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a, 5-1x 25n = 125 d, 25 < 5n:5 < 625
5-1 x 52n = 53 52 < 5n:5 < 54
=> -1+2n=3 => n=4
=>2n = 3--1
=>2n=4
=>n =2
a,\(5^{-1}\times25^n=125 \)
= \(\frac{1}{5}\times25^n=125\)
= \(25^n=125\div\frac{1}{5}\)
= \(25^n=625\)
= \(25^n=25^2\)
\(\Rightarrow n=2\)
\(A=\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}-\frac{5^{10}.7^3+25^5.49^2}{\left(125.7\right)^3+5^9.14^3}\)
\(=\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}-\frac{5^{10}.7^3+5^{10}.7^4}{5^9.7^3+5^9.2^3.7^3}\)
\(=\frac{2^{12}.3^4\left(3-1\right)}{2^{12}.3^5\left(3+1\right)}-\frac{5^{10}.7^3\left(1+7\right)}{5^9.7^3\left(1+2^3\right)}\)
\(=\frac{2}{12}-\frac{5.8}{9}=\frac{1}{6}-\frac{40}{9}=\frac{-77}{18}\)
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 )
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) \(\left(\frac{2}{5}-\frac{1}{2}\right)^2+\left(\frac{1}{2}+\frac{3}{5}\right)^2\)
\(=\left(\frac{-1}{10}\right)^2+\left(\frac{11}{10}\right)^2\)
\(=\frac{1}{100}+\frac{121}{100}=\frac{122}{100}=\frac{61}{50}\)
Bài 2:
a: =>50x+50=0
=>50x=-50
=>x=-1
b: \(\Leftrightarrow5^{2x-1}=5^3\)
=>2x-1=3
=>2x=4
=>x=2
c: \(\Leftrightarrow3^{x-1}+6\cdot3^{x-1}=7\cdot3^6\)
=>3^x-1=3^6
=>x-1=6
=>x=7
a) \(32< 2^x< 128\)
=> \(2^5< 2^x< 2^7\)
=> x = 6
b) \(2^{x-1}+4\cdot2^x=9\cdot2^5\)
=> \(2^{x-1}+2^2\cdot2^x=9\cdot2^5\)
=> \(2^{x-1}+2^{2+x}=9\cdot2^5\)
=> 9.2x-1 = 9.25
=> 2x-1 = \(\frac{9\cdot2^5}{9}=2^5\)
=> x - 1 = 5 => x = 6
c) \(9\cdot27\le3^x\le243\)
=> \(243\le3^x\le243\)
=> x = 5
d) Giống câu b)
e) \(3^{x-1}+5\cdot3^{x-2}=216\)
=> 8.3x-2 = 216
=> 3x-2 = 27
=> 3x-2 = 33
=> x - 2 = 3 => x = 5
f) 27x-3 = 9x+3
=> 27x-3 = 9x+3
=> (33)x-3 = (32)x+3
=> 33x-9 = 32x + 6
=> không thỏa mãn x vì x là phân số mà theo đề bài là số nguyên
g) x2019 = x => x2019 - x = 0 => x(x2018 - 1) = 0 => x = 0 hoặc x = 1
a)
\(2^5< 2^x< 2^7\)
\(5< x< 7\)
\(x=6\)
b)
\(2^{x-1}+2^2\cdot2^x=9\cdot2^5\)
\(2^{x-1}+2^{2+x}=9\cdot2^5\)
\(2^{x-1}\left(1+2^3\right)=9\cdot2^5\)
\(2^{x-1}\cdot9=9\cdot2^5\)
\(2^{x-1}=2^5\)
\(x-1=5\)
\(x=6\)
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)
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. \(\Rightarrow5^2<5^{n-1}<5^4\)
\(\Rightarrow n-1=3\)
\(\Rightarrow n=4\).