a) \(\dfrac{32}{\left(-2\right)^n}=4\)

\(\Rightarrow\left(-2\right)^n=8=\left(-2\right)^3\)

=> n = 3

b) \(\dfrac{8}{2^n}=2\)

\(\Rightarrow2^n=4=2^2\)

=> n = 2

c) \(\left(\dfrac{1}{2}\right)^{2n-1}=\dfrac{1}{8}\)

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

=> 2n - 1 = 3

=> 2n = 4

=> n = 2

Giải:

a) \(\dfrac{32}{\left(-2\right)^n}=4\) 

\(\Rightarrow\left(-2\right)^n=32:4=8\) 

\(\Rightarrow\left(-2\right)^n=8\) 

Vì \(\left(-2\right)^n=2^3\) là ko thể nên n ∈ ∅

b) \(\dfrac{8}{2^n}=2\)

\(\Rightarrow2^n=8:2=4\) 

\(\Rightarrow2^n=4\) 

\(\Rightarrow2^n=2^2\) 

\(\Rightarrow n=2\) 

c) \(\left(\dfrac{1}{2}\right)^{2n-1}=\dfrac{1}{8}\) 

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

\(\Rightarrow2n-1=3\rightarrow n=2\)

\(a=\lim\dfrac{-2n^2}{\sqrt{n^2+2}+\sqrt{n^2+4}}=\lim\dfrac{-2n}{\sqrt{1+\dfrac{2}{n^2}}+\sqrt{1+\dfrac{4}{n^2}}}=\dfrac{-\infty}{2}=-\infty\)

\(b=\lim\dfrac{3-5n^2+10n}{n-2}=\lim\dfrac{-5n+10+\dfrac{3}{n}}{1-\dfrac{2}{n}}=\dfrac{-\infty}{1}=-\infty\)

\(c=\lim\left(\dfrac{1-\dfrac{1}{n}}{\dfrac{\sqrt{3}}{n}-1}-4.2^n\right)=-1-\infty=-\infty\)

\(d=\lim\dfrac{n^3-4n-\left(3n^2+4\right)\left(n-2\right)}{n^2-2n}=\lim\dfrac{-2n^3+6n^2-8n+8}{n^2-2n}\)

\(\lim\dfrac{-2n+6-\dfrac{8}{n}+\dfrac{8}{n^2}}{1-\dfrac{2}{n}}=\dfrac{-\infty}{1}=-\infty\)

\(e=\lim\dfrac{\sqrt{1+\dfrac{1}{n}}-\sqrt{5}}{\sqrt{1+\dfrac{1}{n}}+\sqrt{5}}=\dfrac{1-\sqrt{5}}{1+\sqrt{5}}\)

a) \(2\left(\dfrac{2}{3.5}+\dfrac{4}{5.9}+...+\dfrac{16}{n\left(n+16\right)}\right)=\dfrac{16}{25}\)

\(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{n}-\dfrac{1}{n+16}=\dfrac{8}{25}\)

\(\dfrac{1}{3}-\dfrac{1}{n+16}=\dfrac{8}{25}\)

\(\dfrac{n+13}{3\left(n+16\right)}=\dfrac{8}{25}\)

\(24n+384=25n+325\)

\(25n-24n=384-325\)

\(n=59\)

b) Sai đề nha

\(\left\{{}\begin{matrix}\dfrac{2018}{2019}< 1\\\dfrac{2019}{2020}< 1\\\dfrac{2020}{2021}< 1\\\dfrac{2021}{2022}< 1\end{matrix}\right.\)

\(\Rightarrow\dfrac{2018}{2019}+\dfrac{2019}{2020}+\dfrac{2020}{2021}+\dfrac{2021}{2022}< 4\)

chị ơi hình như chị nhầm rồi P/s cuối phải là 1/n.(n+6)thì phải

\(a. \)

\(\dfrac{-32}{\left(-2\right)^n}=4\)

\(\Rightarrow\left(-2\right)^n=-32:4=-8\)

\(\Rightarrow\left(-2\right)^n=\left(-2\right)^3\)

\(\Rightarrow n=3\)

\(b.\)

\(\dfrac{8}{2^n}=2\)

\(\Rightarrow2^n=4\)

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

\(\Rightarrow n=2\)

\(c.\)

\(\dfrac{16}{\left(-2\right)^n}=-8\)

\(\Rightarrow\left(-2\right)^n=-2\)

\(\Rightarrow n=1\)

\(T=\sqrt{\dfrac{2n^4-4n^3+6n^2-4n+2}{2}}+\sqrt{\dfrac{2n^4+4n^3+6n^2+4n+2}{2}}\)

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

\(=\sqrt{\left(n^2-n\right)^2+2\left(n^2-n\right)+1}+\sqrt{\left(n^2+n\right)^2+2\left(n^2+n\right)+1}\)

\(=\sqrt{\left(n^2-n+1\right)^2}+\sqrt{\left(n^2+n+1\right)^2}\)

\(=n^2-n+1+n^2+n+1\)

\(=2n^2+2\ge2\)

\(T_{min}=2\) khi \(n=0\)

\(a=lim\dfrac{\left(\dfrac{2}{6}\right)^n+1-\dfrac{1}{4}\left(\dfrac{4}{6}\right)^n}{\left(\dfrac{3}{6}\right)^n+6}=\dfrac{1}{6}\)

\(b=\lim\dfrac{\left(n+1\right)^2}{3n^2+4}=\lim\dfrac{n^2+2n+1}{3n^2+4}=\lim\dfrac{1+\dfrac{2}{n}+\dfrac{1}{n^2}}{3+\dfrac{4}{n^2}}=\dfrac{1}{3}\)

\(c=\lim\dfrac{n\left(n+1\right)}{2\left(n^2-3\right)}=\lim\dfrac{n^2+n}{2n^2-6}=\lim\dfrac{1+\dfrac{1}{n}}{2-\dfrac{6}{n^2}}=\dfrac{1}{2}\)

\(d=\lim\left[1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n}-\dfrac{1}{n+1}\right]=\lim\left[1-\dfrac{1}{n+1}\right]=1\)

\(e=\lim\dfrac{1}{2}\left[1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right]\)

\(=\lim\dfrac{1}{2}\left[1-\dfrac{1}{2n+1}\right]=\dfrac{1}{2}\)

A = \(\dfrac{\left(1^4+4\right)\left(5^4+4\right)\left(9^4+4\right)...\left(21^4+4\right)}{\left(3^4+4\right)\left(7^4+4\right)\left(11^4+4\right)...\left(23^4+4\right)}\)

Xét: n⁴ + 4 = (n²+2)² - 4n² = (n²-2n+2)(n²+2n+2) = [(n-1)²+1][(x+1)²+1] nên: A = \(\dfrac{\left(0^2+1\right)\left(2^2+1\right)}{\left(2^2+1\right)\left(4^2+1\right)}.\dfrac{\left(4^2+1\right)\left(6^2+1\right)}{\left(6^2+1\right)\left(8^2+1\right)}.....\dfrac{\left(20^2+1\right)\left(22^2+1\right)}{\left(22^2+1\right)\left(24^2+1\right)}=\dfrac{1}{24^2+1}=\dfrac{1}{577}\)

B = \(\left(\dfrac{n-1}{1}+\dfrac{n-2}{2}+...+\dfrac{2}{n-2}+\dfrac{1}{n-1}\right):\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{n}\right)\)

Đặt C = \(\dfrac{n-1}{1}+\dfrac{n-2}{2}+...+\dfrac{n-\left(n-2\right)}{n-2}+\dfrac{n-\left(n-1\right)}{n-1}\)

= \(\dfrac{n}{1}+\dfrac{n}{2}+...+\dfrac{n}{n-2}+\dfrac{n}{n-1}-1-1-...-1\)

= \(n+\dfrac{n}{2}+\dfrac{n}{3}+...+\dfrac{n}{n-1}-\left(n-1\right)\)

= \(\dfrac{n}{2}+\dfrac{n}{3}+...+\dfrac{n}{n-1}+\dfrac{n}{n}\)

= \(n\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{n}\right)\)

Vậy ...

a)\(\left(\dfrac{1}{2}\right)^n=\dfrac{1}{32}\)

=>\(\left(\dfrac{1}{2}\right)^n=\left(\dfrac{1}{2}\right)^5\)

=>n=5

b)\(\left(\dfrac{343}{125}\right)=\left(\dfrac{7}{5}\right)^n\)

=>\(\left(\dfrac{7}{5}\right)^3=\left(\dfrac{7}{5}\right)^n\)

=>n=3

c)\(\dfrac{16}{2^n}=2\)

=>2ⁿ=\(\dfrac{16}{2}\)

=>2ⁿ=8

=>2ⁿ=2³

=>n=3

d)\(\dfrac{\left(-3\right)^n}{81}=-27\)

=>(-3)ⁿ=-27.81

=>(-3)ⁿ=-2187

=>(-3)ⁿ=(-3)⁷

=>n=7

e)8ⁿ:2ⁿ=4

=>(2³)ⁿ:2ⁿ=4

=>2³ⁿ:2ⁿ=4

=>2^3n-n=4

=>2²ⁿ=4

=>2²ⁿ=2²

=>2n=2

=>n=1

f)3².3ⁿ=3⁵

=>3ⁿ=3⁵:3²

=>3ⁿ=3^5-2

=>3ⁿ=3³

=>n=3

g) (2²:4).2ⁿ=4

=>1.2ⁿ=2²

=>n=2

h)3^-2.3⁴.3ⁿ=3⁷

=>\(\left(\dfrac{1}{3}\right)^2\).3⁴.3ⁿ=3⁷

=>3².3ⁿ=3⁷

=>3²⁺ⁿ=3⁷

=>2+n=7

=>n=5

a: =>x-3/4=1/6-1/2=1/6-3/6=-2/6=-1/3

=>x=-1/3+3/4=-4/12+9/12=5/12

b: =>x(1/2-5/6)=7/2

=>-1/3x=7/2

hay x=-21/2

c: (4-x)(3x+5)=0

=>4-x=0 hoặc 3x+5=0

=>x=4 hoặc x=-5/3

d: x/16=50/32

=>x/16=25/16

hay x=25

e: =>2x-3=-1/4-3/2=-1/4-6/4=-7/4

=>2x=-7/4+3=5/4

hay x=5/8