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a ) \(-\frac{13}{30}+\frac{11}{20}-\frac{7}{15}\)
\(=-\frac{26}{60}+\frac{33}{60}-\frac{28}{60}\)
\(=\frac{-26+33-28}{60}=-\frac{7}{20}\)
b ) \(-\frac{5}{72}:\left(\frac{3}{8}.\frac{7}{9}\right)\)
\(=-\frac{5}{72}:\frac{3.7}{8.9}\)
\(=-\frac{5}{72}:\frac{7}{24}\)
\(=-\frac{5}{72}.\frac{24}{7}=-\frac{5}{21}\)
a) \(\frac{-13}{30}+\frac{11}{20}-\frac{7}{15}=\frac{-26}{60}+\frac{33}{60}-\frac{28}{60}=-\frac{21}{60}=-\frac{7}{20}\)
b) \(\frac{-5}{72}:\left(\frac{3}{8}.\frac{7}{9}\right)=\frac{-5}{72}:\frac{7}{24}=-\frac{5}{21}\)
c) \(\frac{-23}{25}.\frac{10}{13}+\frac{-23}{25}.\frac{3}{13}+\frac{-27}{25}\)
\(=\frac{-23}{25}.\left(\frac{10}{13}+\frac{3}{13}\right)+\frac{-27}{25}\)
\(=\frac{-23}{25}+\frac{-27}{25}\)
\(=\frac{-50}{25}=-2\)
`A=2^{0}+2^{1}+2^{2}+....+2^{99}`
`=(1+2+2^{2}+2^{3}+2^{4})+(2^{5}+2^{6}+2^{7}+2^{8}+2^{9})+......+(2^{95}+2^{96}+2^{97}+2^{97}+2^{99})`
`=(1+2+2^{2}+2^{3}+2^{4})+2^{5}(1+2+2^{2}+2^{3}+2^{4})+.....+2^{95}(1+2+2^{2}+2^{3}+2^{4})`
`=31+2^{5}.31+....+2^{95}.31`
`=31(1+2^{5}+....+2^{95})\vdots 31`
\(A=2^0+2^1+2^2+2^3+2^4+2^5+2^6+...+2^{99}\)
\(=\left(2^0+2^1+2^2+2^3+2^4\right)+2^5\left(2^0+2^1+2^2+2^3+2^4\right)+...+2^{95}\left(2^0+2^1+2^2+2^3+2^4\right)=31+31.2^5+...+31.2^{95}=31\left(1+2^5+...+2^{95}\right)⋮31\)
A = 20 + 21 + 22 + 23 + 24 + 25 … + 299
A=( 20 + 21 + 22 + 23 + 24) +( 25 … + 299)
A= 20.(20 + 21 + 22 + 23 + 24)+25.( 25 … + 299)
A= 1. 31+ 25.31… + 295.31
A= 31. (1+25...+295)
KL: ......
\(A=2^0+2^1+2^2+2^3+2^4+...+2^{99}=\left(2^0+2^1+2^2+2^3+2^4\right)+2^5\left(2^0+2^1+2^2+2^3+2^4\right)+...+2^{95}\left(2^0+2^1+2^2+2^3+2^4\right)=31+31.2^5+...+31.2^{95}=31\left(1+2^5+...+2^{95}\right)⋮31\)
Bài 1
S₂ = 21 + 23 + 25 + ... + 1001
Số số hạng của S₂:
(1001 - 21) : 2 + 1 = 491
⇒ S₂ = (1001 + 21) . 491 : 2 = 250901
--------
S₄ = 15 + 25 + 35 + ... + 115
Số số hạng của S₄:
(115 - 15) : 10 + 1 = 11
⇒ S₄ = (115 + 15) . 11 : 2 = 715
Bài 2
a) 2x - 138 = 2³.3²
2x - 138 = 8.9
2x - 138 = 72
2x = 72 + 138
2x = 210
x = 210 : 2
x = 105
b) 5.(x + 35) = 515
x + 35 = 515 : 5
x + 35 = 103
x = 103 - 35
x = 78
c) 814 - (x - 305) = 712
x - 305 = 814 - 712
x - 305 = 102
x = 102 + 305
x = 407
d) 20 - [7.(x - 3) + 4] = 2
7(x - 3) + 4 = 20 - 2
7(x - 3) + 4 = 18
7(x - 3) = 18 - 4
7(x - 3) = 14
x - 3 = 14 : 7
x - 3 = 2
x = 2 + 3
x = 5
e) 9ˣ⁻¹ = 9
x - 1 = 1
x = 1 + 1
x = 2
a; \(\dfrac{9}{27}\) + \(\dfrac{7}{-49}\)
= \(\dfrac{1}{3}\) - \(\dfrac{1}{7}\)
= \(\dfrac{7}{21}\) - \(\dfrac{3}{21}\)
= \(\dfrac{4}{21}\)
b; - \(\dfrac{12}{10}\) + \(\dfrac{-25}{30}\)
= - \(\dfrac{6}{5}\) - \(\dfrac{5}{6}\)
= -\(\dfrac{36}{30}\) - \(\dfrac{25}{30}\)
= \(\dfrac{-61}{30}\)
c; \(\dfrac{-20}{35}\) + \(\dfrac{-16}{-24}\)
= - \(\dfrac{4}{7}\) + \(\dfrac{2}{3}\)
= - \(\dfrac{12}{21}\) + \(\dfrac{14}{21}\)
= \(\dfrac{2}{21}\)
d; - \(\dfrac{21}{77}\) + \(\dfrac{10}{-35}\)
= - \(\dfrac{3}{11}\) - \(\dfrac{2}{7}\)
= - \(\dfrac{21}{77}\) - \(\dfrac{22}{77}\)
= - \(\dfrac{43}{77}\)
a) 20+21+22+23+24+25
=(20+25)+(21+24)+(22+23)
=45+45+45
=45x3
135
b)
20+21+22+...+29+30
=(20+30)+(21+29)+...(24+26)+259 (tổng có 5 cặp)
=50+50+...+25
=50x5+25
=250+25
=275
#Châu's ngốc
a) 20 + 21 + 22 + 23 + 24 +25
= (20 + 25) + (21 + 24) + (22 + 23)
= 45 + 45 + 45
= 45 . 3 = 135
b) 20 + 21 + 22 +...+ 29 + 30
= (20 + 30) + (21 + 29) +...+ (24 + 26) + 25
= 50 + 50 +...+ 50 + 25
5 số 50
= 50 . 5 + 25
= 250 + 25
= 275
\(a.23.\left(75+25\right)+127.100\)
\(23.100+127.100\)
\(100.\left(127+23\right)\)
\(100.150\)
\(15000\)
\(b.\left(20+30\right)+\left(21+29\right)+22\)
\(50+50+22\)
\(100+22=122\)
\(23.75+25+23+127.100\)
\(=1725+25+23+12700\)
\(=1750+23+12700\)
\(=1773+12700\)
\(=14473\)
\(20+21+22+29+30\)
\(=\left(20+30\right)+\left(21+29\right)+22\)
\(=50+50+22\)
\(=50.2+22\)
\(=100+22\)
\(=122\)
A = 20 + 23 + 25 + .... + 299
4A = 22 + 25 + 27 + .... + 2101
4A - A = (22 + 25 + 27 + .... + 2101) - (20 + 23 + 25 + .... + 299)
3A = 22 + 2101 - 20 - 23
3A = 2101 - 5
A = \(\frac{2^{101}-5}{3}\)