\(A=\frac{4}{3}\times\frac{4}{7}+\frac{4}{7}\times\frac{4}{11}+....+\frac{4}{91}\times\frac{4}{95}+\frac{4}{95}\times\frac{4}{99}\)
hỏi A = ?
K
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TM
4 tháng 3 2016
a) \(\frac{16}{35}+\frac{8}{35}=\frac{24}{35}\)
b)\(\frac{160}{77}-\frac{28}{77}=\frac{132}{77}=\frac{12}{1}=12\)
c)\(\frac{72}{180}=\frac{18}{45}\)
d) \(\frac{90}{360}=\frac{1}{4}\)
TK
21 tháng 5 2019
\(4A=\frac{4}{3.7}+...+\frac{4}{95.99}\)
\(4A=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{95}-\frac{1}{99}\)
\(4A=\frac{1}{3}-\frac{1}{99}=\frac{32}{99}\)
\(\Rightarrow A=\frac{8}{99}\)
21 tháng 5 2019
\(A=\frac{1}{3.7}+\frac{1}{7.11}+...+\frac{1}{95.99}\)
\(A=\frac{1}{4}.\left(\frac{4}{3.7}+\frac{4}{7.11}+...+\frac{4}{95.99}\right)\)
\(A=\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{95}-\frac{1}{99}\right)\)
\(A=\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{99}\right)\)
\(A=\frac{1}{4}.\frac{32}{99}\)
\(A=\frac{8}{99}\)
SY
2 tháng 6 2017
\(a.\) \(\frac{4}{15}.\frac{7}{9}+\frac{4}{15}.\frac{2}{9}\)
\(=\frac{4}{15}\left(\frac{7}{9}+\frac{2}{9}\right)\)
\(=\frac{4}{15}.\frac{9}{9}\)
\(=\frac{4}{15}.1\)
\(=\frac{4}{15}\)
\(b.\) \(\frac{13}{19}.\frac{23}{11}-\frac{13}{19}.\frac{8}{11}-\frac{13}{19}.\frac{4}{11}\)
\(=\frac{13}{19}\left(\frac{23}{11}-\frac{8}{11}-\frac{4}{11}\right)\)
\(=\frac{13}{19}.\frac{11}{11}\)
\(=\frac{13}{19}.1\)
\(=\frac{13}{19}\)
2 tháng 6 2017
a)4/15 x(7/9+2/9)=4/15x1=4/15
b)13/19x(23/11-8/11-4/11)13/19x1=13/19
NT
1
NT
25 tháng 8 2017
\(\frac{-4}{7}\times\frac{5}{11}+\frac{-4}{7}\times\frac{6}{11}+\)\(2017\frac{4}{7}\)
\(=\frac{-4}{7}\times\left(\frac{5}{11}+\frac{6}{11}\right)+2017+\frac{4}{7}\)
\(=\frac{-4}{7}\times1+\frac{4}{7}+2017\)
\(=\left(\frac{-4}{7}+\frac{4}{7}\right)+2017\)
\(=0+2017=2017\)
IA
25 tháng 3 2020
1, =\(\frac{2\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}-\frac{1}{11}\right)}{4\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}-\frac{1}{11}\right)}=\frac{1}{2}\)
2, A=\(\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{99}{100}\)
= \(\frac{1\cdot2\cdot3\cdot....\cdot99}{2\cdot3\cdot4\cdot...\cdot100}=\frac{1}{100}\)
Vậy ......
hok tốt
\(A=\frac{4}{3}\times\frac{4}{7}+\frac{4}{7}\times\frac{4}{11}+...+\frac{4}{91}\times\frac{4}{95}+\frac{4}{95}\times\frac{4}{99}\)
\(=4\left(\frac{1}{3\times7}+\frac{1}{7.11}+...+\frac{1}{91\times95}+\frac{1}{95\times99}\right)\)
\(=4\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{91}-\frac{1}{95}+\frac{1}{95}-\frac{1}{99}\right)\)
\(=4\left(\frac{1}{3}-\frac{1}{99}\right)=4\times\frac{32}{99}=\frac{128}{99}\)