rút gọn: A= 1/3^2+1/3^3+1/3^4+....+1/3^120
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
13 tháng 5 2015
1:
I2x+3I = 5
=> 2x+3 = 5 hoặc 2x+3 = -5
=> 2x = 5 - 3 hoặc 2x = -5 - 3
=> 2x = 2 hoặc 2x = -8
=> x = 2 hoặc x = -4
2:
B = 1/2.2/3.3/4.4/5.....27/28
= 1.2.3.4.5.6...27/2.3.4.5.6...28
= 1/28
3:
2A = 2(1+1/2+1/2^2+1/2^3+1/2^4+...+1/2^2015) = 2+1+1/2+1/2^2+1/2^3+...+1/2^2014
=> 2A-A = ( 2+1+1/2+1/2^2+1/2^3+...+1/2^2014)-(1+1/2+1/2^2+1/2^3+...+1/2^2015)
=> A = 2-1/2^2015
21 tháng 1 2024
Bài 2:
a: \(\dfrac{1}{4};\dfrac{2}{5}\)
\(\dfrac{1}{4}=\dfrac{1\cdot5}{4\cdot5}=\dfrac{5}{20}\)
\(\dfrac{2}{5}=\dfrac{2\cdot4}{5\cdot4}=\dfrac{8}{20}\)
\(\dfrac{2}{3};\dfrac{7}{8}\)
\(\dfrac{2}{3}=\dfrac{2\cdot8}{3\cdot8}=\dfrac{16}{24}\)
\(\dfrac{7}{8}=\dfrac{7\cdot3}{8\cdot3}=\dfrac{21}{24}\)
\(\dfrac{3}{4};\dfrac{5}{6}\)
\(\dfrac{3}{4}=\dfrac{3\cdot3}{4\cdot3}=\dfrac{9}{12}\)
\(\dfrac{5}{6}=\dfrac{5\cdot2}{6\cdot2}=\dfrac{10}{12}\)
b: \(\dfrac{1}{3};\dfrac{7}{9}\)
\(\dfrac{1}{3}=\dfrac{1\cdot3}{3\cdot3}=\dfrac{3}{9}\)
\(\dfrac{7}{9}=\dfrac{7\cdot1}{9\cdot1}=\dfrac{7}{9}\)
\(\dfrac{3}{4};\dfrac{9}{24}\)
\(\dfrac{3}{4}=\dfrac{3\cdot6}{4\cdot6}=\dfrac{18}{24}\)
\(\dfrac{9}{24}=\dfrac{9\cdot1}{24\cdot1}=\dfrac{9}{24}\)
\(\dfrac{7}{10};\dfrac{19}{30}\)
\(\dfrac{7}{10}=\dfrac{7\cdot3}{10\cdot3}=\dfrac{21}{30}\)
\(\dfrac{19}{30}=\dfrac{19\cdot1}{30\cdot1}=\dfrac{19}{30}\)
Bài 1:
\(\dfrac{36}{108}=\dfrac{36:36}{108:36}=\dfrac{1}{3}\)
\(\dfrac{28}{30}=\dfrac{28:2}{30:2}=\dfrac{14}{15}\)
\(\dfrac{42}{98}=\dfrac{42:14}{98:14}=\dfrac{3}{7}\)
\(\dfrac{15}{120}=\dfrac{15:15}{120:15}=\dfrac{1}{8}\)
\(\dfrac{84}{364}=\dfrac{84:28}{364:28}=\dfrac{3}{13}\)
\(\dfrac{120}{100}=\dfrac{120:20}{100:20}=\dfrac{6}{5}\)
\(\dfrac{418}{38}=\dfrac{418:38}{38:38}=\dfrac{11}{1}=11\)
\(\dfrac{96}{1056}=\dfrac{96:96}{1056:96}=\dfrac{1}{11}\)
\(\dfrac{3838}{4040}=\dfrac{3838:101}{4040:101}=\dfrac{38}{40}=\dfrac{38:2}{40:2}=\dfrac{19}{20}\)
\(\dfrac{119119}{123123}=\dfrac{119119:1001}{123123:1001}=\dfrac{119}{123}\)
PB
4
20 tháng 11 2018
\(A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{63}+1\right).\)
\(=\frac{\left(3+1\right)\left(3-1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{63}+1\right)}{2}\)
\(=\frac{\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{63}+1\right)}{2}\)
\(=\frac{\left(3^{64}-1\right)\left(3^{63}+1\right)}{2}\left(\text{bn xem lại chỗ }3^{63}\text{ nhé!! ko thì ko lm đc tiếp đâu}\right)\)
30 tháng 9 2023
\(S=1+3^2+3^4+...+3^{2022}\)
\(3^2S=9S=3^2+3^4+3^6+...+3^{2024}\)
\(S=\dfrac{9S-S}{8}=\left(3^{2024}-1\right):8\)
d, không đáp án nào đúng
AH
Akai Haruma
Giáo viên
30 tháng 9 2023
Lời giải:
$S=1+3^2+3^4+....+3^{2022}$
$9S=3^2S=3^2+3^4+3^6+...+3^{2024}$
$\Rightarrow 9S-S=3^{2024}-1$
$\Rightarrow S=\frac{3^{2024}-1}{8}$
Đáp án D.
AT
29 tháng 5 2021
1.\(A=\left(\sqrt{3}+1\right)\sqrt{\dfrac{14-6\sqrt{3}}{5+\sqrt{3}}}=\left(\sqrt{3}+1\right)\sqrt{\dfrac{\left(14-6\sqrt{3}\right)\left(5-\sqrt{3}\right)}{\left(5+\sqrt{3}\right)\left(5-\sqrt{3}\right)}}\)
\(=\left(\sqrt{3}+1\right)\sqrt{\dfrac{44\left(2-\sqrt{3}\right)}{22}}=\left(\sqrt{3}+1\right)\sqrt{4-2\sqrt{3}}=\left(\sqrt{3}+1\right)\sqrt{\left(\sqrt{3}-1\right)^2}\)
\(=\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)=2\)
AT
29 tháng 5 2021
2.1.a) \(x^2=\left(x-1\right)\left(3x-2\right)\Leftrightarrow x^2=3x^2-5x+2\Leftrightarrow2x^2-5x+2=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x-1\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{2}\end{matrix}\right.\)
b) \(9x^4+5x^2-4=0\Leftrightarrow9x^4+9x^2-4x^2-4=0\)
\(\Leftrightarrow9x^2\left(x^2+1\right)-4\left(x^2+1\right)=0\Leftrightarrow\left(x^2+1\right)\left(9x^2-4\right)=0\)
mà \(x^2+1>0\Rightarrow9x^2=4\Rightarrow x^2=\dfrac{4}{9}\Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{2}{3}\end{matrix}\right.\)
2) Gọi số xe lúc đầu của đội là a(xe) \(\left(a\in N,a>0\right)\)
Theo đề,ta có: \(\left(a-2\right)\left(\dfrac{120}{a}+3\right)=120\Leftrightarrow120+3a-\dfrac{240}{a}-6=120\)
\(\Leftrightarrow\dfrac{3a^2-6a-240}{a}=0\Rightarrow3a^2-6a-240=0\Rightarrow a^2-2a-80=0\)
\(\Leftrightarrow\left(a+8\right)\left(a-10\right)=0\) mà \(a>0\Rightarrow a=10\)
NT
2
G
9 tháng 7 2019
\(A=8\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)-81^{16}\)
\(A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)-81\)
\(A=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)-81^{16}\)
\(A=\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)-81^{16}\)
\(A=\left(3^{16}-1\right)\left(3^{16}+1\right)-81^{16}\)
\(A=3^{32}-1-81^{16}\)
NN
9 tháng 7 2019
A = 8.( 32 + 1 ).( 34 + 1 ).( 38 + 1).( 316 + 1 ) - 8116
A = ( 32 - 1).( 32 + 1 ).( 34 + 1 ).( 38 + 1).( 316 + 1 ) - 8116
A = ( 34 - 1 ).( 34 + 1 ).( 38 + 1).( 316 + 1 ) - 8116
A = ( 38 - 1 ).( 38 + 1).( 316 + 1 ) - 8116
A = ( 316 - 1 ).( 316 + 1 ) - 8116
A = ( 332 - 1 ) - 8116
A = -364
HT
1
14 tháng 7 2021
`(3 + 1) (3^2 + 1) (3^4 + 1) … (3^64 + 1)`
`=(2.(3 + 1) (3^2 + 1) (3^4 + 1) … (3^64 + 1))/2`
`=((3-1)(3 + 1) (3^2 + 1) (3^4 + 1) … (3^64 + 1))/2`
`=((3^2-1)(3^2+1)(3^4+1).....(3^64+1))/2`
`=((3^4-1)(3^4+1)(3^64+1))/2`
`=((3^16-1)....(3^64+1))/2`
`=(3^128-1)/2`
RD
16 tháng 10 2018
Xét số hạng tổng quát:\(\dfrac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}\)
\(=\dfrac{1}{\sqrt{n\left(n+1\right)}\left(\sqrt{n+1}+\sqrt{n}\right)}\)
\(=\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n\left(n+1\right)}}=\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\)
\(\Rightarrow A=1-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{120}}-\dfrac{1}{\sqrt{121}}\)
\(A=1-\dfrac{1}{11}=\dfrac{10}{11}\)
12 tháng 8 2021
Ta có: \(A=\left(\sqrt{6}+\sqrt{10}\right)\cdot\sqrt{4-\sqrt{15}}\)
\(=\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)\)
=5-3=2
\(A=\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+...+\frac{1}{3^{120}}\)
\(3A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{119}}\)
\(3A-A=\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{119}}\right)-\left(\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+...+\frac{1}{3^{120}}\right)\)
\(=\frac{1}{3}-\frac{1}{3^{120}}\)
\(\Rightarrow A=\frac{1}{6}-\frac{1}{2.3^{120}}\)