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a: \(\Leftrightarrow3^n:27^n=\dfrac{1}{9}\)
\(\Leftrightarrow\left(\dfrac{1}{9}\right)^n=\dfrac{1}{9}\)
hay n=1
b: \(\Leftrightarrow3^n\cdot3^2=3^8\)
=>n+2=8
hay n=6
c: \(\Leftrightarrow2^n\cdot\dfrac{9}{2}=9\cdot2^5\)
\(\Leftrightarrow2^n=2^6\)
hay n=6
d: \(\Leftrightarrow8^n=512\)
hay n=3
\(\frac{1}{9}\cdot3^4\cdot3^n=3^8\)
\(=>3^n=3^8:3^4:\frac{1}{9}\)
\(=>3^n=3^8:3^4\cdot9\)
\(=>3^n=3^8:3^4\cdot3^2\)
\(=>3^n=3^6\)
\(=>n=6\)
b) \(\frac{1}{9}.3^4.3^n=3^8\)
\(\Rightarrow\left(\frac{1}{3}\right)^2.3^4.3^n=3^8\)
\(\Rightarrow\frac{1}{3^2}.3^4.3^n=3^8\)
\(\Rightarrow3^2.3^n=3^8\)
\(\Rightarrow3^n=3^8:3^2\)
\(\Rightarrow3^n=3^6\)
\(\Rightarrow n=6\)
Vậy n = 6
a)Ta có: (2x - 1)6 = (2x - 1 )8
=> (2x - 1) . (2x - 1) . (2x - 1) . (2x - 1) . (2x - 1) . (2x - 1) = (2x - 1) . (2x - 1) . (2x - 1) . (2x - 1) . (2x - 1) . (2x - 1) . (2x - 1) . (2x - 1)
=> 2x - 1 = 0; 1
+ Nếu 2x - 1 = 0
=> 2x = 1
=> x = 1/2
+ Nếu 2x - 1 = 1
=> 2x = 2
=> x = 1
\(\frac{1}{2}\cdot2^n+4\cdot2^n=9\cdot2^5\)
\(=>\left(\frac{1}{2}+4\right)\cdot2^n=\frac{9}{2}\cdot2^6\)
\(=>\frac{9}{2}\cdot2^n=\frac{9}{2}\cdot2^6\)
\(=>2^n=2^6\)
\(=>n=6\)
Ta có:
\(\left(\frac{1}{2}\right)^{225}=\left[\left(\frac{1}{2}\right)^9\right]^{25}=\left(\frac{1}{516}\right)^{25}\)
\(\left(\frac{1}{3}\right)^{100}=\left[\left(\frac{1}{3}\right)^4\right]^{25}=\left(\frac{1}{81}\right)^{25}\)
\(\frac{1}{516}< \frac{1}{81}\Rightarrow\left(\frac{1}{516}\right)^{25}< \left(\frac{1}{81}\right)^{25}\Rightarrow\left(\frac{1}{2}\right)^{225}< \left(\frac{1}{3}\right)^{100}\)
Đặt A = 1 +3 +5 +...+(2n-1)
Số số hạng của A là : [(2n-1)-1]:2 +1 = n
Tổng A = [(2n-1)+1]xn:2=n2
=> n2=169
=>n2=132
=>n=13
a)\(\frac{27}{3^{n+1}}=3^2\Leftrightarrow\frac{27}{3^{n+1}}=9\)
\(\Leftrightarrow3^{n+1}=27\div9\)
\(\Leftrightarrow3^{n+1}=3\)
\(\Leftrightarrow3^{n+1}=3^1\)
\(\Leftrightarrow n+1=1\)
\(\Rightarrow n=1-1\)
\(\Rightarrow n=0\)
=> Tích
\(\frac{1}{9}\). 27n=3n
=> 27n :9 =3n
=> 27n: 3n = 9
(33)n : 3n =9
33n : 3n =9
32n = 9
32n= 32
với 2n = 2
=> n=1
vậy n=1
có lộn đề ko bn phải là phép chia chứ