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3A - A = (32 + 33 + 34 + ... + 32007) - (3 + 32 + 33 + ... + 32006)
2A = 32007 - 3\(\Rightarrow\hept{\begin{cases}A=\frac{3^{2007}-3}{2}\\2A+3=3^{2007}\Rightarrow x=2007\end{cases}}\)
\(A=3+3^2+3^3+...+3^{2016}\)
\(\Rightarrow3A=3\left(3+3^2+3^3+...+3^{2016}\right)\)
\(\Rightarrow3A=3^2+3^3+3^4+...+3^{2017}\)
\(\Rightarrow3A-A=\left(3^2+3^3+3^4+...+3^{2017}\right)-\left(3+3^2+3^3+3^{2016}\right)\)
\(\Rightarrow2A=-3+3^{2017}\)
\(\Rightarrow A=\frac{3+3^{2017}}{2}\)
b) \(2A+3=-3+3-3^{2017}=3^{2017}=3^x\)
\(\Rightarrow x=2017\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có: 3A-A = (32+33+34+35+...+32006+32007)-(3+32+33+34+...+32006) = 32007-3
=> 2A= 32007-3 => A=\(\frac{\text{3^{2007}-3}}{2}\)
b) Ta có:
2A= 32007-3 => 2A+3=32007=3x => x=2007
![](https://rs.olm.vn/images/avt/0.png?1311)
b) \(A=3^1+3^2+3^3+...+3^{2006}\)
\(=3+3^2+\left(3^3+3^4+3^5+3^6\right)+....+\left(3^{2003}+3^{2004}+3^{2005}+3^{2006}\right)\)
\(=12+3^3\left(1+3+3^2+3^3\right)+...+3^{2003}\left(1+3+3^2+3^3\right)\)
\(=12+\left(1+3+3^2+3^3\right)\left(3^3+...+3^{2003}\right)\)
\(=12+40\left(3^3+...+3^{2003}\right)\)
\(=12+.....0=.....2\)
Vậy A có tận cùng là chữ số 2
![](https://rs.olm.vn/images/avt/0.png?1311)
a)3A=3(31 + 32 + 33 + ... + 32006)
3A=32+33+...+32007
3A-A=(32+33+...+32007)-(31 + 32 + 33 + ... + 32006)
2A=32007-3
A=\(\frac{3^{2007}-3}{2}\)
b)2A+3=3x
thay 2A=32007-3 vào ta được
<=>32007-3+3=3x
<=>32007=3x
<=>x=2007
\(3A=3^2+3^3+3^4+...+3^{2007}\)
\(3A-A=2A=3^{2007}-3\)
\(A=\frac{3^{2007}-3}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a/ \(A=3+3^2+3^3+....+3^{2006}\)
\(\Leftrightarrow3A=3^2+3^3+.....+3^{2007}\)
\(\Leftrightarrow3A-A=\left(3^2+3^3+...+3^{2007}\right)-\left(3+3^2+...+3^{2006}\right)\)
\(\Leftrightarrow2A=3^{2007}-3\)
\(\Leftrightarrow A=\frac{3^{2007}-3}{2}\)
b/ Ta có :
\(2A=3^{2007}-3\)
\(\Leftrightarrow2A+3=3^{2007}\)
Lại có : \(2A+3=3^x\)
\(\Leftrightarrow3^x=3^{2007}\)
\(\Leftrightarrow x=2007\)
a, A=31 + 32 + 33 + ... + 32006
3A = 32 + 33 + 34 + ... + 32007
3A-A=( 32 + 33 + 34 +...+ 32007 ) - ( 31 + 32 + 33 +...+ 32006)
2A = 32007 - 3
\(\Rightarrow A=\frac{3^{2007}-3}{2}\)
b, 2A + 3 = 3x
\(\Leftrightarrow2.\left(\frac{3^{2007}-3}{2}\right)+3=3^x\)
\(\Leftrightarrow3^{2007}-3+3=3^x\)
\(\Leftrightarrow3^{2007}=3^x\)
\(\Leftrightarrow2007=x\)
Vậy x = 2007
![](https://rs.olm.vn/images/avt/0.png?1311)
3A=\(3^2+3^3+3^4+...+3^{2007}\)
3A-A=2A=\(3^{2007}-3\)
A=\(\frac{3^{2007}-3}{2}\)
b.
2A+3=3^x
3^2007-3+3=3^x
3^2007=3^x
vay x=2007
ta có : 3A=32+33+...+32007
3A-A=32+33+34+....+32007-3-32-33-...-32006
2A=32007-3
A=\(\frac{3^{2007}-3}{2}\)
b,
2A+3=3x
<=>32007-3+3=3x
<=> 32007=32007
<=> x = 2007
vậy x =2007
![](https://rs.olm.vn/images/avt/0.png?1311)
B=3+3^2+...+3^100.
3B=3.3+3^2.3+...+3^100.3
3B=3^2+3^3+...+3^101
3B-B=(3^2+3^3+...+3^101)-(3+3^2+...+3^100)
2B=3^101-3
Mà2B+3=3^n
Suy ra:3^101-3+3=3^n
3^n+3^101
Vậy n=101
Bài 1(b) làm tương tự,còn bài (a) thì bạn tự làm
![](https://rs.olm.vn/images/avt/0.png?1311)
B1:
a, 4(a - 3) - (a2 + 2a) - 5a
= 4a - 12 - a2 - 2a - 5a
= - a2 - 3a - 12
b, a(a+3) - [a(a-3) + 2(a+1)]
= a2 + 3a - [a2 - 3a + 2a + 2]
= a2 + 3a - a2 + 3a - 2a - 2
= 4a - 2
B2:
a, 3(x-4) - (2x+3) = 7-2x
=> 3x - 12 - 2x - 3 = 7 - 2x
=> x - 15 = 7 - 2x
=> 3x = 22
=> x = \(\frac{22}{3}\)
b, x(x2-3) - (x3+x) - 5x = (-8)2
=> x3 - 3x - x3 - x - 5x = 64
=> -9x = 64
=> x = \(\frac{-64}{9}\)
c, x(x-1) - 3(x+2) - x2 = -2.33
=> x2 - x - 3x - 6 - x2 = -54
=> -4x = -48
=> x = 12
![](https://rs.olm.vn/images/avt/0.png?1311)
a)C=3 mũ 2007 - 3 /2
b)x=rỗng
a, Có 3C = 3^2+3^3+....+3^2007
2C = 3C - C = (3^2+3^3+3^4+....+3^2007) - (3+3^2+3^3+....+3^2006)
= 3^2007 - 1
<=> C = (3^2007-2)/2