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a) 100:{250:[450−(4.53−25.4)]}100:{250:[450−(4.53−25.4)]}

=100:{250:[450−(4.125−25.4)]}=100:{250:[450−(4.125−25.4)]}

=100:{250:[450−(500−100)]}=100:{250:[450−(500−100)]}

=100:[250:(450−400)]=100:[250:(450−400)]

=100:(250:50)=100:(250:50)

=100:5=100:5

=20=20

b) 4(18−15)−(5−3).324(18−15)−(5−3).32

=4.3−2.32=4.3−2.32

=4.3−2.9=4.3−2.9

=12−18=12−18

=−6=−6

17 tháng 10 2015

a)25.4:32=25.22:25=25+2-5=22=4

b)3.27:34=3.33:34=34:34=1

c)(29.16+29.32):210

=29.(16+32):210

=48:210-9

=48:2

=24

d)Bạn chữa lại dề nha

(34.57-92.21):35

=(34.57-34.22):35

=34.(57-22):35

=34.35:35

=34

=81

3 tháng 7 2019

a)A=\(\frac{\left(8+100\right).\left[\left(100-8\right):4+1\right]}{2}=\frac{108.242}{2}=13068\) 

b) \(5B=5^2+5^3+...+5^{101}\) 

  \(5B-B=5^{101}-5\) 

\(B=\frac{5^{101}-5}{4}\)

10 tháng 10 2017

ôc cho

11 tháng 10 2017

t k bt làm nên moj ?

a)     205 – [1200 – (42 – 2.3)3] : 40

= 205 – [1200 – ( 16 - 6 ) ] : 40

= 205 – [1200 – ( 10 ) ] : 40

= 205 – [1200 – 1000 ] : 40

= 205 – 200 : 40

= 205 – 50

= 155

b)     177 :[2.(42 – 9) + 32(15 – 10)]

= 177 :[2.(16 – 9) + 9. 5]

= 177 :[2.7 + 9. 5]

= 177 :[ 14 + 45 ]

= 177 : 59

= 3

26 tháng 6 2017

\(A=17^{18}-17^{16}\\ =17^{16}\cdot\left(17^2-1\right)\\ =17^{16}\cdot\left(289-1\right)\\ =17^{16}\cdot288\\ =17^{16}\cdot18\cdot16⋮18\)

Vậy \(A⋮18\)

\(B=1+3+3^2+...+3^{11}\)

Ta có: \(52=4\cdot13\)

\(B=1+3+3^2+...+3^{11}\\ =\left(1+3\right)+\left(3^2+3^3\right)+...+\left(3^{10}+3^{11}\right)\\ =1\cdot\left(1+3\right)+3^2\cdot\left(1+3\right)+...+3^{10}\cdot\left(1+3\right)\\ =\left(1+3\right)\cdot\left(1+3^2+...+3^{10}\right)\\ =4\cdot\left(1+3^2+...+3^{10}\right)⋮4\)

Vậy \(B⋮4\)

\(B=1+3+3^2+...+3^{11}\\ =\left(1+3+3^2\right)+\left(3^3+3^4+3^5\right)+...+\left(3^9+3^{10}+3^{11}\right)\\ =1\cdot\left(1+3+3^2\right)+3^3\cdot\left(1+3+3^2\right)+...+3^9\cdot\left(1+3+3^2\right)\\ =\left(1+3+3^2\right)\cdot\left(1+3^3+...+3^9\right)\\ =13\cdot\left(1+3^3+...+3^9\right)⋮13\)

Vậy \(B⋮13\)

\(4\)\(13\) là hai số nguyên tố cùng nhau nên tao có \(B⋮4\cdot13\Leftrightarrow B⋮52\)

Vậy \(B⋮52\)

\(C=3+3^3+3^5+...3^{31}\)

\(C=3+3^3+3^5+...+3^{31}\\ =\left(3+3^3\right)+\left(3^5+3^7\right)+...+\left(3^{29}+3^{31}\right)\\ =1\cdot\left(3+3^3\right)+3^4\cdot\left(3+3^3\right)+...+3^{28}\cdot\left(3+3^3\right)\\ =\left(3+3^3\right)\cdot\left(1+3^4+...+3^{28}\right)\\ =30\cdot\left(1+3^4+...+3^{28}\right)⋮15\left(\text{vì }30⋮15\right)\)

Vậy \(C⋮15\)

\(D=2+2^2+2^3+...+2^{60}\)

Tao có: \(21=3\cdot7;15=3\cdot5\)

\(D=2+2^2+2^3+...+2^{60}\\ =\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{59}+2^{60}\right)\\ =2\cdot\left(1+2\right)+2^3\cdot\left(1+2\right)+...+2^{59}\cdot\left(1+2\right)\\ =\left(1+2\right)\cdot\left(2+2^3+...+2^{59}\right)\\ =3\cdot\left(2+2^3+...+2^{59}\right)⋮3\)

Vậy \(D⋮3\)

\(D=2+2^2+2^3+...+2^{60}\\ =\left(2+2^3\right)+\left(2^5+2^7\right)+...+\left(2^{57}+2^{59}\right)+\left(2^2+2^4\right)+...+\left(2^{58}+2^{60}\right)\\ =2\cdot\left(1+2^2\right)+2^5\cdot\left(1+2^2\right)+...+2^{57}\cdot\left(1+2^2\right)+2^2\cdot\left(1+2^2\right)+...+2^{58}\cdot\left(1+2^2\right)\\ =\left(1+2^2\right)\cdot\left(2+2^5+...+2^{57}+2^2+...+2^{59}\right)\\ =5\cdot\left(2+2^5+...+2^{57}+2^2+...+2^{59}\right)⋮5\)

Vậy \(D⋮5\)

\(D=2+2^2+2^3+...+2^{60}\\ =\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{58}+2^{59}+2^{60}\right)\\ =2\cdot\left(1+2+2^2\right)+2^4\cdot\left(1+2+2^2\right)+...+2^{58}\cdot\left(1+2+2^2\right)\\ =\left(1+2+2^2\right)\cdot\left(2+2^4+...+2^{58}\right)\\ =7\cdot\left(2+2^4+...+2^{58}\right)⋮7\)

Ta có:

\(D⋮3;D⋮5\Rightarrow D⋮3\cdot5\Leftrightarrow D⋮15\)

\(D⋮3;D⋮7\Rightarrow D⋮3\cdot7\Leftrightarrow D⋮21\)

Vậy \(D⋮15;D⋮21\)

26 tháng 6 2017

Mình chỉ làm mẫu 1 câu thui nha:

\(A=17^{18}-17^{16}\)

\(A=17^{16}.17^2-17^{16}.1\)

\(A=17^{16}\left(17^2-1\right)\)

\(A=17^{16}.288\)

\(A=17^{16}.16.18\)

\(A⋮18\left(đpcm\right)\)

9 tháng 8 2018

B = 1 + 3 + 32 + 33 + 34 + 35 + ... + 311

B = (1 + 3 + 32 + 33 + 34 + 35) + (36 + 37 + 38 + 39 + 310 + 311)

B = 364 + 36.364

B = 364(36 + 1) \(⋮\) 52

12 tháng 6 2017

\(2b)\)

Đặt :


\(S=1+4+4^2+4^3+4^4....................+4^{100}\)

\(4S=4\left(1+4+4^2+4^3+4^4+.............+4^{100}\right)\)

\(4S=4+4^2+4^3+4^4+4^4+.......+4^{101}\)

\(4S-S=\left(4+4^2+4^3+4^4+4^5+.......+4^{101}\right)-\left(1+4+4^2+4^3+4^4+...............+4^{100}\right)\)

\(3S=4^{101}-1\)

\(S=\dfrac{4^{101}-1}{3}\)