x = -11 do ban

a) \(\dfrac{1}{9}.27^n=3^n\)

\(\Leftrightarrow\dfrac{1}{9}=3^n:27^n\)

\(\Leftrightarrow\dfrac{1}{9}=\left(\dfrac{3}{27}\right)^n\)

\(\Leftrightarrow\dfrac{1}{9}=\left(\dfrac{1}{9}\right)^n\)

\(\Leftrightarrow n=1\)

b) \(3^{-2}.3^4.3^n=3^7\)

\(\Leftrightarrow3^2.3^n=3^7\)

\(\Leftrightarrow3^n=3^7:3^2\)

\(\Leftrightarrow3^n=3^5\)

\(\Leftrightarrow n=5\)

c) \(32^{-n}.16^n=2048\)

\(\Leftrightarrow\left(2^5\right)^{-n}.\left(2^4\right)^n=2^{11}\)

\(\Leftrightarrow2^{-5n}.2^{4n}=2^{11}\)

\(\Leftrightarrow2^{-n}=2^{11}\)

\(\Leftrightarrow n=-11\)

a)\(32^{-n}\cdot16^n=2048\)

\(\left(2^5\right)^{-n}\cdot\left(2^4\right)^n\)=2048

\(2^{-5n}\cdot2^{4n}\)=\(2^{11}\)

\(2^{-5n+4n}=2^{11}\)

\(2^{-x}=2^{11}\)

\(\Rightarrow x=-11\)

b)\(2^{-1}\cdot2^n+4\cdot2^n=9\cdot2^5\)

\(\frac{1}{2}\cdot2^n+4\cdot2^n=288\)

\(2^n\left(\frac{1}{2}+4\right)=288\)

\(2^n\cdot\frac{9}{2}=288\)

\(2^n=288:\frac{9}{2}\)

\(2^n=64\)

\(2^n=2^6\)

\(\Rightarrow n=6\)

a) 32^-n . 16ⁿ = 2048

\(\frac{1}{32n}\) . 16ⁿ = 2048

\(\frac{1}{2^n.16^n}\) . 16ⁿ = 2048

\(\frac{1}{2^n}\) = 2048

2^-n = 2048

2^-n = 2¹¹

\(\Rightarrow\) -n = 11

\(\Rightarrow\) n = -11

Vậy n = -11

\(\frac{1}{9}.27^n=3^n\)

\(\Rightarrow\frac{3^n}{27^n}=\frac{1}{9}\)

\(\Rightarrow\left(\frac{3}{27}\right)^n=\frac{1}{9}\)

\(\Rightarrow\left(\frac{1}{9}\right)^n=\frac{1}{9}\)

\(\Rightarrow n=1\)

\(\frac{2030-x}{15}+\frac{2041-x}{13}+\frac{2048-x}{11}+\frac{1961-x}{9}=0\)

\(\Leftrightarrow\frac{2030-x}{15}-1+\frac{2041-x}{13}-2+\frac{2048-x}{11}-3+\frac{1961-x}{9}+6=0\)

\(\Leftrightarrow\frac{2015-x}{15}+\frac{2015-x}{13}+\frac{2015-x}{11}+\frac{2015-x}{9}=0\)

\(\Leftrightarrow\left(2015-x\right)\left(\frac{1}{15}+\frac{1}{13}+\frac{1}{11}+\frac{1}{9}\right)=0\)

Mà \(\frac{1}{15}+\frac{1}{13}+\frac{1}{11}+\frac{1}{9}\ne0\)

\(\Rightarrow2015-x=0\Leftrightarrow x=2015\)