Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\frac{-5}{6}+\frac{8}{3}+\frac{29}{-6}\le x\le\frac{-1}{2}+2+\frac{5}{2}\)
\(\Rightarrow-3\le x\le4\)
\(\Rightarrow x\in\left\{-3;-2;-1;0;1;2;3;4\right\}\)
\(\left(\frac{-2}{3}-\frac{1}{2}\right):\frac{-1}{4}\le x\le\left(\frac{-5}{6}+\frac{9}{4}:\frac{-3}{2}\right)\cdot\frac{-13}{2}\)
\(\Rightarrow\frac{14}{3}\le x\le\frac{91}{6}\)
\(\Rightarrow\frac{28}{6}\le x\le\frac{91}{6}\)
\(\Rightarrow x\in\left\{\frac{28}{6};\frac{29}{6};...;\frac{90}{6};\frac{91}{6}\right\}\)
c. S3 = 165 + 215 chia hết cho 33
ta thấy: 16^5=2^20
=> A=16^5 + 2^15 = 2^20 + 2^15
= 2^15.2^5 + 2^15
= 2^15(2^5+1)
=2^15.33
số này luôn chia hết cho 33
b. S2 = 2 + 22 + 23 + 24 +........... + 2100 chia hết cho 31
= 2(1 + 2 + 22 + 23 + 24 ) + 26( 1 + 2 + 22 + 23 + 24 ) + ....+ (1 + 2 + 22 + 23 + 24 )296
= 2 x 31 + 26 x 31 + ..... + 296 x 31 = 31 x ( 2 + 26 + ..... + 296 )
=> 2 + 22 + 23 + 24 +........... + 2100 chia hết cho 31
=(5^1+5^2)+.....+(5^99+5^100)
=5^1*(1+5)+....+5^99*(1+5)
=5^1*^+....+5^99*6
=6*(5^1+....+5^10)
=>5^1+....+5^100 CHIA HẾT CHO 6 NK
A=1+3+32+33+...+320
A=(1+3)+(32+33)+(34+35)+...+(319+320)
A= 4+32(1+3)+34(1+3)+......+319(1+3)
A=4+32.4+34.4+....+319.4
A=4.(32+34+...+319) =>A chia hết cho 4
0+(
a) S = 5 + 52 + 53 + ... + 5100
=> S = ( 5 + 52 ) + ( 53 + 54 ) + ... + ( 599 + 5100 )
=> S = 5( 1 + 5 ) + 53( 1 + 5 ) + ... + 599( 1 + 5 )
=> S = 5 . 6 + 53 . 6 + ... + 599 . 6
=> S = ( 5 + 53 + ... + 599 ) . 6 chia hết cho 6
=> S chia hết cho 6
b) S1 = 2 + 22 + 23 + ... + 2100
=> S1 = ( 2 + 22 + 23 + 24 + 25 ) + ... + ( 296 + 297 + 298 + 299 + 2100 )
=> S1 = 2( 1 + 2 + 22 + 23 + 24 ) + ... +296( 1 + 2 + 22 + 23 + 24 )
=> S1 = 2 . 31 + ... + 296 . 31
=> S1 = ( 2 + ... + 296 ) . 31 chia hết cho 31
=> S1 chia hết cho 31
c) S2 = 165 + 215
=> S2 = ( 24 )5 + 215
=> S2 = 220 + 215
=> S2 = 220( 1 + 25 )
=> S2 = 220 . 33 chia hết cho 33
=> S2 chia hết cho 33
a: \(S=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot...\cdot\dfrac{-99}{100}=-\dfrac{1}{100}\)
c: \(5S_3=5^6+5^7+...+5^{101}\)
\(\Leftrightarrow4\cdot S_3=5^{101}-5^5\)
hay \(S_3=\dfrac{5^{101}-5^5}{4}\)
d: \(S_4=7\cdot\left(\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}+...+\dfrac{1}{69}-\dfrac{1}{70}\right)\)
\(=7\left(\dfrac{1}{10}-\dfrac{1}{70}\right)=7\cdot\dfrac{6}{70}=\dfrac{6}{10}=\dfrac{3}{5}\)
\(A=2^1+2^2+2^3+2^4+2^5+2^6+2^7+...+2^{99}\)
\(=\left(2^1+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+\left(2^7+2^8+2^9\right)+...+\left(2^{97}+2^{98}+2^{99}\right)\)
\(=2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+2^7\left(1+2+2^2\right)+...+2^{97}\left(1+2+2^2\right)\)
\(=2.7+2^4.7+2^7.7+...+2^{97}.7\)
\(=\left(2+2^4+2^7+...+2^{97}\right).7⋮7\)
\(\Rightarrow A⋮7\)
A = 21 +22 +23 +24 +25 +26 +27 ….+ 299
A = (21 +22 +23) +(24 +25 +26) + ….+ (297+298+299)
A = 14 + (21.23 +22.23 +23.23) + ….+ (21.296+22.296+23.296)
A = 14 + 23(21+22+23) + ...... + 296(21+22+23)
A = 14.1 + 23.14 + ....... + 296.14
A = 14.(1+23+....+296)
14 \(⋮\) 7
=> A \(⋮\) 7 (đpcm)