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TB
2
19 tháng 8 2015
gọi biểu thức tên là A , ta có :
A= (1/21+1/210+1/2010). ( 1/3-1/30-1/5-1/10)
A = (1/21+1/210+1/2010) . 0
A = 0
23 tháng 5 2022
\(=\left(\dfrac{1}{21}+\dfrac{1}{210}+\dfrac{1}{2010}\right)\cdot\dfrac{10-6-3-1}{30}=0\)
C
23 tháng 5 2022
\(\left(\dfrac{1}{3}-\dfrac{1}{5}-\dfrac{1}{10}-\dfrac{1}{30}\right)\times\left(\dfrac{1}{21}+\dfrac{1}{210}+\dfrac{1}{2010}\right)\)
\(=\left(\dfrac{10}{30}-\dfrac{6}{30}-\dfrac{3}{30}-\dfrac{1}{30}\right)\times\left(\dfrac{1}{21}+\dfrac{1}{210}+\dfrac{1}{2010}\right)\)
\(=0\times\left(\dfrac{1}{21}+\dfrac{1}{210}+\dfrac{1}{2010}\right)\)
\(=0\)
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) \(\frac{19}{20}\times\frac{13}{21}+\frac{9}{20}\times\frac{8}{21}\)
\(=\frac{19\times13}{20\times21}+\frac{9\times8}{20\times21}\)
\(=\frac{247}{420}+\frac{72}{420}\)
\(=\frac{319}{420}\)
2 tháng 7 2018
b, 3/5 + 4/7 + 2/8 + 10/25 + 9/21 + 28/16
= 3/5 + 4/7 + 2/8 + 2/5 + 3/7 + 14/8
= (3/5 + 2/5) + ( 4/7 + 3/7) + ( 2/8 + 14/8)
= 1 + 1 + 7/4
= 2 + 7/4 = 15/4
c , 8/7 + 7/6 + 5/8 + 10/12 + 24/28 + 6/16
= c , 8/7 + 7/6 + 5/8 + 5/6 + 6/7 + 1/2
= (8/7 + 6/7) + (7/6 + 5/6) + 5/8 + 1/2
= 14/7 + 12/6 + 5/8 + 1/2
= 2 + 2 + 5/8 + 1/2
= 4 + 9/8 = 41/8
P
1
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=\(\frac{29}{45}\)
DB
19 tháng 5 2017
Ta có:
\(\left(\frac{1}{21}+\frac{1}{210}+\frac{1}{2010}\right)\)\(\times\)\(\left(\frac{1}{3}-\frac{1}{30}-\frac{1}{5}-\frac{1}{10}\right)\)
= \(\left(\frac{1}{21}+\frac{1}{210}+\frac{1}{2010}\right)\)\(\times\)\(\left(\frac{10}{30}-\frac{1}{30}-\frac{6}{30}-\frac{3}{30}\right)\)
= \(\left(\frac{1}{21}+\frac{1}{210}+\frac{1}{2010}\right)\)\(\times\)\(\left(\frac{10-1-6-3}{30}\right)\)
= \(\left(\frac{1}{21}+\frac{1}{210}+\frac{1}{2010}\right)\)\(\times\)\(0\)
= \(0\)
NH
0
NT
1
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.