a, \(\dfrac{1}{9}\). 3\(^4\).3\(^n\) = 3\(^7\)
b, \(\dfrac{1}{2}\).2\(^n\) + 4.2\(^n\)= 9.2\(^5\)
2. tìm số nguyên N
a. \(27^n:3^n=9\)
b. \(\dfrac{25}{5^n}=5\)
c. \(\left(\dfrac{81}{-3}\right)^n=-243\)
d. \(\dfrac{-1}{2}.2^n+4.2^n=9.2^5\)
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
Tìm \(n\in Z\)
a/\(\dfrac{1}{9}.27^n=3^n\)
b/\(3^{-2}.3^4.3^n=3^7\)
c/\(2^{-1}.2n+4.2^n=9.2^5\)
d/\(32^{-n}.16^n=2048\)
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
1.
a, \(^{^2}\left(x-2\right)=9\) b,\(^{^3}\left(3x-1\right)=-8\) c, \(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\) d, \(\left(\dfrac{2}{3}\right)^x=\dfrac{4}{9}\) e, \(\left(\dfrac{1}{2}\right)^{x-1}=\dfrac{1}{16}\) f,\(\left(\dfrac{1}{2}\right)^{2x-1}=8\)
2.tìm số tự nhiên n biết
a, \(3^{n-1}=27\) b, \(3^{n-1}=\dfrac{1}{243}\) c, \(\left(\dfrac{1}{2}\right)^{2n-1}=\dfrac{1}{8}\) d, \(\left(-\dfrac{1}{3}\right)^{n-5}=\dfrac{1}{81}\) e,\(2^{-1}.2^n+4.2^n=9.2^5\)
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 \(x-1=4\Rightarrow x=5\)
f, \(\left(\dfrac{1}{2}\right)^{2x-1}=8\) \(\Rightarrow\left(\dfrac{1}{2}\right)^{2x-1}=\left(\dfrac{1}{2}\right)^{-3}\) Vì \(\dfrac{1}{2}\ne\pm1;\dfrac{1}{2}\ne0\) nên \(2x-1=-3\) \(\Rightarrow2x=-2\Rightarrow x=-1\) Chúc bạn học tốt!!!Tìm x:
\(a\)) \(\dfrac{2}{3}+\left(x-\dfrac{1}{2}\right)^3=\dfrac{19}{27}\)
\(b\)) \(\left(\dfrac{3}{2}\right)^{2x-1}:\left(\dfrac{27}{8}\right)^3=\dfrac{81}{16}\)
\(c\)) \(\dfrac{1}{2}.2^x+4.2^x=9.2^5\)
\(d\)) \(\text{12 - (2x +1)}^2=-69\)
\(a,\Rightarrow\left(x-\dfrac{1}{2}\right)^3=\dfrac{1}{27}=\left(\dfrac{1}{3}\right)^3\\ \Rightarrow x-\dfrac{1}{2}=\dfrac{1}{3}\Rightarrow x=\dfrac{5}{6}\\ b,\Rightarrow\left(\dfrac{3}{2}\right)^{2x-1}:\left(\dfrac{3}{2}\right)^9=\left(\dfrac{3}{2}\right)^4\\ \Rightarrow2x-1-9=4\\ \Rightarrow2x=14\Rightarrow x=7\\ c,\Rightarrow2^{x-1}+2^{x+2}=9\cdot2^5\\ \Rightarrow2^{x-1}\left(1+2^3\right)=9\cdot2^5\\ \Rightarrow2^{x-1}\cdot9=9\cdot2^5\\ \Rightarrow2^{x-1}=2^5\Rightarrow x-1=5\Rightarrow x=6\\ d,\Rightarrow\left(2x+1\right)^2=12+69=81\\ \Rightarrow\left[{}\begin{matrix}2x+1=9\\2x+1=-9\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=-5\end{matrix}\right.\)
\(a,\dfrac{2}{3}+\left(x-\dfrac{1}{2}\right)^3=\dfrac{19}{27}\)
\(\left(x-\dfrac{1}{2}\right)^3=\dfrac{19}{27}-\dfrac{2}{3}\)
\(\left(x-\dfrac{1}{2}\right)^3=\dfrac{1}{27}\)
\(\Rightarrow\left(x-\dfrac{1}{2}\right)^3=\left(\dfrac{1}{3}\right)^3\)
\(\Rightarrow x-\dfrac{1}{2}=\dfrac{1}{3}\)
\(x=\dfrac{1}{2}+\dfrac{1}{3}\)
\(x=\dfrac{1}{5}\)
Tìm STN n biết
a) \(\frac{1}{9}.3^4.3^{n+1}=9^4\)
b) \(\frac{1}{2}.2^n+4.2^n=9.2^5\)
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
\(a)\frac{1}{9}.3^4.3^{n+1}=9^4\)
\(\Rightarrow3^{-2}.3^4.3^{n+1}=3^8\)
\(\Rightarrow3^{-2+4+n+1}=3^{n+3}=3^8\)
\(\Rightarrow n+3=8\Rightarrow n=5\)
\(b)\frac{1}{2}.2^n+4.2^n=9.2^5\)
\(\Rightarrow2^n\left(\frac{1}{2}+4\right)=9.2^5\)
\(\Rightarrow2^n.\frac{9}{2}=9.2^5\Rightarrow2^n=9.2^5.\frac{2}{9}=\frac{9.2^5.2}{9}=2^6\)
\(\Rightarrow n=6\)
Tìm số tụ nhiên n biết: a)1/9 . 34.3n+1=94 b)1/2 .2n+ 4.2n=9.25
Tính tổng đại số
\(A=\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{2}{3}+\dfrac{1}{4}+\dfrac{2}{4}+\dfrac{3}{4}-\dfrac{1}{5}-\dfrac{2}{5}-\dfrac{3}{5}-\dfrac{4}{5}+...+\dfrac{1}{10}+\dfrac{2}{10}+...+\dfrac{9}{10}\)
\(B=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{2}{3}+\dfrac{1}{4}+\dfrac{2}{4}+\dfrac{3}{4}+...+\dfrac{1}{n}+\dfrac{2}{n}+...+\dfrac{n-1}{n}\)\(\left(n\in Z,n\ge2\right)\)
Tìm số nguyên n
a, 3-2.34.33.3n=37
b,2-1.2n+4.2n=9.25
c, 2.16_>2n>4
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
Tìm số nguyên n
a, 3-2. 34. 3n = 37
b, 2-1. 2n+ 4.2n= 9.25
c, 2. 16 >_ 2n>4
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!