\(sosánh\frac{3}{2}và\frac{1212}{3232}\)
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NL
0
S
9 tháng 7 2019
\(1,\frac{1212}{1515}+\frac{1212}{3535}+\frac{1212}{6363}+\frac{1212}{9999}=\frac{12}{15}+\frac{12}{35}+\frac{12}{63}+\frac{12}{99}=6\left(\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}\right)=6\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right).Tacocongthuc:\frac{1}{n}-\frac{1}{n+k}=\frac{k}{n\left(n+k\right)}\Rightarrow\frac{1212}{1515}+\frac{1212}{3535}+\frac{1212}{6363}+\frac{1212}{9999}=6\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-.....-\frac{1}{11}\right)=6\left(\frac{1}{3}-\frac{1}{11}\right)=\frac{48}{33}=\frac{16}{11}\)
\(2,\left(x+1\right)+\left(x+2\right)+.....+\left(x+211\right)=211x+\left(1+2+....+211\right)=211x+\frac{212.211}{2}=211x+22366=23632\Leftrightarrow211x=23632-22366=1266\Leftrightarrow x=6\)
9 tháng 7 2019
a, \(14:\left(4\frac{2}{3}:1\frac{5}{9}\right)+14:\left(\frac{2}{3}+\frac{8}{9}\right)\)
=> \(14:\frac{28}{9}+14:\frac{14}{9}=>14.\frac{9}{28}+14.\frac{9}{14}\)
=> 14. ( \(\frac{9}{28}+\frac{9}{14}\) )
=> \(14.\frac{27}{28}=\frac{419}{28}\)
b, \(\frac{1212}{1515}+\frac{1212}{3535}+\frac{1212}{6363}+\frac{1212}{9999}\)
=> \(\frac{4}{5}+\frac{12}{35}+\frac{4}{21}+\frac{4}{33}\)
=> \(\frac{8}{7}+\frac{24}{77}=\frac{16}{11}\)
bài 2 :
( x + 1 ) + ( x + 2 ) + ... + ( x + 211 ) = 23632
=> ( x + x + x + ... + x ) + ( 1 + 2 + 3 + ... + 211 ) = 23632
=> 211x + 22366 = 23632
=> 211x = 23632 - 22366
=> 211x = 1266
=> x = 1266 : 211
x = 6
TT
3
6 tháng 5 2017
Ta có:
\(\frac{2017^{10}+1}{2017^{10}-1}=1+\frac{2}{2017^{10}-1}\)
Lại có:
\(\frac{2017^{10}-1}{2017^{10}-3}=1+\frac{2}{2017^{10}-3}\)
Vì \(1+\frac{2}{2017^{10}-1}< 1+\frac{2}{2017^{10}-3}\)
Nên \(\frac{2017^{10}+1}{2017^{10}-1}< \frac{2017^{10}-1}{2017^{10}-3}\)
Vậy \(\frac{2017^{10}+1}{2017^{10}-1}< \frac{2017^{10}-1}{2017^{10}-3}\)
LB
6 tháng 5 2017
Ta có
\(\frac{2017^{10}+1}{2017^{10}-1}=\frac{2017^{10}-1+2}{2017^{10}-1}=1+\frac{2}{2017^{10}-1}\)
\(\frac{2017^{10}-1}{2017^{10}-3}=\frac{2017^{10}-3+2}{2017^{10}-3}=1+\frac{2}{2017^{10}-3}\)
\(\Rightarrow1+\frac{2}{2017^{10}-1}< 1+\frac{2}{2017^{10}-1}\)
\(\Rightarrow\frac{2017^{10}+1}{2017^{10}-1}< \frac{2017^{10}-1}{2017^{10}-3}\)
LT
3
20 tháng 6 2018
= \(\frac{12}{15}\) +\(\frac{12}{35}\)+\(\frac{12}{63}\)+\(\frac{12}{99}\)
= 12 x (\(\frac{1}{15}\)+\(\frac{1}{35}\)+\(\frac{1}{63}\)+\(\frac{1}{99}\))
= 12 x ( \(\frac{1}{3x5}\)+\(\frac{1}{5x7}\)+\(\frac{1}{7x9}\)+\(\frac{1}{9x11}\))
= 12 x \(\frac{1}{2}\) x ( \(\frac{1}{3}\)-\(\frac{1}{5}\)+\(\frac{1}{5}\)-\(\frac{1}{7}\)+\(\frac{1}{7}\)-\(\frac{1}{9}\)+\(\frac{1}{9}\)-\(\frac{1}{11}\))
= 6 x ( \(\frac{1}{3}\) - \(\frac{1}{11}\))
= 6 x \(\frac{8}{33}\)
= \(\frac{48}{33}\)=\(\frac{16}{11}\)
Nhớ tk nha
8 tháng 8 2019
\(\frac{1212}{1515}+\frac{1212}{3535}+\frac{1212}{6363}+\frac{1212}{9999}\)
\(=\frac{12}{15}+\frac{12}{35}+\frac{12}{63}+\frac{12}{99}\)
\(=12\cdot\left(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\right)\)
\(=12\cdot\frac{4}{33}\)
\(=\frac{16}{11}\)
L
8 tháng 8 2019
\(\frac{1212}{1515}+\frac{1212}{3535}+\frac{1212}{6363}+\frac{1212}{9999}\)
\(=\frac{12}{15}+\frac{12}{35}+\frac{12}{63}+\frac{12}{99}\)
\(=12\left(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\right)\)
\(=12\left(\frac{1}{3\cdot5}+\frac{1}{3\cdot7}+\frac{1}{7\cdot9}+\frac{1}{9\cdot11}\right)\)
\(=12\cdot\frac{1}{2}\left(\frac{2}{3\cdot5}+\frac{2}{3\cdot7}+\frac{2}{7\cdot9}+\frac{2}{9\cdot11}\right)\)
\(=6\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)\)
\(=6\left(\frac{1}{3}-\frac{1}{11}\right)\)
\(=6\cdot\frac{8}{33}\)
\(=\frac{48}{33}\)
TT
6
\(\frac{3}{2}>1;\frac{1212}{3232}< 1\)\(\Rightarrow\frac{3}{2}>\frac{1212}{3232}\)
Ta có:
\(\frac{3}{2}>\frac{2}{2}=1.\)
Mà \(\frac{1212}{3232}< \frac{3232}{3232}=1.\)
=>\(\frac{3}{2}>1>\frac{1212}{3232}\)
Vậy \(\frac{3}{2}>\frac{1212}{3232.}\)
Ko cần tk mk đâu mk tl cho vui thôi nhá.
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