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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
4 tháng 7 2019
\(\frac{1313}{1212}:x=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5}\)
\(\frac{1313}{1212}:x=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}\)
\(\frac{1313}{1212}:x=\frac{1}{1}-\frac{1}{5}+\frac{1}{5}\)
\(\frac{1313}{1212}:x=\frac{4}{5}+\frac{1}{5}\)
\(\frac{1313}{1212}:x=1\)
\(x=\frac{1313}{1212}:1\)
\(x=\frac{13}{12}\)
4 tháng 7 2019
Lời giải
\(\frac{1313}{1212}:x=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}\)
\(\frac{1313}{1212}:x=\frac{1}{1}-\frac{1}{5}+\frac{1}{5}\)
\(\frac{1313}{1212}:x=\frac{4}{5}+\frac{1}{5}\)
\(x=\frac{1313}{1212}:1\)
\(x=\frac{13}{12}\)
AH
1
TT
10 tháng 9 2018
Tìm x :
\(\left(\dfrac{2}{3}x-\dfrac{4}{9}\right).\left(\dfrac{1}{2}+\dfrac{3}{7}:x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(\dfrac{2}{3}x-\dfrac{4}{9}\right)=0\\\left(\dfrac{1}{2}+\dfrac{3}{7}:x\right)=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\dfrac{2}{3}x=\dfrac{4}{9}\\\dfrac{3}{7}:x=-\dfrac{1}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{6}{7}\end{matrix}\right.\)
Vậy.....
Tính :
\(\dfrac{1313}{1515}+\left(-\dfrac{1011}{5055}\right)=\dfrac{13}{15}-\dfrac{1}{5}=\dfrac{2}{3}\)
HT
0
Lúc nãy, cô còn dạy học nên giờ cô mới giảng cho em được nhé.
B = (1 - \(\dfrac{1}{2}\))\(\times\)(1 - \(\dfrac{1}{3}\))\(\times\)(1 - \(\dfrac{1}{4}\))\(\times\)(1-\(\dfrac{1}{5}\))\(\times\)...\(\times\)(1- \(\dfrac{1}{2003}\))\(\times\)(1-\(\dfrac{1}{2004}\))
B = \(\dfrac{2-1}{2}\)\(\times\)\(\dfrac{3-1}{3}\)\(\times\)\(\dfrac{4-1}{4}\)\(\times\)\(\dfrac{5-1}{5}\)\(\times\)...\(\times\)(\(\dfrac{2003-1}{2003}\))\(\times\)(\(\dfrac{2004-1}{2004}\))
B = \(\dfrac{1}{2}\)\(\times\)\(\dfrac{2}{3}\)\(\times\)\(\dfrac{3}{4}\)\(\times\)\(\dfrac{4}{5}\)\(\times\)...\(\times\)\(\dfrac{2002}{2003}\)\(\times\)\(\dfrac{2003}{2004}\)
B = \(\dfrac{2\times3\times4\times...\times2003}{2\times3\times4\times...\times2003}\)\(\times\) \(\dfrac{1}{2004}\)
B = \(\dfrac{1}{2004}\)