NH
0
K
Khách
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7 tháng 5 2019
A=\(\frac{1}{3}\)+\(\frac{1}{8}\)+\(\frac{1}{15}\)+\(\frac{1}{24}\)+\(\frac{1}{35}\)+\(\frac{1}{48}\)+\(\frac{1}{63}\)+\(\frac{1}{80}\)
A=\(\frac{1}{2}\)(\(\frac{1}{1\cdot3}\)+\(\frac{1}{2\cdot4}\)+\(\frac{1}{3\cdot5}\)+\(\frac{1}{4.6}\)+\(\frac{1}{5.7}\)+\(\frac{1}{6.8}\)+\(\frac{1}{7.9}\)+\(\frac{2}{8.10}\))
A=\(\frac{1}{2}\)(1-1/3 +1/2-1/4 + 1/3 -1/5 +1/4-1/6 +1/5 - 1/7 +1/6 -1/8 +1/7 - 1/9 +1/8 - 1/10)
A= \(\frac{1}{2}\)(1 + 1/2 -1/9 -1/10)
A=\(\frac{29}{45}\)
NC
0
MN
4 tháng 2 2018
a) \(\frac{3}{16}+\frac{4}{15}+\frac{5}{16}+\frac{1}{15}\)
\(=\left(\frac{3}{16}+\frac{5}{16}\right)+\left(\frac{4}{15}+\frac{1}{15}\right)\)
\(=\frac{1}{2}+\frac{1}{3}\)
\(=\frac{5}{6}\)
b) \(\frac{6}{7}\times\frac{8}{15}\times\frac{7}{6}\times\frac{15}{16}\)
\(=\left(\frac{6}{7}\times\frac{7}{6}\right)\times\left(\frac{8}{15}\times\frac{15}{16}\right)\)
\(=1\times\frac{1}{2}=\frac{1}{2}\)
c) \(\frac{19}{20}\times\frac{13}{21}+\frac{9}{20}\times\frac{8}{21}\)
\(=\frac{19\times13}{20\times21}+\frac{9\times8}{20\times21}\)
\(=\frac{247}{420}+\frac{72}{420}\)
\(=\frac{319}{420}\)
NT
10
DN
6
DN
4
KT
14 tháng 8 2018
a) \(D=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+...+\frac{1}{512}+\frac{1}{1024}\)
=> \(2D=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...++\frac{1}{256}+\frac{1}{512}\)
=> \(2D-D=\left(1+\frac{1}{2}+...+\frac{1}{512}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{1024}\right)\)
=> \(D=1-\frac{1}{1024}\)
b) \(Đ=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{18.19}+\frac{1}{19.20}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{18}-\frac{1}{19}+\frac{1}{19}-\frac{1}{20}\)
\(=1-\frac{1}{20}=\frac{19}{20}\)
CM
14 tháng 8 2018
a) D=\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\dots+\frac{1}{512}+\frac{1}{1024}.\)
\(D=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}+\dots+\frac{1}{512}-\frac{1}{1024}\)
\(D=1-\frac{1}{1024}\)
\(D=\frac{1023}{1024}\)
\(Đ=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\dots+\frac{1}{18\cdot19}+\frac{1}{19\cdot20}\)
\(Đ=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\dots+\frac{1}{18}-\frac{1}{19}+\frac{1}{19}-\frac{1}{20}\)
\(Đ=1-\frac{1}{20}\)
\(Đ=\frac{19}{20}\)
Phần c như kiểu sai đề chỗ cuối hay sao ấy.
4 tháng 7 2017
Bài 1 :
\(\frac{2}{19}+\frac{4}{5}+1\frac{7}{9}+\frac{1}{5}\)
\(=\frac{2}{19}+1\frac{7}{9}+\left(\frac{4}{5}+\frac{1}{5}\right)\)
\(=\frac{2}{19}+1\frac{7}{9}+1\)
\(=\frac{493}{171}\)
\(\frac{1}{11}+\frac{2}{11}+\frac{3}{11}+\frac{4}{11}+\frac{5}{11}+\frac{6}{11}+\frac{7}{11}+\frac{8}{11}+\frac{9}{11}+\frac{10}{11}\)
\(=\left(\frac{1}{11}+\frac{10}{11}\right)+\left(\frac{2}{11}+\frac{9}{11}\right)+\left(\frac{3}{11}+\frac{8}{11}\right)+\left(\frac{4}{11}+\frac{7}{11}\right)+\left(\frac{5}{11}+\frac{6}{11}\right)\)
\(=1+1+1+1+1\)
\(=5\)
Bài 2 :
y : (23 + 27) = 9 1000 : (y x 25) = 8
y : 50 = 9 y x 25 = 1000 : 8
y = 9 x 50 y x 25 = 125
y = 450 y = 125 : 25
133 : (190 : y) = 7 y = 5
190 : y = 133 : 7 450 : (y + 42) = 9
190 : y = 19 y + 42 = 450 : 9
y = 190 : 19 y + 42 = 50
y = 10 y = 50 - 42
y = 8