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Câu hỏi 1:=111+1
=>112
Câu hỏi 2:=27+50
=>77
Câu hỏi 3:80-2
=>78
tích đúng nha bạn
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
a)
204-84:12
=204-7
=197
b)
15.23+4.32-5.7
=15.8+4.9-5.7
=120+36-35
=156-35
=121
c)
56:53+23.22
=56-3+23+2
=53+25
=125+32
=157
d)
164.53+47.164
=(164.(53+47)
=164.100
=16400
tìm x:
219-7(x+1)=100
=>7(x+1)=219-100
=>7(x+1)=119
=>x+1=119:7
=>x+1=17
=>x=17-1
=>x=16
(3x-6).3=34
=>(3x-6).3=81
=>3x-6=81:3
=>3x-6=27
=>3x=27+6
=>3x=33
=>x=33:3
=>x=11
a) 960 -[ 50. ( 20 -8 : 2 + 22 )]
= 960 -[ 50. ( 20 -8 : 2 + 4)]
= 960 -[ 50. ( 20 -4+ 4)]
= 960 -[ 50. (16+ 4)]
= 960 -[ 50.20]
= 960 -1000
= -40
b) 28 : 24 + 32 . 33
= 1,166666667+1056
= 1057,166667
c) 56 : 53 + 32 . 33
= 1,056603774+1056
= 1057,056604
d) 136.68 + 16.272
= 9248 + 4352
= 13600
e) 36.333 - 108.111
= 11988 - 11988
= 0
Mình không giỏi môn toán cho lắm, nếu mình làm sai thì thôi nha!
a) 960 - {50 . [(20 - 8 : 2 + 22 )]}
= 960 - { 50 . [( 20 - 8 : 2 + 4 )]}
= 960 - { 50 . [( 20 - 4 + 4 )]}
= 960 - { 50 . [( 16 + 4 )]}
= 960 - { 50 . 20 }
= 960 - 1000
= - 40
b) 28 : 24 + 32 . 33
= 256 : 16 + 9 . 27
= 16 + 243
= 259
c) 56 : 53 + 32 . 33
= 15625 :125 + 9 . 27
= 125 + 243
= 363
d) 136 . 68 + 16 . 272
= 9248 + 4352
= 13600
e) 36 . 333 - 108 . 111
= 11983 - 11983
= 0
4 . 52 - 32 . 24
= 22 . 52 - 23 . 22 . 24
= ( 2 . 5 )2 - 23+2+4
= 102 - 29
= -412
a) 160 - ( 23 . 52 - 6 . 25 ) = 160-( 8 . 25 - 6 . 25 ) = 160 - ( 25 . ( 8-6)) = 160-(25 . 2) = 160- 50 = 110
b) 4 . 52 - 32 : 24 = 4 . 25 - 4 . 8 : 16 = 4 . ( 25 - 8 : 16 ) = 4 . 24,5 = 98
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
\(\frac{1}{2^2}< \frac{1}{1.2}\)
\(\frac{1}{3^2}< \frac{1}{2.3}\)
\(\frac{1}{4^2}< \frac{1}{3.4}\)
\(.....\)
\(\frac{1}{100^2}< \frac{1}{99.100}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)\(< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}< 1\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< 1\left(đpcm\right)\)
nhìu dữ, nhưg liệu có ****?