Tính:
a1/10+-1/20+-1/30+-1/42+-1/56+-1/72+-1/90
1/90-1/72-1/56-1/42-1/30-1/20-1/10-1/6-1/2
9/10-1/90-1/72-1/56-1/42-1/30-1/20-1/12-1/6-1/2
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
9/10-1/90-1/72-1/56-1/42-1/30-1/20-1/12-1/6-1/2
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
9/10-1/90-1/72-1/56-1/42-1/30-1/20-1/12-1/6-1/2 = ?
9/10-1/90-1/72-1/56-1/42-1/30-1/20-1/12-1/6-1/2
=9/10-(1/9*10+1/8*9+...+1/1*2)
=9/10-(1/9-1/10+...+1-1/2)
=9/10-(-1/10+1)=9/10-9/10=0
9/10 +1/90++1/72+1/56+1/42+1/30+1/20+1/12+1/6+1/2
= 9/1.10 + 1/9.10 + 1/8.9 + 1/7.8 + 1/6.7 +1/5.6 + 1/4.5 +1/3.4 +1/2.3 + 1/1.2
= 1/1.2 + 1/2.3 + 1/3.4 + ... + 9/1.10 ( viết ngược lại)
= 1-1/2 + 1/2 -1/3 + 1/3 +....-1/10
= 1 - 1/10
= 9/10
tính:9/10-1/90-1/72-1/56-1/42-1/30-1/20-1/12-1/6-1/2
1)Tinh 9/10-1/90-1/72-1/56-1/42-1/30-1/20-1/12-1/6-1/2
\(\dfrac{9}{10}-\dfrac{1}{90}-\dfrac{1}{72}-\dfrac{1}{56}-\dfrac{1}{42}-\dfrac{1}{30}-\dfrac{1}{20}-\dfrac{1}{12}-\dfrac{1}{6}-\dfrac{1}{2}\)
\(=\dfrac{9}{10}-\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{72}+\dfrac{1}{90}\right)\)
\(=\dfrac{9}{10}-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{8.9}+\dfrac{1}{9.10}\right)\)
\(=\dfrac{9}{10}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\right)\)
\(=\dfrac{9}{10}-\left(1-\dfrac{1}{10}\right)=\dfrac{9}{10}-\dfrac{9}{10}=0\)
\(\dfrac{9}{10}-\dfrac{1}{90}-\dfrac{1}{72}-\dfrac{1}{56}-...-\dfrac{1}{6}-\dfrac{1}{2}=-\left(-\dfrac{9}{10}+\dfrac{1}{90}+\dfrac{1}{72}+\dfrac{1}{56}+...+\dfrac{1}{6}+\dfrac{1}{2}\right)\)
\(=-\left(-\dfrac{9}{10}+\dfrac{1}{2}+\dfrac{1}{6}+...+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\right)\)
\(=-\left(-\dfrac{9}{10}+\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\right)\)
\(=-\left(-\dfrac{9}{10}+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\right)\)
\(=-\left(-\dfrac{9}{10}+1-\dfrac{1}{10}\right)=-\left(-\dfrac{9}{10}+\dfrac{9}{10}\right)=0\)
Ta có :
\(\dfrac{9}{10}-\dfrac{1}{90}-\dfrac{1}{72}-\dfrac{1}{56}-\dfrac{1}{42}-\dfrac{1}{30}-\dfrac{1}{20}-\dfrac{1}{12}-\dfrac{1}{6}-\dfrac{1}{2}\)
\(=\dfrac{9}{10}-\left(\dfrac{1}{90}+\dfrac{1}{72}+\dfrac{1}{56}+\dfrac{1}{42}+\dfrac{1}{30}+\dfrac{1}{20}+\dfrac{1}{12}+\dfrac{1}{6}+\dfrac{1}{2}\right)\)
\(=\dfrac{9}{10}-\left(\dfrac{1}{9.10}+\dfrac{1}{9.8}+\dfrac{1}{7.6}+\dfrac{1}{6.5}+\dfrac{1}{5.4}+\dfrac{1}{4.3}+\dfrac{1}{3.2}+\dfrac{1}{2.1}\right)\)
\(=\dfrac{9}{10}-\left(\dfrac{1}{10}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{8}+..........+\dfrac{1}{2}-\dfrac{1}{1}\right)\)
\(=\dfrac{9}{10}-\left(1-\dfrac{1}{10}\right)\)
\(=\dfrac{9}{10}-\dfrac{9}{10}=0\)
1/2+1/6+1/12+1/20+1/30+1/42+1/56+1/72+1/90+1/10
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+....+\frac{1}{110}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+......+\frac{1}{10.11}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{10}-\frac{1}{11}\)
\(=\frac{1}{1}-\frac{1}{11}=\frac{11}{11}-\frac{1}{11}=\frac{10}{11}\)
Tính: B=1/2+1/6+1/12+1/20+1/30+1/42+1/56+1/72
A=1/90-1/72-1/56-1/42-1/30-1/20-1/12-1/6-1/2
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