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Ta có : \(A=3+3^2+3^3+......+3^{2006}\)
=> \(3A=3^2+3^3+......+3^{2007}\)
=> \(3A-A=3^{2007}-3\)
=> \(2A=3^{2007}-3\)
=> \(A=\frac{3^{2007}-3}{2}\)
b) Ta có : \(2A=3^{2007}-3\) (theo ý a)
=> \(2A+3=3^{2007}\)
=> x = 2007
\(A=3+3^2+3^3+.........+3^{2006}\)
\(\Leftrightarrow3A=3^2+3^3+.........+3^{2007}\)
\(\Leftrightarrow3A-A=\left(3^2+3^3+.......+3^{2007}\right)-\left(3+3^2+.....+3^{2006}\right)\)
\(\Leftrightarrow2A=3^{2007}-3\)
\(\Leftrightarrow A=\frac{3^{2007}-3}{2}\)
\(\Leftrightarrow2A+3=3^{2007}\)
\(\Leftrightarrow3^x=3^{2007}\)
\(\Leftrightarrow x=2007\left(tm\right)\)
a) A = \(3^1+3^2+3^3+.....+3^{2006}\)
=> \(3A=3^2+3^3+3^4+....+3^{2007}\)
=> \(3A-A=3^{2007}-3^1=>2A=3^{2007}-3=>A=\dfrac{3^{2007}-3}{2}\)
b) Thay vào: 2A + 3 = \(3^x\) => \(3^{2007}-3\) + 3 = \(3^x\)
=> \(3^{2007}=3^x=>x=2007\)
Chúc bn học tốt, học 24h ơi chọn cho mình nha
A = 31+32+33+.....+32006
3A = 32+33+34+....+32007
2A = 3A - A = 32007-3
=> A = \(\frac{3^{2007}-3}{2}\)
Vì 2A = 32007-3
=> 2A + 3 = 32007
Mà 2A + 3 = 3x
=> 3x = 32007
=> x = 2007
1/ 3A-A=32007-3 <=> 2A=32007-3 => A=\(\frac{3^{2007}-3}{2}\)
2/ 2A=32007-3 => 2A+3=32007=3x => x=2007
3A = 32 + 33 + 34 + ... + 32007
3A - A = 32007 - 3
2A = 32007 - 3
A = ( 32007 - 3 ) : 2
ta có : 32007 - 3 + 3 = 3x
32007 = 3x
=> x = 2007
a)3A=3(31+32+33+...+32006)
3A=3.31+3.32+3.33+...+3.32006
3A=32+33+...+32007
3A-A=(32+33+...+32007)-(31+32+...+32006)
2A=32007-3
A=(32007-3):2
b)thay A vào ta được
32007-3+3=3x
32007=3x
=>x=2007
\(3A=3^2+3^3+3^4+...+3^{2007}\)
\(\Rightarrow3A-A=2A=3^{2007}-3^1=3.\left(3^{2006}-1\right)\)
Do đó \(A=\frac{3.\left(3^{2006}-1\right)}{2}\)
Ta có : \(2A+3=3^{2007}-3+3=3^{2007}=3^x\)
\(\Rightarrow x=2007\)
\(A=3^1+3^2+...+3^{2006}\)
\(3A=3^2+3^3+...+3^{2007}\)
\(3A-A=\left(3^2+3^3+...+3^{2007}\right)-\left(3^1+3^2+...+3^{2006}\right)\)
\(2A=3^{2007}-3\)
\(A=\frac{3^{2007}-3}{2}\)
\(2A+3=3^x\)
\(\left(3^{2007}-3\right)+3=3^x\)
\(3^{2007}+\left(-3\right)+3=3^x\)
\(3^{2007}+\left[\left(-3\right)+3\right]=3^x\)
\(\Rightarrow3^{2007}=3^x\)
\(\Rightarrow x=2007\)
a) A bằng 31+32+33+34+...+32006
3A bằng 3.(31+32+33+34+...+32006)
3A bằng 32+33+34+35+...+32007
3A-A bằng (32+33+34+35+...+32007) - (31+32+33+34+...+32006)
2A bằng 32007-31
A bằng (32007-3) : 2
b) 2A+3 bằng 3x
Thay 2A bằng 32007-3, ta có :
2A+3 bằng 3x
32007-3+3 bằng 3x
32007 bằng 3x
suy ra x bằng 2007
Vậy x bằng 2007
Bài 1 :
a, \(2^x+2^{x+1}=24\)
\(\Rightarrow2^x.1+2^x.2=24\)
\(\Rightarrow2^x\left(1+2\right)=24\)
\(\Rightarrow2^x=24\div3\)
\(\Rightarrow2^x=8=2^3\)
Vậy : x = 3
b, \(x^2-x=0\)
\(\Rightarrow x.x-x.1=0\)
\(\Rightarrow x\left(x-1\right)=0\)
Để : \(x\left(x-1\right)=0\Rightarrow x-1=0\)
\(\Rightarrow x=1\)
Vậy x = 1
Bài 2 :
a, \(Q=3+3^3+3^5+...+3^{101}\)
\(\Rightarrow9Q=3^3+3^5+3^7+...+3^{103}\)
\(\Rightarrow9Q-Q=\left(3^3+3^5+3^7+...+3^{103}\right)-\left(3+3^3+3^5+...+3^{101}\right)\)
\(\Rightarrow8Q=3^{103}-3\)
\(\Rightarrow Q=\frac{3^{103}-3}{8}\)
b, \(Q=3+3^3+3^5+...+3^{101}\)
\(\Rightarrow Q=\left(3+3^3+3^5\right)+\left(3^7+3^9+3^{11}\right)+...+\left(3^{97}+3^{99}+3^{101}\right)\)
\(\Rightarrow Q=\left(3+3^3+3^5\right)+3^6\left(3+3^3+3^5\right)+...+3^{96}\left(3+3^3+3^5\right)\)
\(\Rightarrow Q=1.273+3^6.273+...+3^{96}.273\)
\(\Rightarrow Q=\left(1+3^6+...+3^{96}\right)273\)
Vì : \(1+3^6+...+3^{96}\in N\) ; \(273=3.91\Rightarrow Q⋮91\)
Vậy ...
Phương An
soyeon_Tiểubàng giải
Võ Đông Anh Tuấn
Nguyễn Huy Tú
Trương Hồng Hạnh
Nguyễn Đình Dũng
Nguyễn Huy Thắng
Trần Quỳnh Mai
Nguyễn Thanh Vân
Nguyễn Thị Thu An
Hoàng Lê Bảo Ngọc
Silver bullet
Nguyễn Anh Duy
Lê Nguyên Hạo
Nguyễn Phương HÀ
1/A=1.21.22.23.24.25 câu 2 làm tương tự
A.2=2.22.23.24.25.26
A.2-A=(2.22.23.24.25.2 mũ 6)-(1.21.22.23.24.25)
A=26-1
3 A=1+3+32+33+...37
3.A=3+32+33+34...+38
2A=38-1
A=(38-1):2
\(A=3+3^2+3^3+.......+3^{2006}\)
\(\Leftrightarrow3A=3^2+3^3+......+3^{2007}\)
\(\Leftrightarrow3A-A=3^{2007}-3\)
\(\Leftrightarrow2A=3^{2007}-3\)
\(\Leftrightarrow A=\frac{3^{2007}-3}{2}\)
\(\Leftrightarrow2A+3=2^{2007}\)
\(\Leftrightarrow2^{2007}=2^x\)
\(\Leftrightarrow x=2007\)
\(3A=3^2+3^3+....+3^{2007}\)
\(3A-A=\left(3^2+3^3+...+3^{2007}\right)-\left(3+3^2+...+3^{2006}\right)\)
\(2A=3^{2007}-3\)
\(A=\frac{3^{2007}-3}{2}\)
b)\(2A+3=3^x\)
\(2A=3^x-3\)
Mà:\(2A=3^{2007}-3\)
\(\Rightarrow x=2007\)