a) 27/82 < 26/75 ( 2025/6250 < 2132\6250)

b) -49/78 > 64/ -95 ( - 3136/7410 > -4992/7410)

c) ta có: \(A=\frac{54.107-53}{53.107}=\frac{53.107+(107-53)}{53.107+54}=\frac{53.107+54}{53.107+54}=1\)

\(B=\frac{135.269-133}{134.269+135}=\frac{134.269+\left(269-133\right)}{134.269+135}=\frac{134.269+136}{134.269+135}>1\)

\(\Rightarrow A< B\)

d) ta có: \(A=\frac{3^{10}+1}{3^9+1}=\frac{3.\left(3^9+1\right)-2}{3^9+1}=\frac{3.\left(3^9+1\right)}{3^9+1}-\frac{2}{3^9+1}=3-\frac{2}{3^9+1}\)

\(B=\frac{3^9+1}{3^8+1}=\frac{3.\left(3^8+1\right)-2}{3^8+1}=\frac{3.\left(3^8+1\right)}{3^8+1}-\frac{2}{3^8+1}=3-\frac{2}{3^8+1}\)

mà \(\frac{2}{3^9+1}< \frac{2}{3^8+1}\Rightarrow3-\frac{2}{3^9+1}< 3-\frac{2}{3^8+1}\)

=> A < B

Cảm ơn Công chúa Ori nhìu nha

a, A = \(\frac{20^8+1}{20^9+1}\)

a.Vì \(\frac{17}{19}< 1\) và \(\frac{19}{17}>1\)

nên \(\frac{17}{19}< 1< \frac{19}{17}\)

hay \(\frac{17}{19}< \frac{19}{17}\)

b) \(\frac{15}{7}=2\frac{1}{7}\) và \(\frac{25}{12}=2\frac{1}{12}\)

Vì \(2\frac{1}{7}>2\frac{1}{12}\) nên \(\frac{15}{7}>\frac{25}{12}\)

\(A=\frac{54.107-53}{53.107+54}\)

\(\Leftrightarrow A=\frac{53.107+107-53}{53.107+54}\)

\(\Leftrightarrow A=\frac{53.107+54}{53.107+54}\)

\(\Leftrightarrow A=1\)

\(B=\frac{135.269-133}{134.269+135}\)

\(\Leftrightarrow B=\frac{134.269+269-133}{134.269+135}\)

\(\Leftrightarrow B=\frac{134.269+135}{134.269+135}\)

\(\Leftrightarrow B=1\)

Vì 1 = 1 nên A =B

\(A=\frac{54\cdot107-53}{53\cdot107+54}=\frac{\left(53+1\right)107-53}{53\cdot107+54}=\frac{53\cdot107+107-53}{53\cdot107+54}=\frac{53\cdot107+54}{53\cdot107+54}=1\)

\(B=\frac{135\cdot268-133}{134\cdot269+135}=\frac{\left(134+1\right)\cdot268-133}{134\cdot269+135}=\frac{134\cdot268+268-133}{34\cdot269+135}=\frac{134\cdot268+135}{134\cdot269+135}=1\)

Vì 1=1 nên A=B

ban tra loi viet de sai roi 269 ma viet thanh 268

bài 2

a, TS= 54 . 107 -53=(53+1) .107-53=53.107+107-53=53.107+ 54

<=> 

\(\frac{TS}{MS}\)=\(\frac{54.107+54}{54.107+54}\)=1

Bài 1 : 

\(a)\) Gọi \(ƯCLN\left(n+1;2n+3\right)=d\)

\(\Rightarrow\)\(\hept{\begin{cases}n+1⋮d\\2n+3⋮d\end{cases}\Rightarrow\hept{\begin{cases}2\left(n+1\right)⋮d\\2n+3⋮d\end{cases}\Rightarrow}\hept{\begin{cases}2n+2⋮d\\2n+3⋮d\end{cases}}}\)

\(\Rightarrow\)\(\left(2n+2\right)-\left(2n+3\right)⋮d\)

\(\Rightarrow\)\(2n+2-2n-3⋮d\)

\(\Rightarrow\)\(\left(-1\right)⋮d\)

\(\Rightarrow\)\(d\inƯ\left(-1\right)\)

Mà \(Ư\left(-1\right)=\left\{1;-1\right\}\)

\(\Rightarrow\)\(d\in\left\{1;-1\right\}\)

Do đó : 

\(ƯCLN\left(n+1;2n+3\right)=\left\{1;-1\right\}\)

Vậy \(\frac{n+1}{2n+3}\) là phân số tối giản với mọi n 

Chúc bạn học tốt ~ 

so sánh hả

ko biết nữa 

\(A>B\)

cách làm