So sánh :
\(\frac{n}{n+1};\frac{n+2}{n+3}\)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) \({u_n} = \frac{{n + 1}}{n}= 1+ \frac{{1}}{n} > 1\).
b) \({u_n} = \frac{{n + 1}}{n}= 1+ \frac{{1}}{n} < 2\).
\(\text{Ta có}:1-\frac{n}{n+1}=\frac{1}{n+1}\)
\(\text{Ta có}:1-\frac{n+1}{n+2}=\frac{1}{n+2}\)
\(\text{Mà }\frac{1}{n+1}>\frac{1}{n+2}\)
\(\text{Nên }\frac{n}{n+1}>n+\frac{n+1}{n+2}\)
a) Ta có: \({u_{n + 1}} = 3\left( {n + 1} \right) - 1 = 3n + 2\).
Suy ra \({u_{n + 1}} > {u_n}\).
b) Ta có: \({v_{n + 1}} = \frac{1}{{{{\left( {n + 1} \right)}^2}}}\).
Suy ra: \({u_{n + 1}} < {u_n}\).
ta thấy:
\(\frac{n}{n+3}< 1\Rightarrow\frac{n}{n+3}< \frac{n+1}{n+4}< \frac{n+1}{n+2}\)
\(\Rightarrow\frac{n}{n+3}< \frac{n+1}{n+2}\)
vậy ...
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 co :
n/n+3=n+3-3/n+3=1-3/n+3
n+1/n+2=n+2-1/n+2=1-1/n+2
vi 3/n+3>1/n+2 nen n/n+3<n+1/n+2
So sánh : \(\frac{n}{n+1}\)và \(\frac{n+2}{n+3}\)
Ta có : \(n\left(n+3\right)\)và \(\left(n+2\right)\left(n+1\right)\)
Ta có : \(n\left(n+3\right)=n^2+3n\)
\(\left(n+2\right)\left(n+1\right)=n\left(n+2\right)+n+2=n^2+2n+n+2=n^2+3n+2\)
Dễ thấy : \(n^2+3n< n^2+3n+2\)
\(\Rightarrow\frac{n}{n+1}< \frac{n+2}{n+3}\)
Phần bù của\(\frac{n}{n+1}\)để =1 là\(\frac{1}{n+1}\)
phần bù của\(\frac{n+2}{n+3}\)để =1 là\(\frac{1}{n+3}\)
\(\frac{1}{n+3}\)< \(\frac{1}{n+1}\)=>phần bù của\(\frac{n}{n+1}\)>\(\frac{n+2}{n+3}\)=>\(\frac{n}{n+1}\)<\(\frac{n+2}{n+3}\)