a,A<B

b,A,<B

c,A<B

a, \(A-B=\frac{3}{8^3}+\frac{7}{8^4}-\frac{7}{8^3}-\frac{3}{8^4}==\left(\frac{7}{8^4}-\frac{3}{8^4}\right)-\left(\frac{7}{8^3}-\frac{3}{8^3}\right)=\frac{4}{8^4}-\frac{4}{8^3}< 0\)

Vậy A < B

b, \(A=\frac{10^7+5}{10^7-8}=\frac{10^7-8+13}{10^7-8}=1+\frac{13}{10^7-8}\)

\(B=\frac{10^8+6}{10^8-7}=\frac{10^8-7+13}{10^8-7}=1+\frac{13}{10^8-7}\)

Vì \(10^7-8< 10^8-7\Rightarrow\frac{1}{10^7-8}>\frac{1}{10^8-7}\Rightarrow\frac{13}{10^7-8}>\frac{13}{10^8-7}\Rightarrow A>B\)

c,Áp dụng nếu \(\frac{a}{b}>1\Rightarrow\frac{a}{b}>\frac{a+n}{a+n}\) có:

 \(B=\frac{10^{1993}+1}{10^{1992}+1}>\frac{10^{1993}+1+9}{10^{1992}+1+9}=\frac{10^{1993}+10}{10^{1992}+10}=\frac{10\left(10^{1992}+1\right)}{10\left(10^{1991}+1\right)}=\frac{10^{1992}+1}{10^{1991}+1}=A\)

Vậy A < B

a) \(< \)

b) \(>\)

c) \(< \)

d) \(>\)

e) \(< \)

g) \(>\)

h) \(>\)

k) \(>\)

a: \(\dfrac{15}{8}-\dfrac{13}{8}=\dfrac{15-13}{8}=\dfrac{2}{8}=\dfrac{1}{4}\)

b: \(\dfrac{7}{15}-\dfrac{2}{15}=\dfrac{7-2}{15}=\dfrac{5}{15}=\dfrac{1}{3}\)

c: \(\dfrac{11}{12}-\dfrac{2}{12}=\dfrac{11-2}{12}=\dfrac{9}{12}=\dfrac{3}{4}\)

d: \(\dfrac{19}{7}-\dfrac{5}{7}=\dfrac{19-5}{7}=\dfrac{14}{7}=2\)

Ta có: \(A=\dfrac{3^{10}+1}{3^9+1}\)

\(\Leftrightarrow A=\dfrac{3^{10}+3-2}{3^9+1}\)

hay \(A=3-\dfrac{2}{3^9+1}\)

Ta có: \(B=\dfrac{3^9+1}{3^8+1}\)

\(\Leftrightarrow B=\dfrac{3^9+3-2}{3^8+1}\)

hay \(B=3-\dfrac{2}{3^8+1}\)

Ta có: \(3^9+1>3^8+1\)

\(\Leftrightarrow\dfrac{2}{3^9+1}< \dfrac{2}{3^8+1}\)

\(\Leftrightarrow-\dfrac{2}{3^9+1}>-\dfrac{2}{3^8+1}\)

\(\Leftrightarrow-\dfrac{2}{3^9+1}+3>-\dfrac{2}{3^8+1}+3\)

hay A>B

a)\(\dfrac{-8}{9}< \dfrac{-7}{9}\\ \dfrac{6}{7}< \dfrac{11}{10}\)

a) Vì -8<-7 nên \(\dfrac{-8}{9}< \dfrac{-7}{9}\)

b) Ta có: \(\dfrac{6}{7}< 1;\dfrac{11}{10}>1\)

nên \(\dfrac{6}{7}< \dfrac{11}{10}\)

Giải:

a) -8/9 < -7/9

b) Ta có: 6/7 = 60/70

                 11/10 = 77/70

Vì 60/70 < 77/70 nên :

    6/7 < 11/10

\(A=\dfrac{8^9+12}{8^9+7}=\dfrac{8^9+7+5}{8^9+7}=\dfrac{8^9+7}{8^9+7}+\dfrac{5}{8^9+7}=1+\dfrac{5}{8^9+7}\left(1\right)\)

\(B=\dfrac{8^{10}+4}{8^{10}-1}=\dfrac{8^{10}-1+5}{8^{10}-1}=\dfrac{8^{10}-1}{8^{10}-1}+\dfrac{5}{8^{10}-1}=1+\dfrac{5}{8^{10}-1}\left(2\right)\)

Từ \(\left(1\right)+\left(2\right)\Leftrightarrow A>B\)

\(A=\dfrac{8^9+12}{8^9+7}=\dfrac{8^9+7+5}{8^9+7}=\dfrac{8^9+7}{8^9+7}+\dfrac{5}{8^9+7}=1+\dfrac{5}{8^9+7}\left(1\right)\)

\(B=\dfrac{8^{10}+4}{8^{10}-1}=\dfrac{8^{10}-1+5}{8^{10}-1}=\dfrac{8^{10}-1}{8^{10}-1}+\dfrac{5}{8^{10}-1}=1+\dfrac{5}{8^{10}-1}\left(2\right)\)

Từ \(\left(1\right)+\left(2\right)\Leftrightarrow A< B\)

a) \(=\left(13\dfrac{2}{7}+2\dfrac{5}{7}\right):\left(-\dfrac{8}{9}\right)\)

\(=16:\dfrac{-8}{9}=\dfrac{-8\cdot\left(-2\right)\cdot9}{-8}=-18\)

b) 

\(=\left(\dfrac{-6}{11}\cdot\dfrac{11}{-6}\right)\cdot\dfrac{7\cdot10\cdot\left(-2\right)}{10}\)

\(=-14\)

c) \(=\dfrac{-1}{2}\cdot\dfrac{4}{3}\cdot\dfrac{-7}{2}\)

\(=\dfrac{-1\cdot2\cdot2\cdot\left(-7\right)}{2\cdot3\cdot2}=\dfrac{7}{3}\)

em trả lời ccaua này hi vọng thầy còn nhớ em

a) -9/4<`1/3

c) 9/-5<7/-10

a) \(\dfrac{-9}{4}< 0\)

\(0< \dfrac{1}{3}\)

Do đó: \(\dfrac{-9}{4}< \dfrac{1}{3}\)