Question

So sánh \(\frac{n}{n+1}\)và \(\frac{n+1}{n+2}\)

Trình bày cách giải nha bạn 

Answer 1

Ta có : 1 - n/n+1 = 1/ n+1 ;  1 - n+1/n+2 = 1/n+2

Vì n + 1 < n + 2 nên 1 /n+1 > 1/n+2 . Vì 1/n+1 > 1/n+2 nên n/n+1 <n/n+2

Answer 2

phần bù đến 1 của n/n+1 là 1-n/n+1=1/n+1

phần bù đến 1 của n+1/n+2 là 1-n+1/n+2=1/n+2

vì 1/n+1>1/n+2 nên n/n+1 < n+1/n+2

Answer 3

Ta có \(\frac{n}{n+1}=\frac{n+1-1}{n+1}=1-\frac{1}{n+1}\)

\(\frac{n+1}{n+2}=\frac{n+2-1}{n+2}=1-\frac{1}{n+2}\)

Ta thấy \(n+1< n+2\)nên \(\frac{1}{n+1}>\frac{1}{n+2}\)do đó \(1-\frac{1}{n+1}< 1-\frac{1}{n+2}\)

Suy ra \(\frac{n}{n+1}< \frac{n+1}{n+2}\)

Vậy \(\frac{n}{n+1}< \frac{n+1}{n+2}\)

Answer 4

Ta co \(\frac{n}{n+1}=\frac{n\times\left(n+2\right)}{\left(n+1\right)\left(n+2\right)}=\frac{n^2+2n}{\left(n+1\right)\left(n+2\right)}\) (1)

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

Tu (1) va (2) => \(\frac{n}{n+1}>\frac{n+1}{n+2}\)

Answer 5

\(\frac{n}{n+1}-1=\frac{n}{n+1}-\frac{n+1}{n+1}=-\frac{1}{n+1}\)

\(\frac{n+1}{n+2}-1=\frac{n+1}{n+2}-\frac{n+2}{n+2}=-\frac{1}{n+2}\)

\(\Rightarrow-\frac{1}{n+1}< -\frac{1}{n+2}\)

\(\Leftrightarrow\frac{n}{n+1}< \frac{n+1}{n+2}\)

Answer 6

Nham dau dau < moi dung nha :))

Answer 7

Ta có \(\frac{n}{n+1}=\frac{n+1-1}{n+1}=1-\frac{1}{n+1}\)

\(\frac{n+1}{n+2}=\frac{n+2-1}{n+2}=1-\frac{1}{n+2}\)

Do \(\frac{1}{n+1}>\frac{1}{n+2}\)nên \(1-\frac{1}{n+1}< 1-\frac{1}{n+2}\)

hay \(\frac{n}{n+1}< \frac{n+1}{n+2}\)

Answer 8

Ta có: \(\frac{n}{n+1}=\frac{n+1-1}{n+1+1-1}=\frac{n+1-1}{n+2-1}\)

Áp dụng bất đẳng thức cơ bản:\(\frac{a-n}{b-n}>\frac{a}{b}\)Với \(\frac{a}{b}< 1\)(Bạn tự chứng minh)

Với \(\frac{a}{b}=\frac{n+1}{n+2}\)mà n + 1 < n + 2 nên \(\frac{a}{b}< 1\). Ta có \(\frac{n+1-1}{n+2-1}< \frac{n+1}{n+2}\)

Answer 9

Thêm câu này nữa nhé:

Do \(\frac{n+1-1}{n+2-1}< \frac{n+1}{n+2}\) mà \(\frac{n+1-1}{n+2-1}=\frac{n}{n+1}\)nên \(\frac{n}{n+1}< \frac{n+1}{n+2}\)