VẬY A>B  

Chúc bạn học tốt ( -_- )

\(a,\dfrac{a}{b}>1\Leftrightarrow a>1\cdot b=b\\ \dfrac{a}{b}< 1\Leftrightarrow a< 1\cdot b=b\\ b,\dfrac{a}{b}=\dfrac{a\left(b+1\right)}{b\left(b+1\right)}=\dfrac{ab+a}{b^2+b}\\ \dfrac{a+1}{b+1}=\dfrac{b\left(a+1\right)}{b\left(b+1\right)}=\dfrac{ab+b}{b^2+b}\\ \forall a=b\Leftrightarrow\dfrac{a}{b}=\dfrac{a+1}{b+1}\\ \forall a>b\Leftrightarrow\dfrac{a}{b}>\dfrac{a+1}{b+1}\\ \forall a< b\Leftrightarrow\dfrac{a}{b}< \dfrac{a+1}{b+1}\)

\(c,\forall a>b\Leftrightarrow\dfrac{a}{b}-1=\dfrac{a-b}{b}>\dfrac{a-b}{b+n}\left(b< b+n;a-b>0\right)=\dfrac{a+n}{b+n}-1\\ \Leftrightarrow\dfrac{a}{b}>\dfrac{a+n}{b+n}\\ \forall a< b\Leftrightarrow1-\dfrac{a}{b}=\dfrac{b-a}{b}>\dfrac{b-a}{b+n}\left(b< b+n;b-a>0\right)=1-\dfrac{a+n}{b+n}\\ \Leftrightarrow1-\dfrac{a}{b}>1-\dfrac{a+n}{b+n}\Leftrightarrow\dfrac{a}{b}>\dfrac{a+n}{b+n}\\ \forall a=b\Leftrightarrow\dfrac{a+n}{b+n}=\dfrac{a}{b}\left(=1\right)\)

A.  Ta có :

1-  n+1/n+2  = 1/n+2   (1)

1 -  n+3/n+4 = 1/n+4   (2)

Từ (1) và (2) ;Ta có :

1/n+2 >1/ n+4

Nên  n+1/n+2 < n+3/n+4 

KL : n+1/n+2 < n+3/n+4

cho tớ l i k e trước nhé rồi tớ sẽ trả lời

Ta có: \(\frac{n}{n+1}=\frac{n\times n+2}{n+1\times n+2}\)
            \(\frac{n+1}{n+2}=\frac{n+1\times n+1}{n+2\times n+1}=\frac{n\times2}{n\times3}\)
=> n + 1/ n + 2 > n/n+1

a, n/n+1 va n+1/n+2

Có n/n+1 + 1/n+1=1

    n+1/n+2 + 1/n+2 = 1

Vì 1/n+1>1/n+2 nên n/n+1<n/n+2 ( Bài này so sanh theo phần bù đơn vị)

c, n/2n+1 va 3n+1/6n+3

Có n/2n+1 = 3n/3.(2n+1) = 3n/6n+3

Vì 3n/6n+3 < 3n+1/6n+3 nên n/2n+1<3n+1/6n+3