tìm n biết: \(\frac{1}{4}.\frac{2}{6}.\frac{3}{8}.\frac{4}{10}.....\frac{30}{64}=2^n\)
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PT
26 tháng 1 2017
1/4.2/6.3/8.4/10.........30/62.31/64=4x
=1/2.1/2.1/2.1/2.............1/2.1/64=4^x
=1/2^30.1/2^6=4^x
=1/2^36=4^x
=1/4^18=4^x
=>x=-18
Y
20 tháng 7 2019
1. Tìm x, biết :
a. ( x - \(\frac{3}{4}\)) \(^2\)= 0
=> x - \(\frac{3}{4}\)= 0
=> x = 0 + \(\frac{3}{4}\)
=> x = \(\frac{3}{4}\)
b. ( x + \(\frac{1}{2}\)) \(^2\)= \(\frac{9}{64}\)
=> ( x + \(\frac{1}{2}\)) \(^2\)= ( \(\frac{3}{8}\)) \(^2\)
=> x + \(\frac{1}{2}\)= \(\frac{3}{8}\)
=> x = \(\frac{3}{8}\)- \(\frac{1}{2}\)
=> x = \(\frac{-1}{8}\)
c. \(\frac{\left(-2\right)^x}{16}=-8\)
=> \(\frac{\left(-2\right)^x}{16}=\frac{-8}{1}=\frac{-128}{16}\)
=> ( -2)\(^x\)= -128
=> ( -2 ) \(^x\)= ( -2) \(^7\)
=> x = 7
H
17 tháng 3 2020
\(\frac{1}{4}.\frac{2}{6}.\frac{3}{8}.\frac{4}{10}.\frac{5}{12}.....\frac{30}{62}.\frac{31}{64}=2^x\)
=>\(\frac{1}{2.2}.\frac{2}{2.3}.\frac{3}{2.4}.\frac{4}{2.5}.\frac{5}{2.6}....\frac{30}{2.31}.\frac{31}{2.32}=2^x\)
=>\(\frac{1.2.3.4.5....30.31}{2.2.2.3.2.4.2.5.2.6...2.31.2.32}=2^x\)
=>\(\frac{2.3.4.5...30.31}{2^{31}.32.\left(2.3.4.5...31\right)}=2^x\)
=>\(\frac{1}{2^{31}.2^5}=2^x\)
=>\(\frac{1}{2^{36}}=2^x\)
=> x=36
Vậy x=36
Chúc bn học tốt nhé!
3 tháng 7 2017
\(\frac{1}{4}.\frac{2}{6}.\frac{3}{8}.\frac{4}{10}...\frac{30}{62}.\frac{31}{64}\)
\(=\frac{1.2.3.4...30.31}{2.2.2.3.2.4.2.5...2.31.2.32}\)
\(=\frac{1.2.3.4...30.31}{2^{31}.\left(2.3.4.5...31\right).32}\)
\(=\frac{1}{2^{31}.32}\)
\(=\frac{1}{2^{31}.2^5}\)
\(=\frac{1}{2^{36}}\)
9 tháng 5 2016
\(\frac{1}{4}.\frac{2}{6}.\frac{3}{8}.\frac{4}{10}.\frac{5}{12}.....\frac{30}{62}.\frac{31}{64}\)
\(=\frac{1.2.3.4....30.31}{4.6.8.10.12....62.64}\)
\(=\frac{1.\left(2.3.4.5....30.31\right)}{2.\left(2.3.4....30.31\right).64}\)
\(=\frac{1}{2.64}\)
\(=\frac{1}{128}\)
12 tháng 5 2016
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+....+\frac{2}{n\left(n+1\right)}=\frac{1999}{2001}\) <=>\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+.......+\frac{2}{n\left(n+1\right)}=\frac{1999}{2001}\)
<=>\(2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{n\left(n+1\right)}\right)=\frac{1999}{2001}\)
<=>\(2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}+\frac{1}{4}+.....\frac{1}{n}-\frac{1}{n-1}\right)=\frac{1999}{2001}\)
<=>\(2\left(\frac{1}{2}-\frac{1}{n+1}\right)=\frac{1999}{2001}\)
<=>\(\frac{1}{2}-\frac{1}{n+1}=\frac{1999}{4002}\)
<=>\(\frac{1}{n+1}=\frac{1}{2001}\)
<=>n+1 =2001
<=>n = 2000
12 tháng 5 2016
ta có:
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{n\left(n+1\right)}=\frac{1999}{2001}\)
\(\frac{1}{2}\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{n\left(n+1\right)}\right)=\frac{1}{2}.\frac{1999}{2001}\)
\(\frac{1}{2.3}+\frac{1}{2.6}+\frac{1}{2.10}+...+\frac{1}{n\left(n+1\right)}=\frac{1999}{4002}\)
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{n\left(n+1\right)}=\frac{1999}{4002}\)
\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{n\left(n+1\right)}=\frac{1999}{4002}\)
\(\frac{1}{2}-\frac{1}{n+1}=\frac{1999}{4002}\)
\(\frac{1}{n+1}=\frac{1}{2}-\frac{1999}{4002}\)
\(\frac{1}{n+1}=\frac{1}{2001}\)
=>\(n+1=2001\)
=>\(n=2000\)
Có thể đề là: \(\frac{1}{4}.\frac{2}{6}.\frac{3}{8}.\frac{4}{10}.....\frac{31}{64}=2^n\)
=> \(\frac{1.2.3.4....31}{\left(2.2\right)\left(2.3\right).\left(2.3\right)\left(2.4\right)\left(2.5\right)...\left(2.31\right).\left(2.32\right)}=2^n\)
=> \(\frac{1.2.3.4...31}{2^{16}.\left(2.3.4.5..31.32\right)}=2^n\) => \(\frac{1}{2^{16}.32}=2^n\) => 216.25.2n = 1
=> 231+n = 1 = 20 => 31 + n = 0 => n = -31
\(=\frac{1.2.3.....30}{4.6.8....64}=\frac{1}{2.2...2.64}=\frac{1}{2^{30}.2^6}=\frac{1}{2^{36}}\) ( 30 số 2)
=> 2^n = 1/2^36
=> n = -36