a: =>9^n=9

=>n=1

b: =>5^n=5

=>n=1

c: \(\Leftrightarrow\left(-27\right)^n=-243\)

=>\(\left(-3\right)^{3n}=\left(-3\right)^5\)

=>3n=5

=>n=5/3

d: =>2^n*9/2=9*2^5

=>2^n=9*2^5:9/2=2^5*2=2^6

=>n=6

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

<=>27ⁿ=3ⁿ:\(\dfrac{1}{9}\)

<=>27ⁿ:3ⁿ=\(\dfrac{1}{9}\)

<=>3³ⁿ:3ⁿ=\(\dfrac{1}{9}\)

<=>3²ⁿ=\(\dfrac{1}{9}\)

<=>9ⁿ=\(\dfrac{1}{9}\)

<=>9ⁿ⁺¹=1

<=>n+1=0

<=>n=-1

vậy n=-1

Bài 1:

a, \(\left(x-2\right)^2=9\)

\(\Rightarrow x-2\in\left\{-3;3\right\}\Rightarrow x\in\left\{-1;5\right\}\)

b, \(\left(3x-1\right)^3=-8\)

\(\Rightarrow3x-1=-2\Rightarrow3x=-1\)

\(\Rightarrow x=-\dfrac{1}{3}\)

c, \(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\)

\(\Rightarrow x+\dfrac{1}{2}\in\left\{-\dfrac{1}{4};\dfrac{1}{4}\right\}\)

\(\Rightarrow x\in\left\{-\dfrac{3}{4};-\dfrac{1}{4}\right\}\)

d, \(\left(\dfrac{2}{3}\right)^x=\dfrac{4}{9}\)

\(\Rightarrow\left(\dfrac{2}{3}\right)^x=\left(\dfrac{2}{3}\right)^2\)

Vì \(\dfrac{2}{3}\ne\pm1;\dfrac{2}{3}\ne0\) nên \(x=2\)

e, \(\left(\dfrac{1}{2}\right)^{x-1}=\dfrac{1}{16}\)

\(\Rightarrow\left(\dfrac{1}{2}\right)^{x-1}=\left(\dfrac{1}{2}\right)^4\)

Vì \(\dfrac{1}{2}\ne\pm1;\dfrac{1}{2}\ne0\) nên \(x-1=4\Rightarrow x=5\)

Mk làm lun, ko viết lại đề bài nữa nhé =))

a) \(\Leftrightarrow\)\(3^2.3^{n+1}=9^4\)

\(\Leftrightarrow3^{n+1}=9^4:3^2\)

\(\Leftrightarrow3^{n+1}=3^6\)

\(\Rightarrow n+1=6\)

\(\Leftrightarrow n=6-1\)

\(\Rightarrow n=5\)

b)\(\Leftrightarrow2^n.\left(\frac{1}{2}+4\right)=9.2^5\)

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

\(\Rightarrow2^n=\left(9.2^5\right):\frac{9}{2}\)

\(\Rightarrow2^n=468:\frac{9}{2}\)

Tự tính nốt KQ giúp mk nha ♥

a)1/9*3⁴*3ⁿ⁺¹=9⁴

3⁴*3ⁿ⁺¹ =9⁴/(1/9)=9⁴*9=9⁵

3⁴*3ⁿ⁺¹=3⁴⁺ⁿ⁺¹=9⁵=(3²)⁵=3¹⁰

=>4+n+1=10

n=10-4-1

Vậy n=5

b)1/2*2ⁿ+4*2ⁿ=9*2⁵

2ⁿ*(1/2+4)=9*2⁵

2ⁿ*4.5=9*2⁵

2ⁿ=9*2⁵/4.5=2⁵*2=2⁶

=> Vậy n=6

\(a)\frac{1}{9}.3^4.3^{n+1}=9^4\)

\(\Rightarrow3^{-2}.3^4.3^{n+1}=3^8\)

\(\Rightarrow3^{-2+4+n+1}=3^{n+3}=3^8\)

\(\Rightarrow n+3=8\Rightarrow n=5\)

\(b)\frac{1}{2}.2^n+4.2^n=9.2^5\)

\(\Rightarrow2^n\left(\frac{1}{2}+4\right)=9.2^5\)

\(\Rightarrow2^n.\frac{9}{2}=9.2^5\Rightarrow2^n=9.2^5.\frac{2}{9}=\frac{9.2^5.2}{9}=2^6\)

\(\Rightarrow n=6\)

a. x=5/6

\(a,\Rightarrow\left(x-\dfrac{1}{2}\right)^3=\dfrac{1}{27}=\left(\dfrac{1}{3}\right)^3\\ \Rightarrow x-\dfrac{1}{2}=\dfrac{1}{3}\Rightarrow x=\dfrac{5}{6}\\ b,\Rightarrow\left(\dfrac{3}{2}\right)^{2x-1}:\left(\dfrac{3}{2}\right)^9=\left(\dfrac{3}{2}\right)^4\\ \Rightarrow2x-1-9=4\\ \Rightarrow2x=14\Rightarrow x=7\\ c,\Rightarrow2^{x-1}+2^{x+2}=9\cdot2^5\\ \Rightarrow2^{x-1}\left(1+2^3\right)=9\cdot2^5\\ \Rightarrow2^{x-1}\cdot9=9\cdot2^5\\ \Rightarrow2^{x-1}=2^5\Rightarrow x-1=5\Rightarrow x=6\\ d,\Rightarrow\left(2x+1\right)^2=12+69=81\\ \Rightarrow\left[{}\begin{matrix}2x+1=9\\2x+1=-9\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=-5\end{matrix}\right.\)

\(a,\dfrac{2}{3}+\left(x-\dfrac{1}{2}\right)^3=\dfrac{19}{27}\)

\(\left(x-\dfrac{1}{2}\right)^3=\dfrac{19}{27}-\dfrac{2}{3}\)

\(\left(x-\dfrac{1}{2}\right)^3=\dfrac{1}{27}\)

\(\Rightarrow\left(x-\dfrac{1}{2}\right)^3=\left(\dfrac{1}{3}\right)^3\)

\(\Rightarrow x-\dfrac{1}{2}=\dfrac{1}{3}\)

\(x=\dfrac{1}{2}+\dfrac{1}{3}\)

\(x=\dfrac{1}{5}\)

a, \(3^{-2}.3^4.3^3.3^n=3^7\)

\(\Rightarrow3^{-2+4+3+n}=3^7\)

\(\Rightarrow3^{5+n}=3^7\)

Vì \(3\ne-1;3\ne0;3\ne1\) nên \(5+n=7\Rightarrow x=2\)

Vậy....

b, \(2^{-1}.2^n+4.2^n=9.2^5\)

\(\Rightarrow2^n.\left(2^{-1}+2^2\right)=9.2^5\)

\(\Rightarrow2^n.4,5=9.32\)

\(\Rightarrow2^n=288:4,5\)

\(\Rightarrow2^n=64=2^6\)

Vì \(2\ne-1;2\ne0;2\ne1\) nên \(n=6\)

Vậy.....

c, \(2.16\ge2^n>4\)

\(\Rightarrow32\ge2^n>2^2\)

\(\Rightarrow2^5\ge2^n>2^2\)

Vì \(2\ne-1;2\ne0;2\ne1\) nên \(5\ge n>2\)

\(\Rightarrow n\in\left\{3;4;5\right\}\)

Vậy.....

Chúc bạn học tốt!!!

Câu hỏi của nguyen lan anh - Toán lớp 7 | Học trực tuyến

a, n=2

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

\(\Rightarrow3^{-2}.3^n=3^7:3^4\)

\(\Rightarrow3^{-2+n}=3^3\)

\(\Rightarrow-2+n=3\)

\(\Rightarrow n=3+2=5\)

Vậy \(n=5.\)

b) \(2^{-1}.2^n+4.2^n=9.2^5\)

\(\Rightarrow2^n\left(2^{-1}+4\right)=9.2^5\)

\(\Rightarrow2^n.9.\dfrac{1}{2}=9.2^5\)

\(\Rightarrow2^n.\dfrac{1}{2}=2^5\)

\(\Rightarrow2^n=2^5.2=2^6\)

\(\Rightarrow n=6.\)

Vậy \(n=6.\)

c) Nhìn cái đề mk chẳng hiểu gì hết, cái dấu sau dấu lớn là dấu gì thế???

a) \(3^{-2}\cdot3^4\cdot3^n=3^7\) (1)

\(\Leftrightarrow3^{n+2}=3^7\)

\(\Leftrightarrow n+2=7\)

\(\Leftrightarrow n=7-2\)

\(\Leftrightarrow n=5\)

Vậy tập nghiệm phương trình (1) là \(S=\left\{5\right\}\)

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

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

\(\Leftrightarrow\left(\dfrac{1}{2}+4\right)\cdot2^n=9\cdot2^5\)

\(\Leftrightarrow\dfrac{9}{2}\cdot2^n=9\cdot2^5\)

\(\Leftrightarrow2^n=2^6\)

\(\Leftrightarrow n=6\)

Vậy tập nghiệm phương trình (2) là \(S=\left\{6\right\}\)

Giải:

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

\(\Leftrightarrow3^{-2+4+n}=3^7\)

Vì \(3=3\)

\(\Rightarrow-2+4+n=7\)

\(\Rightarrow2+n=7\)

\(\Leftrightarrow n=7-2=5\)

Vậy \(n=5\)

c) \(2.16\ge2^n>4\)

\(\Leftrightarrow2.2^4\ge2^n>2^2\)

\(\Leftrightarrow2^5\ge2^n>2^2\)

\(\Rightarrow n\in\left\{3;4;5\right\}\)

Vậy \(n\in\left\{3;4;5\right\}\)

Chúc bạn học tốt!