tìm số tự nhiên n bt
- \(\frac{1}{2}^n\)=\(\frac{1}{32}\)
- \(\frac{343}{125}\)=\(\frac{7}{5}^n\)
- \(\frac{16}{2^n}\)=2
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27 tháng 7 2016
a) \(\left(\frac{1}{2}\right)^m=\frac{1}{32}\)
\(=>\left(\frac{1}{2}\right)^m=\frac{1^5}{2^5}\)
\(=>\left(\frac{1}{2}\right)^m=\left(\frac{1}{2}\right)^5\)
\(=>m=5\)
b) \(\frac{343}{125}=\left(\frac{7}{5}\right)^n\)
\(=>\frac{7^3}{5^3}=\left(\frac{7}{5}\right)^n\)
\(=>\left(\frac{7}{5}\right)^3=\left(\frac{7}{5}\right)^n\)
\(=>n=3\)
27 tháng 7 2016
a) \(\left(\frac{1}{2}\right)^m=\frac{1}{32}\)
\(\Rightarrow\left(\frac{1}{2}\right)^m=\left(\frac{1}{2}\right)^5\)
=> m =5
b) \(\frac{343}{125}=\left(\frac{7}{5}\right)^n\)
\(\Rightarrow\left(\frac{7}{5}\right)^3=\left(\frac{7}{5}\right)^n\)
=> n = 3
IM
14 tháng 9 2016
a) \(\left(\frac{1}{2}\right)^m=\frac{1}{32}\)
\(\Rightarrow\left(\frac{1}{2}\right)^m=\left(\frac{1}{2}\right)^5\)
=> m = 5
Vậy m = 5
b) \(\frac{343}{125}=\left(\frac{7}{5}\right)^n\)
\(\Rightarrow\left(\frac{7}{5}\right)^3=\left(\frac{7}{5}\right)^n\)
=> n = 3
Vậy n = 3
26 tháng 6 2015
a, ( 1/2 ) ^ m = ( 1/2) ^5
=> m = 5
b, ( 7/5) ^n = 343 / 125
=> ( 7/5)^n = (7/5) ^ 3
=> n = 3
Đúng cho tui nha
MT
26 tháng 6 2015
\(a.\left(\frac{1}{2}\right)^m=\frac{1}{32}\)
\(\left(\frac{1}{2}\right)^m=\frac{1^5}{2^5}\)
\(\left(\frac{1}{2}\right)^m=\left(\frac{1}{2}\right)^5\)
=>m=5
\(b.\frac{343}{125}=\left(\frac{7}{5}\right)^n\)
\(\frac{7^3}{5^3}=\left(\frac{7}{5}\right)^n\)
\(\left(\frac{7}{5}\right)^3=\left(\frac{7}{5}\right)^n\)
=>n=3
OI
1
10 tháng 7 2019
a. \(\left(\frac{-1}{5}\right)^n=\frac{-1}{125}\)
<=> \(\left(\frac{-1}{5}\right)^n=\left(\frac{-1}{5}\right)^3\)
<=> n = 3
b. \(\left(\frac{-2}{11}\right)^m=\frac{4}{121}\)
<=> \(\left(\frac{-2}{11}\right)^m=\left(\frac{2}{11}\right)^2\)
<=> m = 2
c. 72n + 72n+2 = 2450
<=> 72n + 72n . 72 = 2450
<=> 72n.(1+72) = 2450
<=> 72n = 72
<=> 2n = 2
<=> n = 1
MT
10 tháng 7 2015
a)\(\left(\frac{1}{2}\right)^m=\frac{1}{32}\)
\(\left(\frac{1}{2}\right)^m=\frac{1^5}{2^5}\)
\(\left(\frac{1}{2}\right)^m=\left(\frac{1}{2}\right)^5\)
=> m=5
b)\(\frac{343}{125}=\left(\frac{7}{5}\right)^n\)
\(\frac{7^3}{5^3}=\left(\frac{7}{5}\right)^n\)
\(\left(\frac{7}{5}\right)^3=\left(\frac{7}{5}\right)^n\)
=>n=3
20 tháng 4 2017
\(\frac{4}{3.5}+\frac{8}{5.9}+\frac{12}{9.15}+...+\frac{32}{n\left(n+16\right)}=\frac{16}{25}\)
\(2\left(\frac{1}{3}-\frac{1}{5}\right)+2\left(\frac{1}{5}-\frac{1}{9}\right)+2\left(\frac{1}{9}-\frac{1}{15}\right)+...+2\left(\frac{1}{n}-\frac{1}{n+16}\right)=\frac{16}{25}\)
\(2\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{15}+...+\frac{1}{n}-\frac{1}{n+16}\right)=\frac{16}{25}\)
\(2\left(\frac{1}{3}-\frac{1}{n+16}\right)=\frac{16}{25}\)
\(\frac{1}{3}-\frac{1}{n+16}=\frac{8}{25}\)
\(\frac{1}{n+16}=\frac{1}{75}\)
\(\Rightarrow n+16=75\)
\(\Rightarrow n=59\)
NT
3
1. \(\left(\frac{1}{2}\right)^n=\frac{1}{32}\)
\(\left(\frac{1}{2}\right)^n=\frac{1^5}{2^5}\)
\(\left(\frac{1}{2}\right)^n=\left(\frac{1}{2}\right)^5\)
Vậy \(n=5\)
2. \(\frac{343}{125}=\left(\frac{7}{5}\right)^n\)
\(\frac{7^3}{5^3}=\left(\frac{7}{5}\right)^n\)
\(\left(\frac{7}{5}\right)^3=\left(\frac{7}{5}\right)^n\)
Vậy \(n=3\)
3. \(\frac{16}{2^n}=2\)
\(2^n=\frac{16}{2}\)
\(2^n=8=2^3\)
Vậy \(n=3\)
1. (1/2)2 = 1/32 <=> (21)n = (25)n <=> 1.n = 5.1 <=> n = 5
=> n = 5
2) 343/125 = (7/5)n <=> (7/5)3 = (7/5)n <=> 3 = n
=> n = 3
3) 16/2n = 2 <=> 16.2n <=> 2n = 2/16 <=> 2n = 1/8 <=> 2n = 8 <=> 2n = 23 <=> n = 3
=> n = 3