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Bài 1:
a) A=1+22+24+.................+2100
2A=(1+22+24+.................+2100)
2A=2+23+...+2101
2A-A=(2+23+...+2101)-(1+22+24+.................+2100)
A=2101-1
b)bạn tự làm
c) C=-1/90-1/72-1/50-1/42-1/30-1/20-1/12-1/6-1/2
\(=-\left(\frac{1}{90}+\frac{1}{72}+...+\frac{1}{2}\right)\)
\(=-\left(\frac{1}{10.9}+\frac{1}{9.8}+...+\frac{1}{2.1}\right)\)
\(=-\left(\frac{1}{10}-\frac{1}{9}+\frac{1}{9}-\frac{1}{8}+...+\frac{1}{2}-1\right)\)
\(=-\left(\frac{1}{10}-1\right)\)
\(=-\left(-\frac{9}{10}\right)=\frac{9}{10}\)
Bài 2:
cứ tính lần lượt là ra
a, ( - 168 ) + 72 . ( - 168 ) + ( - 168 ) . 27
= ( - 168 ) + ( - 12096 ) + ( - 4536 )
= - 12264 + - 4536
= - 16800
b, 22 . ( - 3 ) - ( 110 + 8 ) : ( - 3 )2
= 4 . ( - 3 ) - ( 1 + 8 ) : 9
= ( - 12 ) - 9 : 9
= ( - 12 ) - 1
= - 13
c, ( - 1075 ) - ( 29 - 1075 )
= ( - 1075 ) - ( - 1046 )
= - 29
d, ( - 9 ) + ( - 11 ) + 21 + ( - 1 )
= - 20 + 21 + ( - 1 )
= 1 + - 1
= 0
e, 30 + 12 + ( - 20 ) + ( - 12 ) - ( 30 - 20 ) + ( 12 - 12 )
= 42 + ( - 20 ) + ( - 12 ) - 10 + 0
= 22 + ( - 12 ) - 10 + 0
= 10 - 10 + 0
= 0 + 0
= 0
g, ( 13 - 135 + 49 ) - ( 13 + 49 )
= [( - 122 ) + 49 ] - 62
= ( - 73 ) - 62
= - 135
h, 35 - { 12 - [ ( - 14 ) + ( - 2 ) } ]
= 35 - { 12 - ( - 16 ) }
= 35 - 28
= 7
Bài 2:
a. x - 35 = ( - 12 ) - 3
x - 35 = - 15
x = - 15 + 35
x = 20
b, \(\frac{1}{4}\)+ \(\frac{1}{3}\): 3x = - 5
\(\frac{3}{12}+\frac{4}{12}\): 3x = - 5
\(\frac{7}{12}\): 3x = - 5
3x = \(\frac{7}{12}\): - 5
3x = \(\frac{-7}{60}\)
x = \(\frac{-7}{60}\): 3
x = \(\frac{-7}{180}\)
c,2x-1 = 8
2x-1 = 24
x = 4 + 1
x = 5
\(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\)\(=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\)
\(=\)\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-...-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\)
\(=\)\(1-\dfrac{1}{10}\)
\(=\dfrac{9}{10}\)
b ) \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
= 1 - 1/2 + 1/2 - 1/3 + ... + 1/99 - 1/100
= 1 - 1/100
= 99/100
c ) Đặt A = \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}\)
=> A < \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
=> A < 1 - 1/2 + 1/2 - 1/3 + ... + 1/99 - 1/100= 1 - 1/100 = 99/100 < 1
Vậy \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}\)< 1
b, \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\)\(\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
c,Ta thấy
\(\frac{1}{2^2}< \frac{1}{1.2}\)
\(\frac{1}{3^2}< \frac{1}{2.3}\)
\(\frac{1}{4^2}< \frac{1}{3.4}\)
\(.....\)
\(\frac{1}{100^2}< \frac{1}{99.100}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)\(< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}< 1\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< 1\left(đpcm\right)\)
a)17.85-15.17-120
=17.(85-17)-120
=17.68-12
=1156-120
=1036
b)100:[30-(6-1)^ 2]
=100 :[30-5^ 2]
= 100:[30-25]
=100:5=20
c)/-10/+4+3.2^3+(-14)
=10+4+3.8 +(-14)
=14+24+(-14)
=[14 +(-14)]+24
=0+24=24
d)20-[30-(5-1)^ 2: 2]
=20-[30-4^2:2]
=20-[30-8]
=20-22=-2
e)80-(4.5^2-3.2^3)
= 80-(4.25-3.8)
=80-(4.25-3.2.4)
=80-4.(25-3.2)
= 80-4.19
= 80-76=4
g)35-{12-[-14+6-2)]}
=35 -{12-14-6+2}
=35-12+14+6-2
=23+14+6-2
=37+6-2
=43-2=41
\(A=\frac{1}{2}+\frac{5}{6}+...+\frac{89}{90}=1-\frac{1}{2}+1-\frac{5}{6}+...+1-\frac{1}{90}=\left(1++...+1\right)-\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{9\cdot10}\right)\)\(=9-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)=9-\left(\frac{10}{10}-\frac{1}{10}\right)=9-\frac{9}{10}=\frac{90}{10}-\frac{9}{10}=\frac{89}{10}\)
Bài 1:
A,
(100+121+144)÷(169+196)
=365:365=1
B
,1.2.3...7.8.(9-1-8)
=1.2.3...7.8.0=0
C,
3^2.2^4.2^32/11.2^13.2^22-2^26
=3^2.2^36/11.2^35-2^36
=3^2.2^36/2^35-(11-2)
=9.2/9=2
D,
=1152-374-1152+(-65)+374
=(1152-1152)+(-374+374)+(-65)
=0+0+(-65)=-65
E,
=13-(12-11-10+9)+(8-7-6+5)-(4-3-2+1)
=13-0+0-0=13
Bài 2:
A,(19x+50)÷14=25-16
(19x+50)÷14=9
19x+50=126
19x=76
x=4
B,
31x+(1+2+3+...+30)=1240
31x+465=1240
31x=775
×=25
1.(5.3^11+4.3^12):(3^9.5^2-3^9.2^3)
=(5.3^11+4.3^12):(3^9.5^2-3^9.2^3)
=(5.3^11+4.3^11.3):[3^9.(5^2-2^3)]
=(5.3^11+12.3^11):[3^9.17]
=3^11.(5+12):(3^9.17)
=(3^11.17):(3^9.17)
=17.(3^11:3^9)
=17.3^2
=17.9
=153
2.(12+22+32+...+992+1002).(36.333-108.111)
=2.(12+22+32+...+992+1002).(36.333-36.3.111)
=2.(12+22+32+...+992+1002).(36.333-36.333)
=2.(12+22+32+...+992+1002).0
=0
làm được câu a thui
\(\frac{1}{100}=\frac{1}{10.10}\);\(\frac{1}{90}=\frac{1}{9.10}\);...
Suy ra \(\frac{1}{10.10}-\frac{1}{9.10}-\frac{1}{8.9}-\frac{1}{7.8}-\frac{1}{6.7}-...-\frac{1}{1.2}\)
\(\frac{1}{10}-\frac{1}{10}-\frac{1}{10}-\frac{1}{9}-...-1-\frac{1}{2}\)
\(\frac{1}{10}-\frac{1}{2}\)
\(-\frac{4}{10}\)
Thanks bạn nha