Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
c, \(\frac{-32}{-2^n}=4\)
\(\Rightarrow-2^n=-32:4\)
\(\Rightarrow-2^n=-8\)
\(\Rightarrow-2^n=-2^3\Rightarrow n=3\)
d, \(\frac{8}{2^n}=2\)
\(\Rightarrow2^n=8:2\)
\(\Rightarrow2^n=4\)
\(\Rightarrow2^n=2^2\Rightarrow n=2\)
e, \(\frac{25^3}{5^n}=25\)
\(\Rightarrow5^n=25^3:25\)
\(\Rightarrow5^n=25^2\)
\(\Rightarrow5^n=5^4\Rightarrow n=4\)
i , \(8^{10}:2^n=4^5\)
\(\Rightarrow2^n=8^{10}:4^5\)
\(\Rightarrow2^n=\left(2^3\right)^{10}:\left(2^2\right)^5\)
\(\Rightarrow2^n=2^{30}:2^{10}\)
\(\Rightarrow2^n=2^{20}\Rightarrow n=20\)
k, \(2^n.81^4=27^{10}\)
\(\Rightarrow2^n=27^{10}:81^4\)
\(\Rightarrow2^n=\left(3^3\right)^{10}:\left(3^4\right)^4\)
\(\Rightarrow2^n=3^{30}:3^{16}\)
\(\Rightarrow2^n=3^{14}\)
\(\Rightarrow2^n=4782969\)Không chia hết cho 2 nên ko có Gt n thỏa mãn
Đặt A = \(\frac{n+1}{n+2}\)
=> \(\frac{1}{A}=\frac{n+2}{n+1}\)
=> \(\frac{1}{A}-1=\frac{n+2-n-1}{n+1}=\frac{1}{n+1}\)
Đặt B = \(\frac{n+3}{n+4}\)
=> \(\frac{1}{B}=\frac{n+4}{n+3}\)
=> \(\frac{1}{B}-1=\frac{n+4-n-3}{n+3}=\frac{1}{n+3}\)
Vì \(\frac{1}{n+1}>\frac{1}{n+3}\Rightarrow\frac{1}{A}-1>\frac{1}{B}-1\Rightarrow\frac{1}{A}>\frac{1}{B}\Rightarrow A< B\)
Vậy \(\frac{n+1}{n+2}< \frac{n+3}{n+4}\)
Đặt \(A=\frac{n+1}{n+2}\)
\(\Rightarrow\frac{1}{A}=\frac{n+2}{n+1}\)
\(\Rightarrow\frac{1}{A}-1=\frac{n+2-n+1}{n+1}=\frac{1}{n+1}\)
Đặt \(B=\frac{n+3}{n+4}\)
\(\Rightarrow\frac{1}{B}=\frac{n+4}{n+3}\)
\(\Rightarrow\frac{1}{B}-1=\frac{n+4-n-3}{n+3}=\frac{1}{n+3}\)
Vì \(\frac{1}{n+1}>\frac{1}{n+3}\Rightarrow\frac{1}{A}-1>\frac{1}{B}-1\Rightarrow\frac{1}{A}>\frac{1}{B}\Rightarrow A< B\)
Vậy \(\frac{n+1}{n+2}< \frac{n+3}{n+4}\)
ta có: \(\frac{n+1}{n+2}=\frac{n}{n+2}+\frac{1}{n+2}\) (n>0)
mà \(\frac{n}{n+2}>\frac{n}{n+3}\Rightarrow\frac{n}{n+2}+\frac{1}{n+2}>\frac{n}{n+3}\)
\(\Rightarrow\frac{n+1}{n+2}>\frac{n}{n+3}\)
a) Ta có :
\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}\)
\(< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right)n}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-1}-\frac{1}{n}=1-\frac{1}{n}< 1\)
\(\Rightarrow\)A < 1
b) \(B=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{\left(2n\right)^2}\)
\(B=\frac{1}{2^2}.\left(1+\frac{1}{2^2}+\frac{1}{3^3}+...+\frac{1}{n^2}\right)\)
vì \(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}< 1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{\left(n-1\right)n}< 1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n-1}-\frac{1}{n}< 2-\frac{1}{n}< 2\)
\(\Rightarrow B< \frac{1}{2^2}.2=\frac{1}{2}\)
\(\frac{2n+1}{n+3}=\frac{n+n+1}{n+3}=\frac{n}{n+3}+\frac{n+1}{n+3}\)
Do: \(\frac{n}{n+3}< \frac{n}{n+1};\frac{n+1}{n+3}< \frac{n+1}{n+2}\Rightarrow\frac{n}{n+3}+\frac{n+1}{n+3}< \frac{n}{n+1}+\frac{n+1}{n+2}\Rightarrow\frac{2n+1}{n+3}< \frac{n}{n+1}+\frac{n+1}{n+2}\)