Mình mới lớp 5 nên không biết làm bài này.

Xin lỗi nha! Chúc bạn may mắn......mình chính là Đào Minh Tiến!

a) \(\frac{n}{n+1}\)và \(\frac{n+2}{n+3}\)

\(\frac{n}{n+1}=\frac{n\cdot\left(n+3\right)}{\left(n+1\right)\cdot\left(n+3\right)}\)

\(\frac{n+2}{n+3}=\frac{\left(n+2\right)\cdot\left(n+1\right)}{\left(n+3\right)\cdot\left(n+1\right)}\)

So sánh : \(n\cdot\left(n+3\right)\)và \(\left(n+2\right)\cdot\left(n+3\right)\)

\(n\cdot\left(n+3\right)=n^2+3n\)

\(\left(n+2\right)\cdot\left(n+3\right)=n^2+5n+6\)

\(n^2+3n< n^2+5n+6\)

\(\Leftrightarrow\frac{n}{n+1}< \frac{n+2}{n+3}\)

b) \(\frac{n}{2n+1}\)và \(\frac{3n+1}{6n+3}\)

\(\frac{n}{2n+1}=\frac{n\cdot\left(6n+3\right)}{\left(2n+1\right)\cdot\left(6n+3\right)}\)

\(\frac{3n+1}{6n+3}=\frac{\left(3n+1\right)\cdot\left(2n+1\right)}{\left(6n+3\right)\cdot\left(2n+1\right)}\)

So sánh : \(n\cdot\left(6n+3\right)\)và \(\left(3n+1\right)\cdot\left(2n+1\right)\)

\(n\cdot\left(6n+3\right)=6n^2+3n\)

\(\left(3n+1\right)\cdot\left(2n+1\right)=6n^2+5n+1\)

\(6n^2+3n< 6n^2+5n+1\)

\(\Leftrightarrow\frac{n}{2n+1}< \frac{3n+1}{6n+3}\)

phân số n+1/n+2 lớn hơn

Thiếu đề hay xao ý bn

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

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

Vì \(n+2< n+4\)\(\Rightarrow\frac{1}{n+2}>\frac{1}{n+4}\)

\(\Rightarrow1-\frac{1}{n+2}< 1-\frac{1}{n+4}\)\(\Rightarrow\frac{n+1}{n+2}< \frac{n+3}{n+4}\)

Ta số phân số chung gian là \(\frac{n+1}{n+3}\)

Vì \(\frac{n}{n+3}< \frac{n+1}{n+3}< \frac{n+1}{n+2}\)

Nên \(\frac{n}{n+3}< \frac{n+1}{n+2}\)

Ủng hộ nhé ! 

1: B là số nguyên

=>n-3 thuộc {1;-1;5;-5}

=>n thuộc {4;2;8;-2}

3:

a: -72/90=-4/5
b: 25*11/22*35

\(=\dfrac{25}{35}\cdot\dfrac{11}{22}=\dfrac{5}{7}\cdot\dfrac{1}{2}=\dfrac{5}{14}\)

c: \(\dfrac{6\cdot9-2\cdot17}{63\cdot3-119}=\dfrac{54-34}{189-119}=\dfrac{20}{70}=\dfrac{2}{7}\)

a) \(\frac{5}{9}=\frac{20}{36};\frac{1}{4}=\frac{9}{36}\)

\(\frac{20}{36}>\frac{9}{36}\Rightarrow\frac{5}{9}>\frac{1}{4}\)

\(\frac{72}{73}=\frac{4248}{4307};\frac{58}{59}=\frac{4234}{4307}\)

\(\frac{4248}{4307}>\frac{4234}{4307}\Rightarrow\frac{72}{73}>\frac{58}{59}\)

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

\(\Rightarrow\frac{n}{n+3}=\frac{n+1}{n+2}\)

a,>

b,>

c,<

\(\dfrac{n}{n+3}-\dfrac{n-1}{n+4}\)

\(=\dfrac{n^2+4n-n^2-2n+3}{\left(n+4\right)\left(n+3\right)}=\dfrac{2n+3}{\left(n+4\right)\left(n+3\right)}>0\)

=>n/n+3>(n-1)/(n+4)

\(\frac{n}{n+1}\)<\(\frac{n+2}{n+3}\) với n>=0