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Vì câu a có dấu x mk ko hiểu nên mk làm câu b nhé:
có 12 - 22 = -3
32 - 42 = -7
...................
992 - 1002 = -199
vậy chúng cách nhau 4 đơn vị
⇒ -((199 + 3).((199 - 3):4 + 1):2))) = -5050 vậy A = -5050
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a) 52 . x = 62 + 82
\(5^2\cdot x=36+64\)
\(5^2\cdot x=100\)
\(x=100\div5^2\)
\(x=100\div25\)
\(x=4\)
b) ( 22 + 42 ) . x + 24 . 5 . x = 102
\(\left(4+16\right)\cdot x+16\cdot5\cdot x=100\)
\(x\cdot\left(20+80\right)=100\)
\(x\cdot100=100\)
\(x=100\div100\)
\(x=1\)
c ) 24 . x = 26
\(x=2^6\div2^4\)
\(x=2^{6-4}\)
\(x=2^2\)
\(x=4\)
d) 33 . x + 23 . x = 102
\(x\cdot\left(23+27\right)=100\)
\(x\cdot50=100\)
\(x=100\div50\)
\(x=2\)
e) 78 . x = 710
\(x=7^{10}\div7^8\)
\(x=7^{10-8}\)
\(x=7^2\)
\(x=49\)
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A = 2o + 21 + 22 + ... + 22010
=> 2A = 21 + 22 + 23 + ... + 22010 + 22011
Mà A = 20 + 21 + 22 + ... + 22010
=> 2A - A = A = 1 + 22011
B = 1 + 3 + 32 + ... + 3100
=> 3B = 3 + 32 + 33 + ... + 3100 + 3101
Mà B = 1 + 3 + 32 + ... + 3100
=> 3B - B = 2B = 2 + 3101
=> B = ( 2 + 3101 ) : 2
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Đặt \(S=1+2^1+2^2+2^3+2^4+...+2^{20}\)
\(=2^0+2^1+2^2+2^3+2^4+2^{19}\) ( tong cap so nhan co 20 so hang. cong boi q=2.u1=1)
\(\Rightarrow s=\frac{u1.\left(1-q^{20}\right)}{\left(1-q\right)}=\frac{\left(1-2^{20}\right)}{\left(1-2\right)}=10485...\)
A=1+2^1+2^2+...+2^20
=>2A=2+2^2+2^3+.....+2^20+2^21
=>2A - A=(2+2^2+...+2^21)-(1+2+2^2+...+2^20)
hay A=2^21-1
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Đặt \(A=1+2+2^2+2^3+...+2^{20}\)
\(2A=2+2^2+2^3+2^4+...+2^{21}\)
\(2A-A=\left(2+2^2+2^3+...+2^{21}\right)-\left(1+2+2^2+...+2^{20}\right)\)
\(A=2^{21}-1\)
Ta đặt
A= 1+2^1+2^2+2^3+....2^20
2A= 21+22+23+....+221
=>2A-A=(2^1+2^2+2^3+...+2^21)-(1+2^2+2^3+...)
1A=2^21-1
Vậy A=2^21-1
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Câu 1.
C = 5 + 42 + 43 + ... + 42020
a) Xét A = 42 + 43 + ... + 42020
=> 4A = 43 + 44 + ... + 42021
=> 4A - A = 3A
= 43 + 44 + ... + 42021 - ( 42 + 43 + ... + 42020 )
= 43 + 44 + ... + 42021 - 42 - 43 - ... - 42020
= 42021 - 42
=> A = \(\frac{4^{2021}-4^2}{3}\)
Thế vào C ta được : \(C=5+\frac{4^{2021}-4^2}{3}=\frac{15}{3}+\frac{4^{2021}-4^2}{3}=\frac{4^{2021}+15-16}{3}=\frac{4^{2021}-1}{3}\)
b) D = 42021 => \(\frac{D}{3}=\frac{4^{2021}}{3}\)
Vì 42021 - 1 < 42021 => \(\frac{4^{2021}-1}{3}< \frac{4^{2021}}{3}\)
=> C < D/3
c) Dùng kết quả ý a) ta được :
3C + 1 = 42x-6
<=> \(3\cdot\frac{4^{2021}-1}{3}+1=4^{2x-6}\)
<=> 42021 - 1 + 1 = 42x-6
<=> 42021 = 42x-6
<=> 2021 = 2x - 6
<=> 2x = 2027
<=> x = 2027/2
Câu 2.
( x - 1 )( 4 + 22 + 23 + ... + 220 ) = 222 - 221
Xét A = 22 + 23 + ... + 220
=> 2A = 23 + 24 + ... + 221
=> A = 2A - A
= 23 + 24 + ... + 221 - ( 22 + 23 + ... + 220 )
= 23 + 24 + ... + 221 - 22 - 23 - ... - 220
= 221 - 4
Thế vô đề bài ta được
( x - 1 )( 4 + 221 - 4 ) = 222 - 221
<=> ( x - 1 ).221 = 221( 2 - 1 )
<=> x - 1 = 1
<=> x = 2
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a) 3x = 33.35 = 38
=> x = 8
b) 2x.2 = 222
2x+1 = 22
=> x + 1 = 22
x = 21
c) (7x-11)3 = 25.52 + 200 = 800 + 200 = 1 000 = 103
=> 7x -11 = 10
7x = 21
x = 3
d) 6x = 362 = (62)2 = 64
=> x = 4
e) 64.4x = 49
4x = 49/64
f) (21-1)3 = 203 ( xem lại đề)
h) 64.4x.2 = 45
4x.128 = 45
4x = 45/128
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(2a3x2y).(8a2x3y4).(16a3x3y3)
= 2a3 . x3 . 16a3 . x3 .y3 . x2.8a2 .y4
= a3x3a3 . x3 .y3 .2.16.8. x2.a2y4
=(axaxy)3 .y4 44 .(xy)2
=(axaxy)3 .(4y)4 (xy)2
\(2^2+2^2+2^3+2^4+...+2^x=2^{21}\)
Đặt \(2^2+2^3+2^4+...+2^x\)là \(B\), ta có :
\(2B=2.\left(2^2+2^3+2^4+...+2^x\right)\)
\(\Rightarrow2B=2^3+2^4+2^5+...+2^{x+1}\)
\(\Rightarrow2B-B=\left(2^3+2^4+2^5+...+2^{x+1}\right)-\left(2^2-2^3+2^4+...+2^x\right)\)
\(\Rightarrow B=2^{x+1}-2^2\)
Thay \(B\)vào (1) ta có :
\(2^2+\left(2^{x+1}-2^2\right)=2^{21}\)
\(\Rightarrow2^{x+1}=2^{21}\)
\(\Rightarrow x+1=21\)
\(\Rightarrow x=21-1\)
\(\Rightarrow x=20\)
Vậy \(x=20\)