{7}{6}+4\cdot\dfrac{9}{8}=\dfrac{5}{2}-\dfrac{7}{2}+\dfrac{9}{2}=\dfrac{7}{2}\)
b: \(=18-16\cdot\dfrac{1}{2}+\dfrac{1}{16}\cdot\dfrac{3}{4}\)
=10+3/64
=643/64
c: \(=\dfrac{2}{3}\cdot\dfrac{9}{4}-\dfrac{3}{4}\cdot\dfrac{8}{3}+\dfrac{7}{5}\cdot\dfrac{5}{14}=\dfrac{3}{2}-2+\dfrac{1}{2}=2-2=0\)