\(a,\dfrac{11}{49}< \dfrac{11}{46};\dfrac{11}{46}< \dfrac{13}{46}\\ Nên:\dfrac{11}{49}< \dfrac{13}{46}\\ b,\dfrac{62}{85}< \dfrac{62}{80};\dfrac{62}{80}< \dfrac{73}{80}\\ Nên:\dfrac{62}{85}< \dfrac{73}{80}\\ c,\dfrac{n}{n+3}< \dfrac{n}{n+2};\dfrac{n}{n+2}< \dfrac{n+1}{n+2}\\ Nên:\dfrac{n}{n+3}< \dfrac{n+1}{n+2}\)

1,

Ta có:
\(\dfrac{73}{75}=1-\dfrac{2}{75}\)

\(\dfrac{77}{79}=1-\dfrac{2}{79}\)
So sánh phân số \(\dfrac{2}{75}\) và \(\dfrac{2}{79}\)
Vì \(75< 79\) nên \(\dfrac{1}{75}>\dfrac{1}{79}\)
Vậy \(1-\dfrac{2}{75}< 1-\dfrac{2}{79}\)
Hay \(\dfrac{73}{75}< \dfrac{77}{79}\)

2,

Vì \(\dfrac{53}{100}>\dfrac{47}{100}>\dfrac{47}{106}\) nên \(\dfrac{53}{100}>\dfrac{47}{106}\)

3,

Ta có:
\(\dfrac{81}{79}=1+\dfrac{2}{79}\)

\(\dfrac{65}{63}=1+\dfrac{2}{63}\)
So sánh phân số \(\dfrac{2}{79}\) và \(\dfrac{2}{63}\)
Vì \(79>63\) nên \(\dfrac{81}{79}< \dfrac{65}{63}\)
Hay \(\Rightarrow1+\dfrac{2}{79}< 1+\dfrac{2}{63}\)
Vậy \(\dfrac{81}{79}< \dfrac{65}{63}\)

4,

\(\dfrac{48}{47}>1>\dfrac{84}{85}\)

Vậy \(\dfrac{48}{47}>\dfrac{84}{85}\)

giúp mình với ạ

cảm ơn ạ

a)\(12< 13;49>47\)

\(\Rightarrow\dfrac{12}{49}< \dfrac{13}{47}\)

b)\(\dfrac{64}{85}>\dfrac{43}{85}\Rightarrow\dfrac{64}{85}>\dfrac{1}{2}\)

\(\dfrac{17}{35}< \dfrac{17}{34}\Rightarrow\dfrac{17}{35}< \dfrac{1}{2}\)

\(\Rightarrow\dfrac{17}{35}< \dfrac{64}{85}\)

c) \(\dfrac{19}{31}>\dfrac{16}{31}\Rightarrow\dfrac{19}{31}>\dfrac{1}{2}\)

\(\dfrac{17}{35}< \dfrac{17}{34}\Rightarrow\dfrac{17}{35}< \dfrac{1}{2}\)

\(\Rightarrow\dfrac{17}{35}< \dfrac{19}{31}\)

d)

\(1-\dfrac{67}{77}=\dfrac{10}{77}\)

\(1-\dfrac{73}{83}=\dfrac{10}{83}\)

\(\dfrac{10}{77}>\dfrac{10}{83}\Rightarrow\dfrac{67}{77}< \dfrac{73}{83}\)

e)\(1-\dfrac{456}{461}=\dfrac{5}{461}\)

\(1-\dfrac{123}{128}=\dfrac{5}{128}\)

\(\dfrac{5}{461}< \dfrac{5}{128}\Rightarrow\dfrac{456}{461}>\dfrac{123}{128}\)

\(a,\dfrac{12}{49}< \dfrac{12}{47}< \dfrac{13}{47}\Rightarrow\dfrac{12}{49}< \dfrac{12}{47}\)

b, Ta có: \(\dfrac{17}{35}=\dfrac{51}{105}\)

\(\dfrac{64}{85}>\dfrac{64}{105}>\dfrac{51}{105}\Rightarrow\dfrac{64}{85}>\dfrac{51}{105}\) hay \(\dfrac{64}{85}>\dfrac{17}{85}\)

c,\(\dfrac{19}{31}>\dfrac{17}{31}>\dfrac{17}{35}\Rightarrow\dfrac{19}{31}>\dfrac{17}{35}\)

d, \(\dfrac{67}{77}+\dfrac{10}{77}=1\)

\(\dfrac{73}{83}+\dfrac{10}{83}=1\)

\(\dfrac{10}{77}>\dfrac{10}{83}\Rightarrow\dfrac{67}{77}< \dfrac{73}{83}\)

e, \(\dfrac{456}{461}+\dfrac{5}{461}=1\)

\(\dfrac{123}{128}+\dfrac{5}{128}=1\)

\(\dfrac{5}{461}< \dfrac{5}{128}\Rightarrow\dfrac{456}{461}>\dfrac{123}{128}\)

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

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

Lời giải:
a.

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

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

b.

\(10A=\frac{10^{12}-10}{10^{12}-1}=\frac{(10^{12}-1)-9}{10^{12}-1}=1-\frac{9}{10^{12}-1}<1\)

\(10B=\frac{10^{11}+10}{10^{11}+1}=\frac{(10^{11}+1)+9}{10^{11}+1}=1+\frac{9}{10^{11}+1}>1\)

$\Rightarrow 10A< 10B\Rightarrow A< B$

Lời giải:

a) Xét hiệu \(\frac{a+n}{b+n}-\frac{a}{b}=\frac{(a+n).b-a(b+n)}{b(b+n)}=\frac{n(b-a)}{b(b+n)}\)

Nếu $b>a$ thì $\frac{a+n}{b+n}-\frac{a}{b}>0\Rightarrow \frac{a+n}{b+n}>\frac{a}{b}$

Nếu $b<a$ thì $\frac{a+n}{b+n}-\frac{a}{b}<0\Rightarrow \frac{a+n}{b+n}<\frac{a}{b}$

Nếu $b=a$ thì $\frac{a+n}{b+n}-\frac{a}{b}=0\Rightarrow \frac{a+n}{b+n}=\frac{a}{b}$

b) Rõ ràng $10^{11}-1< 10^{12}-1$. 

Đặt $10^{11}-1=a; 10^{12}-1=b; 11=n$ thì: $a< b$; $A=\frac{a}{b}$ và $B=\frac{10^{11}+10}{10^{12}+10}=\frac{a+n}{b+n}$

Áp dụng kết quả phần a:

$b>a\Rightarrow \frac{a+n}{b+n}>\frac{a}{b}$ hay $B>A$

A = \(\dfrac{n^9+1}{n^{10}+1}\) 

\(\dfrac{1}{A}\) = \(\dfrac{n^{10}+1}{n^9+1}\) = n -  \(\dfrac{n-1}{n^9+1}\)

B = \(\dfrac{n^8+1}{n^9+1}\)

\(\dfrac{1}{B}\) = \(\dfrac{n^9+1}{n^8+1}\) =  n - \(\dfrac{n-1}{n^8+1}\)

Vì n > 1 ⇒ n - 1> 0

       \(\dfrac{n-1}{n^9+1}\) < \(\dfrac{n-1}{n^8+1}\)

⇒ n - \(\dfrac{n-1}{n^9+1}\) > n - \(\dfrac{n-1}{n^8+1}\)⇒ \(\dfrac{1}{A}>\dfrac{1}{B}\)

⇒ A < B 

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

-------------------------

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

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