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Ta có:\(5^n.2,5-30.5^n-6.5^n-1=5^n.\left(25-30-6\right)-1=5^n.\left(-11\right)-1\)-1
\(a, 10^{n+1} -6.10 ^n\)
= \(10^n (10-6)=4.10^n\)
\(B/ 2^{n+3} + 2^{n+2} - 2^{n+1} +2^n\)
= \(2^n (2^3+2^2-2+1)\)
= \(2^n (8+4-2+1)\)
\(= 11.2^n\)
\(C/ 90.10^k - 10^{k +2} + 10^{k +1} \)
\(= 10^k(90-2+1)\)
= \(89.10^k\)
\(D/ 2,5 . 5^{n-3} . 10+5^n -6 .5^{n-1}\)
\(= 5.5.5^{n-3} +5^n-6.5^{n-1}\)
= \(5^2 .5^{n-3}+5^n-6.5^{n-1} \)
= \(5^{n-3+2}+5^n -6.5^{n-1}\)
\(= 5^{n-1}(1+5-6)\)
= \(5^{n-1}.0\)
= 0
1.1
x=(3/5)^7:(3/5)^5=(3/5)^7-5=(3/5)62=6/5=1,2
1.2
x=5+7/10+3/10=5+10/10=5+1=6
1.3
x=\(\frac{18}{23}\) :\(\frac{6}{7}\) =\(\frac{18}{23}\) . \(\frac{7}{6}\) = \(\frac{21}{23}\)
\(2,5.5^{n-3}.10+5^n-6.5^{n-1}\)
=25.\(5^n\):3+\(5^n\)\(-\)6.\(5^n\):5
=\(\dfrac{25}{3}\).\(5^n\)+\(5^n\)\(-\)\(\dfrac{6}{5}\).\(5^n\)
=\(5^n\).\(\left(\dfrac{25}{3}+1-\dfrac{6}{5}\right)\)
=\(5^n\).\(\dfrac{158}{15}\)
c, \(\frac{-32}{-2^n}=4\)
\(\Rightarrow-2^n=-32:4\)
\(\Rightarrow-2^n=-8\)
\(\Rightarrow-2^n=-2^3\Rightarrow n=3\)
d, \(\frac{8}{2^n}=2\)
\(\Rightarrow2^n=8:2\)
\(\Rightarrow2^n=4\)
\(\Rightarrow2^n=2^2\Rightarrow n=2\)
e, \(\frac{25^3}{5^n}=25\)
\(\Rightarrow5^n=25^3:25\)
\(\Rightarrow5^n=25^2\)
\(\Rightarrow5^n=5^4\Rightarrow n=4\)
i , \(8^{10}:2^n=4^5\)
\(\Rightarrow2^n=8^{10}:4^5\)
\(\Rightarrow2^n=\left(2^3\right)^{10}:\left(2^2\right)^5\)
\(\Rightarrow2^n=2^{30}:2^{10}\)
\(\Rightarrow2^n=2^{20}\Rightarrow n=20\)
k, \(2^n.81^4=27^{10}\)
\(\Rightarrow2^n=27^{10}:81^4\)
\(\Rightarrow2^n=\left(3^3\right)^{10}:\left(3^4\right)^4\)
\(\Rightarrow2^n=3^{30}:3^{16}\)
\(\Rightarrow2^n=3^{14}\)
\(\Rightarrow2^n=4782969\)Không chia hết cho 2 nên ko có Gt n thỏa mãn
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,\(5^{-1}\times25^n=125 \)
= \(\frac{1}{5}\times25^n=125\)
= \(25^n=125\div\frac{1}{5}\)
= \(25^n=625\)
= \(25^n=25^2\)
\(\Rightarrow n=2\)