cho a= 3^2+3^3+3^4+...+3^2005+3^2006 .
tìm x để:2A+3=3x
1.Tìm x, biết:
1+3+5+...+x=81
2.A= 3+3^2+3^3+...+3^2006
a, Tính A.
b, Tìm x để: 2A+3=3x
a) Ta có : \(3A=3^{2007}+3^{2006}+...+3^3+3^2\)
A = \(3^{2006}+...+3^3+3^2+3\)
\(\Rightarrow2A=3^{2007}-3\)
\(\Rightarrow A=\frac{3^{2007}-3}{2}\)
b) Ta có \(2A=3^{2007}-3\)\(\Rightarrow2A+3=3^{2007}\)
Theo bài ta có: \(2A+3=3x\)
\(\Rightarrow3^{2007}=3x\)
\(\Rightarrow3.3^{2006}=3x\)
\(\Rightarrow x=3^{2006}\)
cho A=3^1 +3^2 +3^3+....+3^2006 Thu gọn A b,tìm x để 2A+3 =3^x
3A=3^2+3^3+...+3^2007
=>3a-A=(3^2+3^3+...+3^2007)-(3^1+3^2+...+3^2006)
=>2A=3^2007-3^1=3^2007-3
=>2A+3=3^2007-3+3=3^2007=3^x
=>x=2007
Cho A=3^1+3^2+3^3........+3^2006
A:thu gon A
B tìm x để 2A+3=3^x
CHO A = 3^+3^2+3^3+........+3^2006
a) thu gọn A
b) tìm X để 2A+3=3^x
cho A=3^1 + 3^2 +..........+3^2006
a; thu gọn A
b; tìm x để 2A+3=3^x
Cho A=3^1+3^2+3^3+...+3^2006
Thu gọn A
Tìm x để 2A+3=3^x
\(A=3+3^2+3^3+...+3^{2006}\)
\(\Leftrightarrow3A=3\left(3+3^2+3^3+....+3^{2006}\right)\)
\(\Leftrightarrow3A=3^2+3^3+3^4+....+3^{2007}\)
\(\Leftrightarrow3A-A=\left(3^2+3^3+3^4+...+3^{2007}\right)-\left(3+3^2+3^3+...+3^{2006}\right)\)
\(\Leftrightarrow2A=3^{2007}-3\)
\(\Leftrightarrow A=\frac{3^{2007}-3}{2}\)
Ta có \(2A=3^{2007}-3\)
=> 2A+3=\(3^{2007}-3+3=3^{2007}\)
=> x=2007
A=3^1+3^2+3^3+....+3^2006
3A=3^2+3^3+...+3^2007
=>2A=3^2007-3
=>2A+3=3^x
3^2007-3+3=3^x
3^2007=3^x
=>x=2007
Vậy x=2007
Cho A = 31+ 32+ 33+....+ 32006. Tìm x để 2A+3=3x
Cho A=31+32+33+34+.......+32006
Tìm x để 2A+3=3x
\(A=3+3^2+3^3+...3^{2006}\)
\(3A=3^2+3^3+...+3^{2007}\)
\(3A-A=\left(3^2-3^2\right)+....+\left(3^{2006}-3^{2006}\right)+3^{2007}-3\)
\(2A=3^{2007}-3\Rightarrow2A+3=3^{2007}-3+3=3^{2007}=3^x\)
Vậy x = 2007
A=3+3^2+....+3^2006
=>3A=3^2+3^3+....+3^2007
=>3A-A=(3^2+3^3+....+3^2007)-(3+3^2+....+3^2006)
=>2A=3^2007-3
khi đó 2A+3=3^2007-3+3=3^2007=3^x
=>x=2007
ta co: A=31+32+33+...+32006
3A=3(31+32+33+...+32006)
3A=32+33+34+...+32007
3A-A=(32+33+34+...+32007)-(31+32+33+...+32006)
2A=32007-3
=>2A+3=32007-3+3=32007=3X
=>x=2007
Cho A= 31+32+33+....+32006
tìm x để 2A+3=3x
Câu hỏi của Yuki Yudai - Toán lớp 6 - Học toán với OnlineMath
Em tham khảo nhé!
Ta có:
A=\(3^1+3^2+....+3^{2006}\)
=>3A=\(3^2+3^3+3^4+...+3^{2007}\)
=>3A-A=\(\left(3^2+3^3+...+3^{2007}\right)-\left(3^1+3^2+...+3^{2006}\right)\)
2A=\(3^{2007}-3\)
=>2A+3=\(3^x\)
<=>\(3^{2007}-3+3\)=\(3^x\)
<=>\(3^{2007}=3^x\)
=>x=2007
Vậy x=2007 thì...
dễ này mình giai cho
tính; A= 3^1+3^2 +....+3^2006
3A =3^2+3^3+....+3^2007
3A-A =2A=[3^2+3^3+...+3^2007]-[3^1+3^2+...+3^2006]
2A=3^2007-3^1
Vậy 2A+3=3^x
3^2007-3+3=3^x
Nên 3^x = 3^2007
x= 2007