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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
\(A=\dfrac{8^9+12}{8^9+7}=\dfrac{8^9+7+5}{8^9+7}=\dfrac{8^9+7}{8^9+7}+\dfrac{5}{8^9+7}=1+\dfrac{5}{8^9+7}\left(1\right)\)
\(B=\dfrac{8^{10}+4}{8^{10}-1}=\dfrac{8^{10}-1+5}{8^{10}-1}=\dfrac{8^{10}-1}{8^{10}-1}+\dfrac{5}{8^{10}-1}=1+\dfrac{5}{8^{10}-1}\left(2\right)\)
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 \(\left(1\right)+\left(2\right)\Leftrightarrow A< B\)
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
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)
1) \(\dfrac{5}{6}-\dfrac{6}{7}+\dfrac{7}{8}-\dfrac{8}{9}+\dfrac{10}{9}-\dfrac{5}{6}+\dfrac{6}{7}-\dfrac{7}{8}+\dfrac{8}{9}\)
\(=\left(\dfrac{5}{6}-\dfrac{5}{6}\right)-\left(\dfrac{6}{7}+\dfrac{6}{7}\right)+\left(\dfrac{7}{8}-\dfrac{7}{8}\right)-\left(\dfrac{8}{9}+\dfrac{8}{9}\right)+\dfrac{10}{9}\)
\(=0-0+0-0+\dfrac{10}{9}\)
\(=\dfrac{10}{9}\)
2) \(\dfrac{1}{13}+\dfrac{16}{7}+\dfrac{3}{105}-\dfrac{9}{7}-\dfrac{-12}{13}\)
\(=\left(\dfrac{1}{13}-\left(-\dfrac{12}{13}\right)\right)+\left(\dfrac{16}{7}-\dfrac{9}{7}\right)+\dfrac{3}{105}\)
\(=1+1+\dfrac{3}{105}\)
\(=\dfrac{213}{105}=\dfrac{71}{35}\)
\(A=\dfrac{8^9+13}{8^9+7}=\dfrac{8^9+7+6}{8^9+7}=1+\dfrac{6}{8^9+7}\)
\(B=\dfrac{8^{10}-1+5}{8^{10}-1}=1+\dfrac{5}{8^{10}-1}\)
Vì \(1+\dfrac{6}{8^9+7}>1+\dfrac{5}{8^{10}-1}\) \(\Rightarrow A>B\)
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