TD
2
K
Khách
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VN
3
25 tháng 3 2018
A=\(\frac{6}{4}+\frac{6}{28}+\frac{6}{70}+\frac{6}{130}+\frac{6}{208}\)
A=\(\frac{6}{1\cdot4}+\frac{6}{4\cdot7}+\frac{6}{7\cdot10}+\frac{6}{10\cdot13}+\frac{6}{13\cdot16}\)
A:2=\(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+\frac{3}{10\cdot13}+\frac{3}{13\cdot16}\)
A:2=\(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\)\(\frac{1}{16}\)
A:2=\(1-\frac{1}{16}\)
A:2=\(\frac{15}{16}\)
A=\(\frac{15}{8}\)
vậy ...
3 tháng 5 2017
Bạn ơi, có chắc là 6/280 ở cuối không.
Trả lời nhanh để mink giải câu này cho
NH
0
T
14 tháng 4 2019
c) \(A=\frac{6}{4}+\frac{6}{28}+\frac{6}{70}+\frac{6}{130}+\frac{6}{208}\)
\(=\frac{6}{1.4}+\frac{6}{4.7}+\frac{6}{7.10}+\frac{6}{10.13}+\frac{6}{13.16}\)
\(=2\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}\right)\)
\(=2\left(1-\frac{1}{16}\right)\)
\(=2.\frac{15}{16}\)
\(=\frac{15}{8}\)
Vậy A=\(\frac{15}{8}\)
KN
14 tháng 4 2019
a) \(\frac{3^2}{1.4}+\frac{3^2}{4.7}+\frac{3^2}{7.10}+...+\frac{3^2}{97.100}\)
\(=3\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{97.100}\right)\)
\(=3\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\right)\)
\(=3\left(1-\frac{1}{100}\right)\)
\(=3.\frac{99}{100}=\frac{297}{100}\)
12 tháng 8 2018
A = \(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{56}\)
\(=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{7.8}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{7}-\frac{1}{8}\)
\(=\frac{1}{2}-\frac{1}{8}=\frac{3}{8}\)
B = \(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{11.13}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{11}-\frac{1}{13}\)
\(=1-\frac{1}{13}=\frac{12}{13}\)
19 tháng 3 2019
\(A=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{56}\)
\(=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{7.8}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{7}-\frac{1}{8}\)
\(=\frac{1}{2}-\frac{1}{8}=\frac{3}{8}\)
\(B=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{11.13}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{11}-\frac{1}{13}\)
\(=1-\frac{1}{13}=\frac{12}{13}\)
BT
1
TH
22 tháng 3 2017
1/2A=1/2(6/4+6/28+6/70+6/130+6/208)
= 3/4+3/28+3/70+3/130+3/208
= 1-1/4+1/4-1/7+..................-1/16
=1-1/16
=15/16 => A=15/8
11 tháng 8 2019
Đặt P = ... ( biểu thức đề bài )
Nhận xét: Với \(k\inℕ^∗\) ta có:
\(\frac{k+2}{k!+\left(k+1\right)!+\left(k+2\right)!}=\frac{k+2}{k!+\left(k+1\right).k!+\left(k+2\right).k!}=\frac{k+2}{2.k!\left(k+2\right)}=\frac{1}{2.k!}\)
\(\Rightarrow\)\(P=\frac{1}{2.1!}+\frac{1}{2.2!}+...+\frac{1}{2.6!}=\frac{1}{2}\left(1+\frac{1}{2}+...+\frac{1}{720}\right)=...\)
15 tháng 3 2018
\(a)\) \(A=\frac{5\left(2^2.3^2\right)^9.\left(2^2\right)^6-2\left(2^2.3\right)^{14}.3^4}{5.2^{28}.3^{18}-7.2^{29}.3^{18}}\)
\(A=\frac{2^{30}.3^{18}.5-2^{29}.3^{18}}{2^{28}.3^{18}.5-2^{29}.3^{18}.7}\)
\(A=\frac{2^{29}.3^{18}\left(2.5-1\right)}{2^{28}.3^{18}\left(5-2.7\right)}\)
\(A=\frac{2\left(10-1\right)}{5-14}\)
\(A=\frac{2.9}{-9}\)
\(A=-2\)
Vậy \(A=-2\)
\(b)\) \(B=81.\left[\frac{12-\frac{12}{7}-\frac{12}{289}-\frac{12}{85}}{4-\frac{4}{7}-\frac{4}{289}-\frac{4}{85}}:\frac{5+\frac{5}{13}+\frac{5}{169}+\frac{5}{91}}{6+\frac{6}{13}+\frac{6}{169}+\frac{6}{91}}\right].\frac{158158158}{711711711}\)
\(B=81.\left[\frac{12\left(1-\frac{1}{7}-\frac{1}{289}-\frac{1}{85}\right)}{4\left(1-\frac{1}{7}-\frac{1}{289}-\frac{1}{85}\right)}:\frac{5\left(1+\frac{1}{13}+\frac{1}{169}+\frac{1}{91}\right)}{6\left(1+\frac{1}{13}+\frac{1}{169}+\frac{1}{91}\right)}\right].\frac{158158158}{711711711}\)
\(B=81.\left[\frac{12}{4}:\frac{5}{6}\right].\frac{2}{9}\)
\(B=81.\frac{18}{5}.\frac{2}{9}\)
\(B=\frac{324}{5}\)
Vậy \(B=\frac{324}{5}\)
Chúc bạn học tốt ~ ( mỏi tay qué >_< )
7 tháng 4 2017
\(A=81.\left[\frac{12-\frac{12}{7}-\frac{12}{289}-\frac{12}{85}}{4-\frac{4}{7}-\frac{4}{289}-\frac{4}{85}}:\frac{5+\frac{5}{13}+\frac{5}{169}+\frac{5}{91}}{6+\frac{6}{13}+\frac{6}{169}+\frac{6}{91}}\right].\frac{158158158}{711711711}\)
\(A=81.\left[\frac{12.\left(1-\frac{1}{7}-\frac{1}{289}-\frac{1}{85}\right)}{4.\left(1-\frac{1}{7}-\frac{1}{289}-\frac{1}{85}\right)}:\frac{5.\left(1+\frac{1}{13}+\frac{1}{169}+\frac{1}{91}\right)}{6.\left(1+\frac{1}{13}+\frac{1}{169}+\frac{1}{91}\right)}\right].\frac{158}{711}\)
\(A=81.\left(\frac{12}{4}:\frac{5}{6}\right).\frac{2}{9}\)
\(A=81.3.\frac{6}{5}.\frac{2}{9}\)
\(A=\frac{324}{5}\)
Nhớ là: THANKS YOU VERY "MUCH" chứ không phải là THANKS YOU VERY "MATH"!!!
7 tháng 4 2017
\(A=81.\frac{158158158}{711711711}.\frac{12.\left(\frac{1}{7}-\frac{1}{289}-\frac{1}{85}\right)}{4.\left(\frac{1}{7}-\frac{1}{289}-\frac{1}{85}\right)}:\frac{5.\left(\frac{1}{13}+\frac{1}{169}+\frac{1}{91}\right)}{6.\left(\frac{1}{13}+\frac{1}{169}+\frac{1}{91}\right)}\)
\(=81.\frac{158}{711}.\frac{12}{4}:\frac{5}{6}=\frac{1422}{79}.3.\frac{6}{5}=\frac{1422.3.6}{79.5}=\frac{25596}{395}\)
\(\frac{6}{4}+\frac{6}{28}+\frac{6}{70}+\frac{6}{130}\)
\(=\frac{6}{1x4}+\frac{6}{4x7}+\frac{6}{7x10}+\frac{6}{10x13}\)
\(=2\left(\frac{3}{1x4}+\frac{3}{4x7}+\frac{3}{7x10}+\frac{3}{10x13}\right)\)
\(=2\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}\right)\)
\(=2\left(1-\frac{1}{13}\right)\)
\(=2x\frac{12}{13}\)
\(=\frac{24}{13}\)
\(\frac{6}{4}+\frac{6}{28}+\frac{6}{70}+\frac{6}{130}\)
\(=\frac{6}{1.4}+\frac{6}{4.7}+\frac{6}{7.10}+\frac{6}{10.13}\)
\(=2\left(\frac{3}{1.4}+\frac{3}{1.7}+\frac{3}{7.10}+\frac{3}{10.13}\right)\)
\(=2\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}\right)\)
\(\Leftrightarrow2=\left(1-\frac{1}{13}\right)\)
\(=2.\frac{12}{13}\)
\(=\frac{24}{13}\)