a) \(\frac{1}{8}\cdot16^n=2^n\)
giúp mk z ạ
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\(\frac{1}{8}.16^n=2^n\)
\(16^n=2^n:\frac{1}{8}\)
\(16^n=2^n.8\)
\(16^n=2^n.2^3\)
\(\left(2^4\right)^n=2^{n+3}\)
\(2^{4n}=2^{n+3}\)
\(\Rightarrow4n=n+3\)
\(4n-n=3\)
\(3n=3\)
\(n=1\)
\(KL:n=1\)
CHÚC BN HỌC TỐT!!!!!
\(\frac{1}{8}.16^n=2^n\)
\(\frac{16^n}{8}=2^n\)
\(\Rightarrow\frac{2^{4n}}{2^3}=2^n\)
\(2^{4n-3}=2^n\)
\(4n-3=1\)
\(\Rightarrow n=1\)
Vậy n = 1
\(\frac{1}{8}.16^n=2^n\Rightarrow\frac{1}{8}=\frac{2^n}{16^n}\Rightarrow\frac{1}{8}=\left(\frac{2}{16}\right)^n\Rightarrow\frac{1}{8}=\left(\frac{1}{8}\right)^n\Rightarrow n=1\)
vậy n=1
\(A=\frac{2n-1}{n+8}-\frac{n-14}{n+8}=\frac{2n-1-\left(n-14\right)}{n+8}=\frac{n+13}{n+8}\)
Để A thuộc Z thì \(n+13⋮n+8\Rightarrow n+13-\left(n+8\right)⋮n+8\)
\(\Rightarrow5⋮n+8\Rightarrow n+8\inƯ\left(5\right)=\left\{1;5;-1;-5\right\}\)
\(\Leftrightarrow n\in\left\{-7;-3;-9;-13\right\}\)
OK
\(\dfrac{1.2.4+2.4.8+4.8.16+8.16.32}{1.3.4+2.6.8+4.12.16+8.24.32}\)
\(=\dfrac{8.\left(1+8+4.16+16.32\right)}{12.\left(1+8+4.16+16.32\right)}\)
\(=\dfrac{8}{12}=\dfrac{2}{3}\)
\(=\dfrac{8+8\cdot8+8\cdot64+8\cdot512}{12+12\cdot8+12\cdot64+12\cdot512}=\dfrac{8}{12}=\dfrac{2}{3}\)