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TD
26 tháng 1 2017
Ta có: \(\frac{9}{10}-\frac{1}{90}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)
= \(\frac{9}{10}-\frac{1}{10.9}-\frac{1}{9.8}-\frac{1}{8.7}-\frac{1}{7.6}-\frac{1}{6.5}-\frac{1}{5.4}-\frac{1}{4.3}-\frac{1}{3.2}-\frac{1}{2.1}\)
= \(\frac{9}{10}-\frac{1}{10}-\frac{1}{9}-...-\frac{1}{2}-\frac{1}{1}\)
= \(\frac{9}{10}+\frac{1}{10}-\frac{1}{1}\)
= 1 - 1 = 0
Vậy kết quả của phép tính là 0
HP
0
NT
4
25 tháng 6 2017
\(\dfrac{9}{10}-\dfrac{1}{90}-\dfrac{1}{72}-\dfrac{1}{56}-\dfrac{1}{42}-\dfrac{1}{30}-\dfrac{1}{20}-\dfrac{1}{12}-\dfrac{1}{6}-\dfrac{1}{2}\)
\(=\dfrac{9}{10}-\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{72}+\dfrac{1}{90}\right)\)
\(=\dfrac{9}{10}-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{8.9}+\dfrac{1}{9.10}\right)\)
\(=\dfrac{9}{10}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\right)\)
\(=\dfrac{9}{10}-\left(1-\dfrac{1}{10}\right)=\dfrac{9}{10}-\dfrac{9}{10}=0\)
25 tháng 6 2017
\(\dfrac{9}{10}-\dfrac{1}{90}-\dfrac{1}{72}-\dfrac{1}{56}-...-\dfrac{1}{6}-\dfrac{1}{2}=-\left(-\dfrac{9}{10}+\dfrac{1}{90}+\dfrac{1}{72}+\dfrac{1}{56}+...+\dfrac{1}{6}+\dfrac{1}{2}\right)\)
\(=-\left(-\dfrac{9}{10}+\dfrac{1}{2}+\dfrac{1}{6}+...+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\right)\)
\(=-\left(-\dfrac{9}{10}+\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\right)\)
\(=-\left(-\dfrac{9}{10}+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\right)\)
\(=-\left(-\dfrac{9}{10}+1-\dfrac{1}{10}\right)=-\left(-\dfrac{9}{10}+\dfrac{9}{10}\right)=0\)
HM
8 tháng 9 2016
em lớp 6 nha
B= 1/2 + 1/6 + 1/12 +1/20 + 1/30 + 1/42 + 1/56 + 1/72
B= 1/1*2 + 1/2*3 + 1/3*4 + 1/4*5 + 1/5*6 + 1/6*7 + 1/7*8 + 1/8*9
B=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9
B=1+0-0-0-0-0-0-0-1/9
B=1-1/9
B=8/9
k em nha
15 tháng 3 2019
\(=\frac{9}{10.11}-\frac{1}{9.10}-\frac{1}{8.9}-\frac{1}{7.8}-\frac{1}{6.7}-\frac{1}{5.6}-\frac{1}{4.5}-\frac{1}{3.4}-\frac{1}{2.3}-\frac{1}{1.2}\)
\(=\frac{9}{10.11}-\frac{10-9}{9.10}-\frac{9-8}{8.9}-...-\frac{2-1}{1.2}\)
\(=\frac{9}{10.11}-\frac{10}{9.10}+\frac{9}{9.10}-...-\frac{2}{1.2}+\frac{1}{1.2}\)
\(=\frac{9}{10.11}-\frac{1}{9}+\frac{1}{10}-\frac{1}{8}+\frac{1}{9}-\frac{1}{7}+\frac{1}{8}-...-\frac{1}{2}+\frac{1}{3}-1+\frac{1}{2}\)
\(=\frac{9}{10.11}+\frac{1}{10}-1\)
\(=-\frac{9}{11}\)
SN
1
Q
7 tháng 8 2017
\(\dfrac{9}{10}-\dfrac{1}{90}-\dfrac{1}{72}-\dfrac{1}{56}-\dfrac{1}{42}-\dfrac{1}{30}-\dfrac{1}{20}-\dfrac{1}{12}-\dfrac{1}{6}-\dfrac{1}{2}\)
\(=\dfrac{9}{10}-\left(\dfrac{1}{90}+\dfrac{1}{72}+\dfrac{1}{56}+\dfrac{1}{42}-\dfrac{1}{30}-\dfrac{1}{20}-\dfrac{1}{12}-\dfrac{1}{6}-\dfrac{1}{2}\right)\)
\(=\dfrac{9}{10}-\left(\dfrac{1}{9\cdot10}+\dfrac{1}{9\cdot8}+\dfrac{1}{7\cdot8}+\dfrac{1}{7\cdot6}+\dfrac{1}{5\cdot6}-\dfrac{1}{5\cdot4}-\dfrac{1}{3\cdot4}-\dfrac{1}{3\cdot2}-\dfrac{1}{1\cdot2}\right)\)
\(=\dfrac{9}{10}-\left(\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{8}-\dfrac{1}{9}+...+1-\dfrac{1}{2}\right)\)
\(=\dfrac{9}{10}-\left(1-\dfrac{1}{10}\right)\)
\(=\dfrac{9}{10}-\dfrac{9}{10}\)
\(=0\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
23 tháng 6
a; [6.(- \(\dfrac{1}{3}\))3 - 3.(- \(\dfrac{1}{3}\) + 1)] - ( - \(\dfrac{1}{3}\) - 1)
= [6. \(\dfrac{-1}{3^3}\) - 3.\(\dfrac{2}{3}\)] - ( - \(\dfrac{1}{3}\) - \(\dfrac{3}{3}\))
= [\(\dfrac{-2}{9}\) - 2] + \(\dfrac{4}{3}\)
= [\(\dfrac{-2}{9}\) - \(\dfrac{18}{9}\)] + \(\dfrac{12}{9}\)
= - \(\dfrac{20}{9}\) + \(\dfrac{12}{9}\)
= \(\dfrac{-8}{9}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
23 tháng 6
b; (63 + 3.62 + 33): 13
= (216 + 3.36 + 27) : 13
= (216 + 108 + 27): 13
= (324 + 27): 13
= 351 : 13
= 27
3 tháng 9 2016
\(\text{Ta có: }\)\(\frac{1}{90}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)
\(=\frac{1}{90}-\left(\frac{1}{72}+\frac{1}{56}+\frac{1}{42}+\frac{1}{30}+\frac{1}{20}+\frac{1}{12}+\frac{1}{6}+\frac{1}{2}\right)\)
\(=\frac{1}{90}-\left(\frac{1}{9.8}+\frac{1}{8.7}+\frac{1}{7.6}+\frac{1}{6.5}+\frac{1}{5.4}+\frac{1}{4.3}+\frac{1}{3.2}+\frac{1}{2.1}\right)\)
\(=\frac{1}{90}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{8}-\frac{1}{9}\right)\)
\(=\frac{1}{90}-\left(1-\frac{1}{9}\right)\)
\(=\frac{1}{90}-\frac{8}{9}=-\frac{81}{90}=-\frac{9}{10}\)
3 tháng 9 2016
<=>
D = 1/90+1/72+1/56+1/42+1/30+1/20+1/12+1/6+1/2
D = 1/(1x2) + 1/(2x3) + 1/(3x4) + 1/(4x5) + 1/(5x6) + … + 1/(9x10)
D = 1 – 1/2 + 1/2 – 1/3 + 1/3 – 1/4 + …. + 1/9 – 1/10
D = 1 – 1/10
D = 9/10
Ta có :
9/10-1/90-1/72-1/56-1/42-1/30-1/20-1/12-1/6-1/2
= 9/10 -( 1/90 + 1/72 + ... + 1/2)
= 9/10 - { 1/( 9.10) + 1/(9.8) + ... + 1/( 2.1)}
= 9/10 - ( 1/9 - 1/10 + 1/8 - 1/9 + ...+ 1 - 1/2) ( 1/90 = 1/(9.10) = 1/9 - 1/10)
= 9/10 - ( 1 - 1/10)
= 9/10 - 9/10
= 0