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A
29 tháng 6 2019
\(\frac{12}{7}>1\)
\(\frac{89}{1112}< 1\)
\(\Rightarrow\frac{12}{7}>\frac{89}{1112}\) ( So sánh với 1)
HL
29 tháng 6 2019
Các so sánh ta có là so sánh tất cả các phân số với 1.
Ta có: \(\frac{12}{7}>1;\frac{89}{1112}< 1\) nên ta có \(\frac{12}{7}>\frac{89}{1112}\).
Vậy \(\frac{12}{7}>\frac{89}{1112}\).
~ Hok tốt ~
TT
28 tháng 2 2020
Ta có : \(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+...+\frac{89}{90}\)
\(=1-\frac{1}{2}+1-\frac{1}{6}+1-\frac{1}{12}+...+1-\frac{1}{90}\)
\(=9-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\right)\)
\(=9-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=9-\left(1-\frac{1}{10}\right)\)
\(=\frac{81}{10}\)
DD
5 tháng 8 2015
\(\frac{1}{2}+\frac{5}{6}+...+\frac{89}{90}=1-\frac{1}{2}+1-\frac{1}{6}+...+1-\frac{1}{90}=9-\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{90}\right)\)
\(=9-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{9.10}\right)=9-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=9-\left(1-\frac{1}{10}\right)=8+\frac{1}{10}=\frac{81}{10}\)
16 tháng 8 2015
\(\frac{1}{2}+\frac{5}{6}+...+\frac{89}{90}\)
\(=1-\frac{1}{2}+1-\frac{1}{6}+...+1-\frac{1}{90}\)
\(=9-\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{90}\right)\)
\(=9-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{9.10}\right)\)
=\(9-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=9-\left(1-\frac{1}{10}\right)\)
\(=8+\frac{1}{10}=\frac{81}{10}\)
\(=8+\frac{1}{10}=\frac{81}{10}\)
DN
1
TD
23 tháng 2 2015
t=(1-1/2+1-1/6+1-1/12+..........+1-1/90+1-1/110)-10/11
t=(1+1+1+...........+1+1)-(1/2+1/6+1/12+.......+1/90+1/110)-10/11
có 10 số 1 vì từ 1/2 đến 109/110 có 10 p/số
t=1.10-(1/1.2+1/2.3+1/3.4+.........+1/9.10+1/10.11)-10/11
t=10-(1-1/2+1/2_1/3+1/3_1/4+.............+1/9-1/10+1/10-1/11)-10/11
t=10-(1-1/11)-10/11
t=10-10/11-10/11
t=10-(10/11+10/11)
t=10-10/11.2
t=10-20/11
t=110/11-20/11
t=90/11
KT
28 tháng 4 2018
\(\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+\frac{29}{30}+\frac{41}{42}+\frac{55}{56}+\frac{71}{72}+\frac{89}{90}\)
\(=1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+1-\frac{1}{30}+1-\frac{1}{42}+1-\frac{1}{56}+1-\frac{1}{72}+1-\frac{1}{90}\)
\(=8-\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\right)\)
\(=8-\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{!}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\right)\)
\(=8-\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+.....+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\)
\(=8-\left(\frac{1}{2}-\frac{1}{10}\right)\)
\(=8-\frac{2}{5}=\frac{38}{5}\)
13 tháng 9 2020
1/2+5/6+11/12+19/20+29/30+41/42+55/56+71/72+89/90
=1-1/2+1-1/6+1-1/12+1-1/20+1-1/30+1-1/42+1-1/56+1-1/72+1-1/90
=9 – (1/2+1/6+1/12+1/20+1/30+1/42+1/56+1/72+1/90)
=9 – [1/(1x2)+1/(2x3)+1/(3x4)+1/(4x5)+1/(5x6)+1/(6x7)+1/(7x8)+1/(8x9)+1/(9x10)]
=9 – ( 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+1/9-1/10)
=9 – (1 – 1/10) = 9 – 9/10
= 81/10
25 tháng 6 2016
\(y=\frac{1+5+11+19+29+41+55+71+89}{2+6+12+20+30+42+56+72+90}\)
\(y=\frac{1x2-1+2x3-1+3x4-1+4x5-1+5x6-1+6x7-1+7x8-1+8x9-1+9x10-1}{1x2+2x3+3x4+4x5+5x6+6x7+7x8+8x9+9x10}\)
\(y=\frac{\left(1x2+2x3+...+9x10\right)-\left(1+1+1+1+1+1+1+1+1\right)}{1x2+2x3+...+9x10}\)
\(y=\frac{1x2+2x3+...+9x10}{1x2+2x3+...+9x10}-\frac{9}{1x2+2x3+...+9x10}\)
\(y=1-\frac{9}{1x2+2x3+...+9x10}\)
HT
3 tháng 9 2015
T = \(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+...+\frac{109}{110}+\frac{10}{11}\)
T = \(\frac{10}{11}+\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{6}\right)+\left(1-\frac{1}{12}\right)+...+\left(1-\frac{1}{110}\right)\)
T = \(\frac{10}{11}+\left(1+1+...+1\right)+\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{110}\right)\)
T = \(\frac{10}{11}+10+\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}\right)\)
T = \(\frac{120}{11}+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\right)\)
T = \(\frac{120}{11}-\left(1-\frac{1}{11}\right)\)
T = \(\frac{120}{11}-\frac{9}{11}\)
T = \(\frac{111}{11}\)
3 tháng 9 2015
bài của bạn Giang tính nhầm bước thứ hai từ cuối lên
Sửa: T = \(\frac{120}{11}-\frac{10}{11}=10\)
20 tháng 7 2017
\(=1-\frac{1}{1\cdot2}+1-\frac{1}{2\cdot3}+1-\frac{1}{3\cdot4}+...+1-\frac{1}{9\cdot10}\)
\(=\left[1+1+1+...+1\right]-\left[\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{9\cdot10}\right]\)
\(=9-\left[\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right]=9-\left[1-\frac{1}{10}\right]\)
\(=9-\frac{9}{10}=\frac{81}{10}\)
KT
15 tháng 10 2018
\(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+...+\frac{89}{90}\)
\(=1-\frac{1}{2}+1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+...+1-\frac{1}{90}\)
\(=9-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\right)\)
\(=9-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)\)
\(=9-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{9}-\frac{1}{10}\right)\)
\(=9-\left(1-\frac{1}{10}\right)\)
\(=9-\frac{9}{10}=\frac{81}{10}\)
\(\frac{12}{5}:\frac{89}{7}\)
= \(\frac{12}{5}\) x \(\frac{7}{89}\)
= \(\frac{84}{445}\)
12/5 : 89/7
=12/5 x 7/8
=84 /445