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

thank ly ly nha

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

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 \(\frac{1998}{1999^{1998}+1}>\frac{1998}{1999^{1999}+1}\Rightarrow1999-\frac{1998}{1999^{1998}+1}< 1999-\frac{1998}{1999^{1999}+1}\)

                                                                   \(\Rightarrow A< B\)

\(A=\dfrac{1999^{1999}+1}{1999^{1998}+1}\)

\(\dfrac{1}{1999}A=\dfrac{1999^{1999}+1}{1999^{1999}+1999}\)

\(\dfrac{1}{1999}A=\dfrac{1999^{1999}}{1999^{1999}}-\dfrac{1998}{1999^{1999}+1999}\)

\(\dfrac{1}{1999}A=1-\dfrac{1998}{1999^{1999}+1999}\)

\(B=\dfrac{1999^{2000}+1}{1999^{1999}+1}\)

\(\dfrac{1}{1999}B=\dfrac{1999^{2000}+1}{1999^{2000}+1999}\)

\(\dfrac{1}{1999}B=\dfrac{1999^{2000}}{1999^{2000}}-\dfrac{1998}{1999^{2000}+1999}\)

\(\dfrac{1}{1999}B=1-\dfrac{1998}{1999^{2000}+1999}\)

Vì  \(\dfrac{1998}{1999^{1999}+1999}>\dfrac{1998}{1999^{2000}+1999}=>\dfrac{1}{1999}A< \dfrac{1}{1999}B=>A< B\)

\(A=\dfrac{1999^{1999}+1}{1999^{1998}+1}=\dfrac{\left(1999^{1999}+1\right)^2}{\left(1999^{1998}+1\right)\left(1999^{1999}+1\right)}\)

\(A=\dfrac{\left(1999^{1999}\right)^2+2.1999^{1999}+1}{\left(1999^{1998}+1\right)\left(1999^{1999}+1\right)}\left(1\right)\)

\(B=\dfrac{1999^{2000}+1}{1999^{1999}+1}=\dfrac{\left(1999^{2000}+1\right)\left(1999^{1998}+1\right)}{\left(1999^{1998}+1\right)\left(1999^{1999}+1\right)}\)

\(B=\dfrac{\left(1999.1999^{1999}+1\right)\left(\dfrac{1}{1999}.1999^{1999}+1\right)}{\left(1999^{1998}+1\right)\left(1999^{1999}+1\right)}\)

\(B=\dfrac{\left(1999^{1999}\right)^2+1999.1999^{1999}+\dfrac{1}{1999}.1999^{1999}+1}{\left(1999^{1998}+1\right)\left(1999^{1999}+1\right)}\)

\(B=\dfrac{\left(1999^{1999}\right)^2+\left(1999+\dfrac{1}{1999}\right).1999^{1999}+1}{\left(1999^{1998}+1\right)\left(1999^{1999}+1\right)}\left(2\right)\)

mà \(\left(1999+\dfrac{1}{1999}\right)>2\)

\(\left(1\right).\left(2\right)\Rightarrow A< B\)

Sửa dòng cuối chỗ ''Vì phần mẫu của \(A< B\)'' thành ''Vì phần mẫu của \(\dfrac{1998}{1999^{1999}+1999}< \dfrac{1998}{1999^{2000}+1999}\)'' nhé.

mk ko bt 123

buồn quá lúc sáng lại bị cô phê bình vì bài này

\(C=\frac{1999^{2000}+1}{1999^{1999}+1}< \frac{1999^{1999}+1+1998}{1999^{2000}+1+1998}\)

\(=\frac{1999^{1999}+1999}{1999^{2000}+1999}\)

\(=\frac{1999\cdot(1999^{1998}+1)}{1999\cdot(1999^{1999}+1)}\)

\(=\frac{1999^{1999}+1}{1999^{1998}+1}=D\)

Vậy...