Question

4, so sánh A và B:

a,A=\(\dfrac{3}{8^3}+\dfrac{7}{8^4}\);B=\(\dfrac{7}{8^3}+\dfrac{3}{8^4}\)

b,A=\(\dfrac{10^7+5}{10^7-8}\);B=\(\dfrac{10^8+6}{10^8-7}\)

c,A=\(\dfrac{10^{1992}+1}{10^{1991}+1}\);B=\(\dfrac{10^{1993}+1}{10^{1992}+1}\)

Answer 1

a,A<B

b,A,<B

c,A<B

Answer 2

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