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a,2A=8+23+24+25+............+221
2A-A=(8+23+24+25+...........+221)-(4+22+23+24+.............+220)
A=(8+23+221)-(4+22)
A=8+8+221-4-4
A=4+4+221
A=23+221
b,(x+1)+(x+2)+.............+(x+100)=5750
=>x+1+x+2+...........+x+100=5750
=>100x+(1+2+............+100)=5750
=>100x+\(\frac{100.\left(100+1\right)}{2}\)=5750
=>100x+5050=5750
=>100x=5750-5050
=>100x=700
=>x=700:100
=>x=7
A=4+22+23+24+...+220
=22+22+23+24+...+220
=>2A=23+23+24+...+221
=>2A-A=23+23+24+...+221-22-22-23-24-...-220
=>A(2-1)=23+221-22-22
=>A=8+221-4-4
=>A=221
mình bik giải câu b thôi
TA CÓ
(x+1)+(x+2)+.................+(x+100)=5750
x+1+x+2+..............+x+100=5750
(x+x+....+x)+(1+2+3+.......+100)=5750
100.x + 5050 =5750
100.x=5750-5050
100x=700
=>x=700:100=7
VẬY X = 7
a,
A = 4 + 22 + 23 + 24 + .. + 220
Đặt A1 = 22 + 23 + 24 + .. + 220
2A1 = 2.( 22 + 23 + 24 + .. + 220)
= 23 + 24 + 25 + ... + 22
2A1 - A1 = (22 + 23 + 24 + .. + 220) - (23 + 24 + 25 + ... + 22 )
A1 = 221 - 22
= 221 - 4
=> A = 4 + 221 - 4
=> A = 221
S=30+32+34+36+...+3200
6S=32+34+36+...+3202
6S-S=(32+34+36+...+3202)-(1+32+34+...+3200)
5S=1+(32-32)+(34-34)+...+(3200-3200)+3202
S=(3200+1):5\(\frac{ }{ }\)
b, Ta có : ( x + 1 ) + ( x + 2 ) +... + ( x + 100) = 5750
<=> ( x + x + ... + x ) + ( 1 + 2 + ... + 100 ) =5750
<=> 100.x + 5050 = 5750
<=> 100.x = 5750 - 5050
<=> 100. x = 700
<=> x =700 : 100 = 7
Vậy x = 7
a, A=4+2^2+2^3+...+2^20
2A=2(4+2^2+2^3+...+2^20)
2A=8+2^3+2^4+...+2^21
2A-A=(8+2^3+2^4+...+2^21)-(4+2^2+2^3+...+2^20)
A=2^21+8-4-2^2
A=2^21
Vay
a) A=\(2^{21}\)
b)
(x+1)+(X+2)+...+(x+100)=5750
=> 100x+(1+2+3+...+100)=5750
=> 100x+\(\frac{\left(100+1\right).100}{2}=5750\)
=> 100x+5050=5750
=>100x=700
=>x=7
a, \(2A=8+2^3+2^4+...+2^{21}\)
\(\Rightarrow2A-A=2^{21}+8-\left(4+2^2\right)+\left(2^3-2^3\right)+...+\left(2^{20}-2^{20}\right)=2^{21}\)
b, (x+1)+(x+2)+(x+3)+...+(x+100)=5750
= 100x + (1+2+...+100) = 5750
=> x = [ 5750 - (1+2+...+100) ] :100
=> x = (5750 - 5050) : 100
=> x = 7