\(\frac{1}{\sqrt{1}+\sqrt{2}}\)+\(\frac{1}{\sqrt{2}+\...">
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8 tháng 11 2016

Ta có:

\(A=\frac{1}{\sqrt{2}+\sqrt{1}}+\frac{1}{\sqrt{3}+\sqrt{2}}+...+\frac{1}{\sqrt{25}+\sqrt{24}}\)

\(=\frac{\sqrt{2}-\sqrt{1}}{2-1}+\frac{\sqrt{3}-\sqrt{2}}{3-2}+...+\frac{\sqrt{25}-\sqrt{24}}{25-24}\)

\(=\frac{\sqrt{2}-\sqrt{1}}{1}+\frac{\sqrt{3}-\sqrt{2}}{1}+...+\frac{\sqrt{25}-\sqrt{24}}{1}\)

\(=5-1=4\)

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

\(\frac{1}{\sqrt{1}\sqrt{2}}+\frac{1}{\sqrt{2}\sqrt{3}}+...+\frac{1}{\sqrt{24}\sqrt{25}}\)

\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{24}}-\frac{1}{\sqrt{25}}\)

\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{25}}\)

30 tháng 10 2016

lớp 9 mà

30 tháng 10 2016

lớp 7 mak bn

29 tháng 10 2020

a) \(\left(\frac{2^2}{5}\right)+5\frac{1}{2}.\left(4,5-2,5\right)+\frac{2^3}{-4}\)

\(=\frac{4}{5}+\frac{11}{2}.2+\frac{-8}{4}\)

\(=\frac{4}{5}+11-2\)

\(=\frac{4}{5}+9\)

\(=\frac{49}{9}\)

b) \(\left(-2^3\right)+\frac{1}{2}:\frac{1}{8}-\sqrt{25}+\left|-64\right|\)

\(=-8+4-5+64\)

= 55

c) \(\frac{\sqrt{3^2+\sqrt{39}^2}}{\sqrt{91^2}-\sqrt{\left(-7\right)^2}}\)

\(=\frac{\sqrt{9+39}}{91-\sqrt{49}}\)

\(=\frac{\sqrt{48}}{91-7}\)

\(=\frac{4\sqrt{3}}{84}\)

\(=\frac{\sqrt{3}}{41}\)

d) Xem lại đề nhé em!

e) \(\sqrt{25}-3\sqrt{\frac{4}{9}}\)

\(=5-3.\frac{2}{3}\)

= 5 - 2

= 3

h) \(\left(-3^2\right).\frac{1}{3}-\sqrt{49}+\left(5^3\right):\sqrt{25}\)

\(=-9.\frac{1}{3}-7+125:5\)

\(=-3-7+25\)

= 15

21 tháng 7 2016

Câu a)
\(A=\sqrt{20+1}+\sqrt{40+2}+\sqrt{60+3}\)
\(=\sqrt{1\left(20+1\right)}+\sqrt{2\left(20+1\right)}+\sqrt{3\left(20+1\right)}\)
\(=\sqrt{20+1}\left(\sqrt{1}+\sqrt{2}+\sqrt{3}\right)\)

\(B=\sqrt{1}+\sqrt{2}+\sqrt{3}+\sqrt{20}+\sqrt{40}+\sqrt{60}\)
\(=1\left(\sqrt{1}+\sqrt{2}+\sqrt{3}\right)+\left(\sqrt{1}\cdot\sqrt{20}+\sqrt{2}\cdot\sqrt{20}+\sqrt{3}\cdot\sqrt{20}\right)\)
\(=\sqrt{1}\left(\sqrt{1}+\sqrt{2}+\sqrt{3}\right)+\sqrt{20}\left(\sqrt{1}+\sqrt{2}+\sqrt{3}\right)\)
\(=\left(\sqrt{20}+\sqrt{1}\right)\left(\sqrt{1}+\sqrt{2}+\sqrt{3}\right)\)

Ta thấy: \(\hept{\begin{cases}\left(\sqrt{20+1}\right)^2=20+1\\\left(\sqrt{20}+\sqrt{1}\right)^2=20+1+2\sqrt{20}\end{cases}}\)
\(\Rightarrow\left(\sqrt{20+1}\right)^2< \left(\sqrt{20}+\sqrt{1}\right)^2\Rightarrow\sqrt{20+1}< \sqrt{20}+\sqrt{1}\)
Vậy A < B.

21 tháng 7 2016

a) A<B

4 tháng 8 2019

a) \(\sqrt{\frac{4}{81}}:\sqrt{\frac{25}{81}}-1\frac{2}{5}\)

\(=\frac{2}{9}:\frac{5}{9}-\frac{7}{5}\)

\(=\frac{2}{5}-\frac{7}{5}\)

\(=-1.\)

b) \(\sqrt{36}.\sqrt{\frac{25}{16}}+\frac{1}{4}\)

\(=6.\frac{5}{4}+\frac{1}{4}\)

\(=\frac{15}{2}+\frac{1}{4}\)

\(=\frac{31}{4}.\)

c) \(1\frac{1}{2}+\frac{4}{7}:\left(-\frac{8}{9}\right)\)

\(=\frac{3}{2}+\frac{4}{7}:\left(-\frac{8}{9}\right)\)

\(=\frac{3}{2}+\left(-\frac{9}{14}\right)\)

\(=\frac{6}{7}.\)

d) \(1,17-0,4.\left(\frac{1}{2}\right)^2-\frac{1}{-5}\)

\(=\frac{117}{100}-\frac{2}{5}.\frac{1}{4}-\left(-\frac{1}{5}\right)\)

\(=\frac{117}{100}-\frac{1}{10}+\frac{1}{5}\)

\(=\frac{107}{100}+\frac{1}{5}\)

\(=\frac{127}{100}.\)

Chúc bạn học tốt!

4 tháng 8 2019

a, \(\frac{4}{81}:\sqrt{\frac{25}{81}-1\frac{2}{5}}\)

\(\Rightarrow\frac{4}{81}:\frac{5}{9}-\frac{7}{5}\)

\(\Rightarrow\frac{4}{81}.\frac{9}{5}-\frac{7}{5}\)

\(\Rightarrow\frac{4}{9}.\frac{1}{5}-\frac{7}{5}\)

\(\Rightarrow\frac{-59}{45}\)

b,\(\sqrt{36}.\sqrt{\frac{25}{16}+\frac{1}{4}}\)

\(\Rightarrow6.\frac{5}{4}+\frac{1}{4}\)

\(\Rightarrow\frac{15}{2}+\frac{1}{4}\)

\(\Rightarrow\frac{31}{4}\)

c,\(1\frac{1}{2}+\frac{4}{7}:\frac{-8}{9}\)

\(\Rightarrow\frac{3}{2}-\frac{4}{7}.\frac{-8}{9}\)

\(\Rightarrow\frac{3}{2}-\frac{9}{14}\)

\(\Rightarrow\frac{6}{7}\)

d, \(1,17-\left(0,4.\frac{1}{2}\right)^2-\frac{1}{5}\)

\(\Rightarrow\frac{117}{100}-\left(\frac{1}{5}\right)^2-\frac{1}{5}\)

\(\Rightarrow\frac{117}{100}-\frac{1}{25}-\frac{1}{5}\)

\(\Rightarrow\frac{93}{100}\)

2 tháng 12 2019

Ta có:

\(\frac{1}{\sqrt{1}}>\frac{1}{\sqrt{100}}\)

\(\frac{1}{\sqrt{2}}>\frac{1}{\sqrt{100}}\)

\(.............\)

\(\frac{1}{\sqrt{99}}>\frac{1}{\sqrt{100}}\)

Khi đó:

\(A=\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+.....+\frac{1}{\sqrt{100}}\)

\(>\frac{1}{\sqrt{100}}+\frac{1}{\sqrt{100}}+.......+\frac{1}{\sqrt{100}}\left(100sohang\right)\)

\(=10\)

2 tháng 12 2019

Có BĐT sau:

\(\sqrt{\left(n-1\right)\left(n+1\right)}< n\)

\(\Leftrightarrow\left(n-1\right)\left(n+1\right)< n^2\)

\(\Leftrightarrow n^2-1< n^2\)

\(\Leftrightarrow-1< 0\left(true!!\right)\)

Áp dụng vào ta có:

\(\sqrt{2019\cdot2021}< 2020\Rightarrowđpcm\)

23 tháng 12 2016

a là âm 2.671428571

5 tháng 4 2017

a, \(-\frac{187}{70}\)

b,\(\frac{27}{70}\)

c,\(\frac{53}{14}\)

d,\(\frac{27}{4}\)

e,1

f,\(\frac{23}{4}\)

g,-1

i,6

k,315

l,\(\frac{9}{2}\)