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) \(>\)

b: \(A=\dfrac{10^7-8+13}{10^7-8}=1+\dfrac{13}{10^7-8}\)

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

mà \(10^7-8< 10^8-7\)

nên A>B

c: \(\dfrac{1}{10}A=\dfrac{10^{1992}+1}{10^{1992}+10}=1-\dfrac{9}{10^{1992}+10}\)

\(\dfrac{1}{10}B=\dfrac{10^{1993}+1}{10^{1993}+10}=1-\dfrac{9}{10^{1993}+10}\)

mà \(\dfrac{9}{10^{1992}+10}>\dfrac{9}{10^{1993}+10}\)

nên 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

2/

a/ \(\dfrac{7}{10}=\dfrac{7.15}{10.15}=\dfrac{105}{150}\)

\(\dfrac{11}{15}=\dfrac{11.10}{15.10}=\dfrac{110}{150}\)

-Vì \(\dfrac{105}{150}< \dfrac{110}{150}\)(105<110)nên \(\dfrac{7}{10}< \dfrac{11}{15}\)

b/ \(\dfrac{-1}{8}=\dfrac{-1.3}{8.3}=\dfrac{-3}{24}\)

-Vì \(\dfrac{-3}{24}>\dfrac{-5}{24}\left(-3>-5\right)\)nên\(\dfrac{-1}{8}>\dfrac{-5}{24}\)

c/\(\dfrac{25}{100}=\dfrac{25:25}{100:25}=\dfrac{1}{4}\)

\(\dfrac{10}{40}=\dfrac{10:10}{40:10}=\dfrac{1}{4}\)

-Vì \(\dfrac{1}{4}=\dfrac{1}{4}\)nên\(\dfrac{25}{100}=\dfrac{10}{40}\)

a/ \(\dfrac{7}{10}< \dfrac{11}{15}\)

c/ \(\dfrac{25}{100}=\dfrac{10}{40}\)

a) 

b) 

c) \(\dfrac{25}{100}=\dfrac{10}{40}\)

Em quy đồng rồi so sánh

quy đồng rồi so sánh phân số nhá 

\(\dfrac{5}{6}< \dfrac{9}{10}>\dfrac{7}{8}\)

Lời giải:

\(A=\frac{10^7-5}{10^7-8}=\frac{10^7-8+3}{10^7-8}=1+\frac{3}{10^7-8}\)

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

Ta thấy: \(\frac{3}{10^7-8}=\frac{30}{10^8-80}> \frac{30}{10^8-7}> \frac{13}{10^8-7}\)

\(\Rightarrow 1+\frac{3}{10^7-8}> 1+\frac{13}{10^8-7}\Rightarrow A>B\)

đề sai ko

xem lại giùm cái

\(A=\dfrac{10^7+5}{10^7-8}=\dfrac{10^7-8+13}{10^7-8}=1+\dfrac{13}{10^7-8}< 2\)

\(B=\dfrac{10^8+6}{10^5-7}>\dfrac{10^8}{10^5}=10^3\)

B >>> A

d, Vì B=10^1993+1/10^1992+1 > 1 =>10^1993+1/10^1992+1>10^1993+1+9/10^1992+1+9 = 10^1993+10/10^1992+10= 10. (10^1992+1)/10. (10^1991+1) = 10^1992+1/10^1991+1=A Vậy A=B

cau d B>1 ta co tinh chat (\(\dfrac{a}{b}>\dfrac{a+m}{b+m}\) ) B> \(\dfrac{10^{1993}+1+9}{10^{1992}+1+9}\)\(=\dfrac{10^{1993}+10}{10^{1992}+10}\)=\(\dfrac{10\left(10^{1992}+1\right)}{10\left(10^{1991}+1\right)}\)=\(\dfrac{10^{1992}+1}{10^{1991}+1}\)=A

Suy ra B>A(chuc ban hoc goi nhe)