so sánh: A=\(\dfrac{10^{15}+1}{10^{16}+1}\) và B = \(\dfrac{10^{16}+1}{10^{17}+1}\)
So sánh
A = 10^15+1/10^16+1 và B = 10^16+1/10^17+1
So sánh A và B
A= 10^15+1 / 10^16+1
B= 10^16+1 / 10^17+1
Ta có:
10A=1016+10/1016+1=1+(9/1016+1)
10B=1017+10/1017+1=1+(9/1017+1)
Vì 9/1016+1 > 9/1017+1 nên 10A>10B,do đó A>B
Ta có:
10A=10^16+10/10^16+1=1+﴾9/10^16+1﴿
10B=10^17+10/10^17+1=1+﴾9/10^17+1﴿
Vì 9/10^16+1 > 9/10^17+1 nên 10A>10B,do đó A>B
Ta có:
10A= 10^16+10 / 10^16+1
=1+ 9 / 10^16 + 1
10B= 10^17+10 / 10^17+1
=1+ 9 / 10^17 + 1
Vì 9 / 10^16 + 1 > 9 / 10^17 + 1 nên 10A>10B
Do đó A > B
so sánh A=10^15+1/10^16+1
B=10^16+1/10^17+1
TRƯỚC TIÊN TA SO SÁNH 10 VỚI 10B
10A=10^16+10/10^16+1=1\(\frac{9}{16+1}\)
10B=10^17+10/10+17+1=1\(\frac{9}{17+1}\)
VÌ 9/16+1>9/17+1
=>10A>10B
=>A>B
AI TÍCH MK ;MK TÍCH LẠI
so sánh A và B biết A=10^15+11/10^16+1 và B=10^17+1/10^18+1
1. So sánh
a) A=\(\dfrac{10^{15}.11}{10^{16}+1}\) với B=\(\dfrac{10^{16}+10}{10^{16}+1}\)
b) C+\(\dfrac{10^{10}+1}{10^{10}-1}\) với D=\(\dfrac{10^{10}-1}{10^{13}-3}\)
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
A= 1015+1/1016+1
B= 1016+1/1017+1
So sánh A và B
A=10^15+1/10^16+1
=>10A=1+9/10^16+1
B=10^16+1/10^17+1
=>10B=1+9/10^17+1
=>10A>10B=>A>B
Vậy:A>B
So sánh A và B biết:\(A=\frac{10^{15}+1}{10^{16}+1}vàB=\frac{10^{16}+1}{10^{17}+1}\)
So sánh A và B biết:
\(A=\frac{10^{15}+1}{10^{16}+1}\)và \(B=\frac{10^{16}+1}{10^{17}+1}\)
\(10A=\frac{10^{16}+10}{10^{16}+1}=\frac{10^{16}+1+9}{10^{16}+1}=1+\frac{9}{10^{16}+1}\)
\(10B=\frac{10^{17}+10}{10^{17}+1}=\frac{10^{17}+1+9}{10^{17}+1}=1+\frac{9}{10^{17}+1}\)
Nhận thấy: \(\frac{9}{10^{17}+1}< \frac{9}{10^{16}+1}\)=> 10B < 10A
=> A > B
A = ( 10^15+1 ) / ( 10^16+1 ) => 10A = ( 10^16+10 ) / ( 10^16+1 ) = 1 + ( 9/10^15+1 )
B = ( 10^16+1 ) / ( 10^17+1 ) => 10B = ( 10^17+10 ) / ( 10^17+1 ) = 1 + ( 9/10^16+1 )
Vì 10^15+1 < 10^16+1 nên 9/10^15+1 > 9/10^16+1 => 1 + ( 9/10^15+1 ) > 1 + ( 9/10^16+1 )
Vậy A > B