SO sánh A và B Biết A=1999^1999+1/1999^2000 +1va B=1999^1998+1/1999^1999+1
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VN
Võ Ngọc Phương
VIP
3 tháng 8 2023
So sánh
\(A=\dfrac{1999^{1999}+1}{1999^{1998}+1}\) ; \(B=\dfrac{1999^{2000}+1}{1999^{1999}+1}\)
Ta có: \(B=\dfrac{1999^{2000}+1}{1999^{1999}+1}>1\) ( vì tử > mẫu )
Do đó: \(B=\dfrac{1999^{2000}+1}{1999^{1999}+1}>\dfrac{1999^{2000}+1+1998}{1999^{1999}+1+1998}=\dfrac{1999^{2000}+1999}{1999^{1999}+1999}=\dfrac{1999.\left(1999^{1999}+1\right)}{1999.\left(1999^{1998}+1\right)}=\dfrac{1999^{1999}+1}{1999^{1998}+1}=A\)
Vậy B > A
Chúc bạn học tốt
NK
1
26 tháng 7 2017
\(\frac{1999^{1999+1}}{1999^{2000+1}}=1-\frac{1}{1999^{2000+1}};\)\(\frac{1999^{1998+1}}{1999^{1999+1}}=1-\frac{1}{1999^{1999+1}}\)
Vì \(1-\frac{1}{1999^{2000+1}}< 1-\frac{1}{1999^{1999+1}}\)nên \(\frac{1999^{1999+1}}{1999^{2000+1}}>\frac{1999^{1998+1}}{1999^{1999+1}}\)
VN
5
TP
1
8 tháng 5 2018
ta có: \(A=\frac{1999^{1999}+1}{1999^{1998}+1}=\frac{1999.\left(1999^{1998}+1\right)-1998}{1999^{1998}+1}=\frac{1999.\left(1999^{1998}+1\right)}{1999^{1998}+1}-\frac{1998}{1999^{1998}+1}\)
\(=1999-\frac{1998}{1999^{1998}+1}\)
\(B=\frac{1999^{2000}+1}{1999^{1999}+1}=\frac{1999.\left(1999^{1999}+1\right)-1998}{1999^{1999}+1}=\frac{1999.\left(1999^{1999}+1\right)}{1999^{1999}+1}-\frac{1998}{1999^{1999}+1}\)
\(=1999-\frac{1998}{1999^{1999}+1}\)
mà \(\frac{1998}{1999^{1998}+1}>\frac{1998}{1999^{1999}+1}\Rightarrow1999-\frac{1998}{1999^{1998}+1}< 1999-\frac{1998}{1999^{1999}+1}\)
\(\Rightarrow A< B\)
LK
0
TU
3
27 tháng 7 2023
\(A=\dfrac{1999^{1999}+1}{1999^{1998}+1}\)
\(\dfrac{1}{1999}A=\dfrac{1999^{1999}+1}{1999^{1999}+1999}\)
\(\dfrac{1}{1999}A=\dfrac{1999^{1999}}{1999^{1999}}-\dfrac{1998}{1999^{1999}+1999}\)
\(\dfrac{1}{1999}A=1-\dfrac{1998}{1999^{1999}+1999}\)
\(B=\dfrac{1999^{2000}+1}{1999^{1999}+1}\)
\(\dfrac{1}{1999}B=\dfrac{1999^{2000}+1}{1999^{2000}+1999}\)
\(\dfrac{1}{1999}B=\dfrac{1999^{2000}}{1999^{2000}}-\dfrac{1998}{1999^{2000}+1999}\)
\(\dfrac{1}{1999}B=1-\dfrac{1998}{1999^{2000}+1999}\)
Vì \(\dfrac{1998}{1999^{1999}+1999}>\dfrac{1998}{1999^{2000}+1999}=>\dfrac{1}{1999}A< \dfrac{1}{1999}B=>A< B\)
26 tháng 7 2023
\(A=\dfrac{1999^{1999}+1}{1999^{1998}+1}=\dfrac{\left(1999^{1999}+1\right)^2}{\left(1999^{1998}+1\right)\left(1999^{1999}+1\right)}\)
\(A=\dfrac{\left(1999^{1999}\right)^2+2.1999^{1999}+1}{\left(1999^{1998}+1\right)\left(1999^{1999}+1\right)}\left(1\right)\)
\(B=\dfrac{1999^{2000}+1}{1999^{1999}+1}=\dfrac{\left(1999^{2000}+1\right)\left(1999^{1998}+1\right)}{\left(1999^{1998}+1\right)\left(1999^{1999}+1\right)}\)
\(B=\dfrac{\left(1999.1999^{1999}+1\right)\left(\dfrac{1}{1999}.1999^{1999}+1\right)}{\left(1999^{1998}+1\right)\left(1999^{1999}+1\right)}\)
\(B=\dfrac{\left(1999^{1999}\right)^2+1999.1999^{1999}+\dfrac{1}{1999}.1999^{1999}+1}{\left(1999^{1998}+1\right)\left(1999^{1999}+1\right)}\)
\(B=\dfrac{\left(1999^{1999}\right)^2+\left(1999+\dfrac{1}{1999}\right).1999^{1999}+1}{\left(1999^{1998}+1\right)\left(1999^{1999}+1\right)}\left(2\right)\)
mà \(\left(1999+\dfrac{1}{1999}\right)>2\)
\(\left(1\right).\left(2\right)\Rightarrow A< B\)
PT
1
CM
19 tháng 4 2018
Đặt A=1998/1999+1999/2000 B=1998+1999/1999+2000 =1998/1999+2000 + 1999/1999+2000 Vì 1998/1998>1998/1999+2000 1999/2000>1999/1999+2000 Nên A>B
ta thấy 19991999 + 1 / 19992000 + 1 < 1 và 1998 > 0
nên ta có: A < 19991999 + 1 + 1998 / 19992000 + 1 + 1998
< 19991999 + 1999 / 19992000 + 1999
< 1999(19991998 + 1) / 1999(19991999 + 1)
< 19991998 + 1 / 19991999 + 1
< B
Vậy A < B
để tui xem lại đã hink như tui làm bài này zùi