Bằng nhau nha

Ta có: \(\dfrac{n+1}{n+5}-\dfrac{n+2}{n+3}\)

\(=\dfrac{n^2+4n+3-n^2-7n-10}{\left(n+5\right)\left(n+3\right)}\)

\(=\dfrac{-3n-7}{\left(n+5\right)\left(n+3\right)}\)

A = \(\dfrac{n}{n+3}\)và B = \(\dfrac{n+1}{n+2}\)

Ta có : A giữ nguyên

           B = \(\dfrac{n+1}{n+2}=\dfrac{n}{n+2}+\dfrac{1}{n+2}\)

          \(\Rightarrow\) \(\dfrac{n}{n+3}< \dfrac{n}{n+2}+\dfrac{1}{n+2}\)

          \(\Rightarrow\) \(\dfrac{n}{n+3}< \dfrac{n+1}{n+2}\)

         \(\Rightarrow A< B\) 

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

Lời giải:

$\frac{n+3}{n+4}=\frac{(n+4)-1}{n+4}=1-\frac{1}{n+4}$

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

Vì $n+4> n+2$ nên $\frac{1}{n+4}< \frac{1}{n+2}$

Suy ra $1-\frac{1}{n+4}> 1-\frac{1}{n+2}$

Hay $\frac{n+3}{n+4}> \frac{n+1}{n+2}$

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$\frac{n-1}{n+4}< \frac{n-1}{n+2}=\frac{(n+2)-3}{n+2}=1-\frac{3}{n+2}$

$<1-\frac{n+3}=\frac{n}{n+3}$

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

Ta có: \(\frac{n+1}{n+4}=\frac{n+4-3}{n+4}=\frac{n+4}{n+4}-\frac{3}{n+4}=1-\frac{3}{n+4}\)

\(\frac{n}{n+3}=\frac{n+3-3}{n+3}=\frac{n+3}{n+3}-\frac{3}{n+3}=1-\frac{3}{n+3}\)

Vì \(\frac{3}{n+4}< \frac{3}{n+3}\Rightarrow1-\frac{3}{n+4}>1-\frac{3}{n+3}\Rightarrow\frac{n+1}{n+4}>\frac{n}{n+3}\)

Vậy \(\frac{n+1}{n+4}>\frac{n}{n+3}\)

\(\frac{n+1}{n+2}\)và \(\frac{n}{n+3}\)
\(\orbr{\begin{cases}\frac{n+1}{n+2}=\frac{\left(n+1\right)\left(n+3\right)}{\left(n+2\right)\left(n+3\right)}=\frac{n^2+3n+n+3}{\left(n+2\right)\left(n+3\right)}=\frac{n^2+4n+3}{\left(n+2\right)\left(n+3\right)}\\\frac{n}{n+3}=\frac{n\left(n+2\right)}{\left(n+2\right)\left(n+3\right)}=\frac{n^2+2n}{\left(n+2\right)\left(n+3\right)}\end{cases}}\)\(\Rightarrow\frac{n^2+4n+3}{\left(n+2\right)\left(n+3\right)}>\frac{n^2+2n}{\left(n+2\right)\left(n+3\right)}\Rightarrow\frac{n+1}{n+2}>\frac{n}{n+3}\)

Ta nhân chéo (n+1) x (n+3)=n^2+n+3n+3 (1)

n x (n+2)=n^2+2n (2)

Ta thấy (1)>(2) do n^2+n+3n+3 >  n^2+2n  nên (n+1) x (n+3) > n x (n+2)

Từ đó suy ra n+1/n+2 > n/n+3 ( tính chất )