Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a, Ta có : \(10^{15}\cdot11=10^{15}\left(10+1\right)=10^{16}+10^{15}\)
Vì \(10^{16}+10^{15}>10^{16}+10\)
\(\Rightarrow\dfrac{10^{16}+10^{15}}{10^{16}+1}>\dfrac{10^{16}+10}{10^{16}+1}\)
Hay A>B
b, Ta có : \(C=\dfrac{10^{10}+1}{10^{10}-1}=\dfrac{10^{10}}{10^{10}-1}+\dfrac{1}{10^{10}-1}\)
\(D=\dfrac{10^{10}-1}{10^{13}-3}=\dfrac{10^{10}}{10^{13}-3}+\dfrac{-1}{10^{13}-3}\)
Vì \(\dfrac{10^{10}}{10^{10}-1}>\dfrac{10^{10}}{10^{13}-3};\dfrac{1}{10^{10}-1}>\dfrac{-1}{10^{13}-3}\)
\(\Rightarrow\dfrac{10^{10}+1}{10^{10}-1}>\dfrac{10^{10}-1}{10^{13}-3}\)
Hay C > D
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, \(A-B=\frac{3}{8^3}+\frac{7}{8^4}-\frac{7}{8^3}-\frac{3}{8^4}==\left(\frac{7}{8^4}-\frac{3}{8^4}\right)-\left(\frac{7}{8^3}-\frac{3}{8^3}\right)=\frac{4}{8^4}-\frac{4}{8^3}< 0\)
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
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)
Ta có: \(A=\dfrac{10^{10}+1}{10^{10}-1}=\dfrac{10^{10}-1+2}{10^{10}-1}=1+\dfrac{2}{10^{10}-1}\)
\(B=\dfrac{10^{10}-1}{10^{10}-3}=\dfrac{10^{10}-3+2}{10^{10}-3}=1+\dfrac{2}{10^{10}-3}\)
Vì \(\dfrac{2}{10^{10}-1}< \dfrac{2}{10^{10}-3}\Rightarrow1+\dfrac{2}{10^{10}-1}< 1+\dfrac{2}{10^{10}-3}\)
\(\Rightarrow A< B\)
Vậy A < B
\(A=\dfrac{10^{10}+1}{10^{10}-1}=\dfrac{10^{10}-1+2}{10^{10}-1}=1+\dfrac{2}{10^{10}-1}\)
\(B=\dfrac{10^{10}-1}{10^{10}-3}=\dfrac{10^{10}-3+2}{10^{10}-3}=1+\dfrac{2}{10^{10}-3}\)
Vì \(10^{10}-1>10^{10}-3\) nên ta có
\(\dfrac{2}{10^{10}-1}< \dfrac{2}{10^{10}-3}\)
Vậy \(A< B\)
\(A=\dfrac{-9}{10^{2010}}+\dfrac{-19}{10^{2011}}=\dfrac{-90}{10^{2011}}+\dfrac{-19}{10^{2011}}=\dfrac{\left(-90\right)+\left(-19\right)}{10^{2011}}=\dfrac{-109}{10^{2011}}\)\(B=\dfrac{-9}{10^{2011}}+\dfrac{-19}{10^{2010}}=\dfrac{-9}{10^{2011}}+\dfrac{-190}{10^{2011}}=\dfrac{\left(-9\right)+\left(-190\right)}{10^{2011}}=\dfrac{-199}{10^{2011}}\)\(\text{Vì }\dfrac{-109}{10^{2011}}>\dfrac{-199}{10^{2011}}\text{ nên }A>B\)
Vì 18/91 < 18/90 =1/5
23/114>23115=1/5
vậy 18/91<1/5<23/114
suy ra 18/91<23/114
vì 21/52=210/520
Mà 210/520=1-310/520
213/523=1-310/523
310/520>310/523
vậy 210/520<213/523
suy ra 21/52<213/523
Ta có :
\(A=\dfrac{10^{10}+1}{10^{10}-1}=\dfrac{10^{10}-1+1+1}{10^{10}-1}=\dfrac{\left(10^{10}-1\right)+1}{10^{10}-1}=1+\dfrac{2}{2016^{10}-1}\) \(\left(1\right)\)
\(B=\dfrac{10^{10}-1}{10^{10}-3}=\dfrac{10^{10}-3-1+3}{10^{10}-3}=\dfrac{\left(10^{10}-3\right)+2}{10^{10}-3}=1+\dfrac{2}{10^{10}-3}\) \(\left(2\right)\)
Từ \(\left(1\right)+\left(2\right)\) \(\Rightarrow\) \(A< B\)
Chúc bn học tốt!!
Ta có A=\(\dfrac{10^{10}+1}{10^{10}-1}=\dfrac{10^{10}-1+2}{10^{10}-1}=\dfrac{10^{10}-1}{10^{10}-1}+\dfrac{2}{10^{10}-1}\)
\(=1+\dfrac{2}{10^{10}-1}\)
B=\(\dfrac{10^{10}-1}{10^{10}-3}=\)\(\dfrac{10^{10}-3+2}{10^{10}-3}\)=\(\dfrac{10^{10}-3}{10^{10}-3}+\dfrac{2}{10^{10}-3}\)
=\(1+\dfrac{2}{10^{10}-3}\)