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

32^n / 16^n = 2048

(32/16)^n = 2048

2^n = 2048

2^n = 2^11

n = 11

Vậy n = 11.

\(\frac{32^n}{16^n}=2048\)

\(\Rightarrow\left(\frac{32}{16}\right)^n=2048\)

\(\Rightarrow2^n=2048\)

\(\Rightarrow2^n=2^{11}\)

\(\Rightarrow n=11\)

Vậy n = 11

_Chúc bạn học tốt_

a)1/9.27^n=3^n

             3^n=3^n

         =>n={0;1;2;3;...}     

a) n= 2;3;5;7;...(n là số nguyên)

<=>2(n+3)+13 chia hết n+3

=>13 chia hết n+3

=>n+3\(\in\){-13,-1,1,13}

=>n\(\in\){-16,-4,-2,11}

vì x \(\in\)Z => x \(\in\){-16,-4,-2}

vậy x  \(\in\){-16,-4,-2}

giải giúm mình đáp số thôi ! Mau lên mấy bạn

Không thích khai triển hằng đẳng thức bậc 5 thì có thể làm thế này, dễ hiểu dễ biến đổi:

\(sin^6x+cos^6x=\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)=1-\dfrac{3}{4}sin^22x\)

\(=1-\dfrac{3}{4}\left(\dfrac{1}{2}-\dfrac{1}{2}cos4x\right)=\dfrac{5}{8}+\dfrac{3}{8}cos4x\)

\(sin^4x+cos^4x=\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x=1-\dfrac{1}{2}sin^22x\)

\(=1-\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{2}cos4x\right)=\dfrac{3}{4}+\dfrac{1}{4}cos4x\)

\(sin^{10}x+cos^{10}x=\left(sin^6x+cos^6x\right)\left(sin^4x+cos^4x\right)-sin^4x.cos^4x\left(sin^2x+cos^2x\right)\)

\(=\left(\dfrac{5}{8}+\dfrac{3}{8}cos4x\right)\left(\dfrac{3}{4}+\dfrac{1}{4}cos4x\right)-\dfrac{1}{16}sin^42x\)

\(=\dfrac{15}{32}+\dfrac{3}{8}cos4x+\dfrac{3}{32}cos^24x-\dfrac{1}{16}\left(\dfrac{1}{2}-\dfrac{1}{2}cos4x\right)^2\)

\(=\dfrac{15}{32}+\dfrac{3}{8}cos4x+\dfrac{3}{32}\left(\dfrac{1}{2}+\dfrac{1}{2}cos8x\right)-\dfrac{1}{64}\left(1-2cos4x+cos^24x\right)\)

\(=\dfrac{15}{32}+\dfrac{3}{8}cos4x+\dfrac{3}{64}+\dfrac{3}{64}cos8x-\dfrac{1}{64}+\dfrac{1}{32}cos4x-\dfrac{1}{64}\left(\dfrac{1}{2}+\dfrac{1}{2}cos8x\right)\)

\(=\dfrac{63}{128}+\dfrac{13}{32}cos4x+\dfrac{5}{128}cos8x\)

oke anh

32^-n.16^n=2048=>1/32^n.16^n=2048

=>1/(16^n.2^n).16^N=2048

=>1/2^n=2048=>n= -11

a) Ta có: \(\frac{1}{9}\cdot27^n=3^n\)

\(\Leftrightarrow\frac{1}{3^2}\cdot\left(3^3\right)^n=3^n\)

\(\Leftrightarrow3^{3n}=3^{n+2}\)

\(\Rightarrow3n=n+2\)

\(\Rightarrow n=1\)

b) Ta có: \(3^2.3^4.3^n=3^7\)

\(\Rightarrow3^n=3\)

\(\Rightarrow n=1\)

c) Ta có: \(2^{-1}.2^n+4.2^n=9.2^5\)

\(\Leftrightarrow2^n\cdot\frac{9}{2}=9.2^5\)

\(\Rightarrow2^n=2^6\)

\(\Rightarrow n=6\)

d) Ta có: \(32^{-n}.16^n=2048\)

\(\Leftrightarrow\frac{1}{2^{5n}}\cdot2^{4n}=2^{11}\)

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

\(\Rightarrow4n=5n+11\)

\(\Rightarrow n=-11\)