Tìm số nguyên n ,biết:
a,\(5^{-1}\times25^n=125\)
b,\(3^{-1}\times3^n+6\times3^{n-1}=7\times3^6\)
c,\(3^4<\frac{1}{9}\times27^n<3^{10}\)
d,\(25<5^n:5<625\)
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22 tháng 1 2017
\(A=\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}-\frac{5^{10}.7^3+25^5.49^2}{\left(125.7\right)^3+5^9.14^3}\)
\(=\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}-\frac{5^{10}.7^3+5^{10}.7^4}{5^9.7^3+5^9.2^3.7^3}\)
\(=\frac{2^{12}.3^4\left(3-1\right)}{2^{12}.3^5\left(3+1\right)}-\frac{5^{10}.7^3\left(1+7\right)}{5^9.7^3\left(1+2^3\right)}\)
\(=\frac{2}{12}-\frac{5.8}{9}=\frac{1}{6}-\frac{40}{9}=\frac{-77}{18}\)
b ) 3n+2 - 2n+2 + 3n - 2n
= ( 3n+2 + 3n ) - ( 2n+2 + 2n )
= 3n ( 32 + 1 ) - 2n ( 22 + 1 )
= 3n.10 - 2n-1.2.5
= 3n.10 - 2n-1.10
= ( 3n - 2n-1 ).10 chia hết cho 10 ( đpcm )
NC
16 tháng 8 2020
a) \(\frac{7^3.5^8}{49.25^4}=\frac{7^3.5^8}{7^2.5^8}=7\)
b) \(\frac{3^9.25.5^3}{15.625.3^8}=\frac{3^9.5^2.5^3}{3.5.5^4.3^8}=\frac{3^9.5^5}{3^9.5^5}=1\)
c) \(\frac{2^{50}.3^{61}+2^{90}.3^{16}}{2^{51}.3^{61}+2^{91}.3^{16}}=\frac{2^{50}.3^{16}\left(3^{45}+2^{40}\right)}{2^{51}.3^{16}\left(3^{45}+2^{40}\right)}=\frac{1}{2}\)
d) \(\left(\frac{2}{5}-\frac{1}{2}\right)^2+\left(\frac{1}{2}+\frac{3}{5}\right)^2\)
\(=\left(\frac{-1}{10}\right)^2+\left(\frac{11}{10}\right)^2\)
\(=\frac{1}{100}+\frac{121}{100}=\frac{122}{100}=\frac{61}{50}\)
T
14 tháng 1 2017
a) A = \(\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}-\frac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}\)
=> A = \(\frac{2^{12}.3^5-\left(2^2\right)^6.\left(3^2\right)^2}{\left(2^2\right)^6.3^6+\left(2^3\right)^4.3^5}-\frac{5^{10}.7^3-\left(5^2\right)^5.\left(7^2\right)^2}{125^3.7^3+5^9.\left(2.7\right)^3}\)
=> A = \(\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}-\frac{5^{10}.7^3-5^{10}.7^4}{\left(5^3\right)^3.7^3+5^9.2^3.7^3}\)
=> A = \(\frac{2^{12}.3^4\left(3-1\right)}{2^{12}.3^5\left(3+1\right)}-\frac{5^{10}.7^3\left(1-7\right)}{5^9.7^3+5^9.2^3.7^3}\)
=> A = \(\frac{3-1}{3\left(3+1\right)}-\frac{5^{10}.7^3.\left(-6\right)}{5^9.7^3\left(1+2^3\right)}\)
=> A = \(\frac{2}{3.4}-\frac{5.\left(-6\right)}{9}\)
A = \(\frac{1}{3.2}-\frac{-30}{9}\)
A = \(\frac{1}{6}-\frac{-10}{3}\)
A = \(\frac{1}{6}+\frac{10}{3}=\frac{1}{6}+\frac{20}{6}=\frac{21}{6}\)
=> A = \(\frac{7}{2}=3\frac{1}{2}\)
vậy A = \(3\frac{1}{2}\)
b) ta có:
3n+2-2n+2+3n-2n = (3n+2+3n) - (2n+2-2n)
= 3n(9+1) - 2n(4+1)
= 3n.10 - 2n.5
ta thấy: 3n.10 \(⋮\) 10
2n là một số chẵn mà 1 số chẵn nhân vs 5 luôn ra kết quả có tận cùng bằng 0 => 2n.5 \(⋮\) 10
=> 3n. 10 - 2n.5 \(⋮\) 10
=> 3n+2-2n+2+3n-2n \(⋮\) 10 vs mọi số nguyên dương n ( đpcm)
9 tháng 11 2014
\(\frac{1}{8}=12,5\%\) ; \(\frac{1}{16}=6,25\%\) ; \(\frac{1}{2}=50\%\) ; \(\frac{1}{4}=25\%\)
Thay vào trên mà tính.
= \(1+\left(\frac{3\left(1x2+2x4x2\right)}{3\left(5+5x3x25\right)}+1\right)-\left(1+\frac{18}{54}\right)-1\) = \(\frac{18}{380}-\frac{18}{54}\)
25<5^n:5<625
=>5^2<5^n-1<5^4
=>2<n-1<4
=>n-1=3
=>n=4
a. \(\Rightarrow5^{-1}.5^{2n}=5^3\)
\(\Rightarrow5^{2n-1}=5^3\)
=> 2n-1=3
=> 2n=4
=> n=2
b. \(\Rightarrow3^{n-1}+6.3^{n-1}=7.3^6\)
\(\Rightarrow\left(1+6\right).3^{n-1}=7.3^6\)
\(\Rightarrow7.3^{n-1}=7.3^6\)
=> n-1=6
=> n=7
c. \(\Rightarrow3^4<3^{-2}.3^{3n}<3^{10}\)
\(\Rightarrow3^4<3^{3n-2}<3^{10}\)
\(\Rightarrow3n-2\in\left\{5;6;7;8;9\right\}\)
\(\Rightarrow3n\in\left\{7;8;9;10;11\right\}\)
\(\text{Mà n là số nguyên}\Rightarrow n=3\).
d. \(\Rightarrow5^2<5^{n-1}<5^4\)
\(\Rightarrow n-1=3\)
\(\Rightarrow n=4\).