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b/, 3x + 2 + 3x = 90
=> 3x.32 + 3x = 90
=> 3x.9 + 3x = 90
=> 3x.(9+1) = 90
=> 3x.10 = 90
=> 3x = 90:10
=> 3x = 9 = 32
=> x = 2

\(A=2^1+2^2+.....+2^{2010}\)
\(=\left(2^1+2^2\right)+\left(2^3+2^4\right)+.....+\left(2^{2009}+2^{2010}\right)\)
\(=2\left(1+2\right)+2^3\left(1+2\right)+.....+2^{2009}\left(1+2\right)\)
\(=\left(1+2\right)\left(2+2^3+.....+2^{2009}\right)\)
\(=3\left(2+2^3+....+2^{2009}\right)\)chia hết cho 3

\(C=2+2^2+2^3+...+2^{100}\)
\(2C=2^2+2^3+...+2^{100}+2^{101}\)
\(2C-C=\left(2^2+2^3+....+2^{101}\right)-\left(2+2^2+...+2^{100}\right)\)
\(C=2^{101}-2\)
Thay \(C=2^{101}-2\)vào \(2^{2x-1}-2\)ta được:
\(2^{2x-1}-2=2^{101}-2\)
\(\Rightarrow2^{2x-1}=2^{101}-2+2\)
\(\Rightarrow2^{2x-1}=2^{101}\)
\(\Rightarrow2x-1=101\)
\(\Rightarrow2x=102\)
\(\Rightarrow x=102\div2\)
\(\Rightarrow x=51\)
Vậy x = 51.
Nhớ k nhé!
2C = 2\(^2\)+ 2\(^3\)+ 2\(^4\)+ ... + 2\(^{101}\)
2C - C = ( 2\(^2\)+ 2\(^3\)+ 2\(^4\)+ ... + 2\(^{101}\)) - ( 2 + 2\(^2\)+ 2\(^3\)+ ... + 2\(^{100}\))
C = 2\(^2\)+ 2\(^3\)+ 2\(^4\)+ ... + 2\(^{101}\)- 2 - 2\(^2\)- 2\(^3\)- ... - 2\(^{100}\)
C = ( 2\(^2\)+ 2\(^2\)) + ( 2\(^3\)- 2\(^3\)) + ( 2\(^4\)+ 2\(^4\)) + ... + (2\(^{100}\)- 2\(^{100}\)) + ( 2\(^{101}\)- 2)
C = 2\(^{101}\)- 2
Ta có :
2\(^{2x-1}\) - 2 = 2\(^{101}\)- 2
2\(^{2x-1}\)= 2\(^{101}\)
2x - 1 = 101
2x = 101 + 1 = 102
x = 102 : 2 = 51
Vậy x = 51

A) TA CÓ 2 TRƯỜNG HỢP XẢY RA: CHÚNG = 0 HOẶC BẰNG 1.
TH1: \(\left(x-6\right)^3=\left(x-6\right)^2=0\)
\(\Rightarrow\left(x-6\right)^3=\left(x-6\right)^2=0\)
\(\Rightarrow x-6=x-6=0\Rightarrow x=6\)
TH2:
\(\left(x-6\right)^3=\left(x-6\right)^2=1\)
\(\Rightarrow\left(x-6\right)^3=\left(x-6\right)^2=1\)
\(\Rightarrow x-6=x-6=1\Rightarrow x=7\)
giúp mk lên 200 điểm nhé các bạn .
a, (x - 6)3 = (x - 6)2
vì 3 > 2 nên không có giá trị nào thỏa mãn (x - 6)3 = (x - 6)2
vậy x \(\in\varnothing\)
b, {[(2x + 14) : 22 - 3] : 2} - 1 = 0
[(2x + 14) : 4 - 3] : 2 = 0 + 1
[(2x + 14) : 4 - 3] : 2 = 1
[(2x + 14) : 4 - 3] = 1.2
(2x + 14) : 4 - 3 = 2
(2x + 14) : 4 = 2 + 3
(2x + 14) : 4 = 5
2x + 14 = 5.4
2x + 14 = 20
2x = 20 - 14
2x = 6
x = 6 : 2
x = 3

2a/ 2x - 3 = 16 => 2x - 3 = 24 => x - 3 = 4 => x = 7
b/ {x2 - [82 - (52 - 8.3)3 - 7.9]3 - 4.12}3 = 1
=> x2 - [82 - (52 - 8.3)3 - 7.9]3 - 4.12 = 1
=> x2 - [64 - (25 - 24)3 - 63]3 - 48 = 1
=> x2 - (64 - 1 - 63)3 = 1 + 48
=> x2 - 0 = 49
=> x2 = 49
=> x = 7

Ta có: 1/22 < 1/1.2
1/32 < 1/2.3
.......................
........................
1/1002 < 1/99.100
=> 1/22+1/32+1/42+......+1/1002 < 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/99 .100
= > 1/22+1/32+1/42+......+1/1002 < 1-1/2 + 1/2 -1/3 + .... + 1/99 - 1/100
=> 1/22+1/32+1/42+......+1/1002 < 1 - 1/100
=>1/22+1/32+1/42+......+1/1002 < 99/100
\(\frac{1}{2}.\left(x-\frac{2}{3}\right)^2=\frac{2}{18}\)
\(\left(x-\frac{2}{3}\right)^2=\frac{2}{18}:\frac{1}{2}\)
\(\left(x-\frac{2}{3}\right)^2=\frac{2}{9}\)
@_@ sr bn nhak đến đoạn này mik tịt r