Ta có :

\(999999^{1999}=999999^{1998}\cdot999999=\left(999999^2\right)^{1998}\cdot999999=...1^{1998}\cdot999999=...1\cdot999999=...9\)

\(555553^3=........7\)

Mà ...9 - ...7 = ...2 ko chia hết cho 10.

=> Ko chứng minh đc

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

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

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

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

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

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

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

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

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

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

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 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

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

                                                                   \(\Rightarrow A< B\)

\(A=\left(1+\dfrac{1999}{1}\right)\left(1+\dfrac{1999}{2}\right)...\left(1+\dfrac{1999}{1000}\right)\)

\(=\dfrac{2000}{1}.\dfrac{2001}{2}.\dfrac{2002}{3}...\dfrac{2999}{1000}\)\(=\dfrac{2000.2001.2002...2999}{1.2.3...1000}\)

\(B=\left(1+\dfrac{1000}{1}\right)\left(1+\dfrac{1000}{2}\right)...\left(1+\dfrac{1000}{1999}\right)\)

\(=\dfrac{1001}{1}.\dfrac{1002}{2}.\dfrac{1003}{3}...\dfrac{2999}{1999}\) \(=\dfrac{1001.1002.1003...2999}{1.2.3...1999}\)

\(\Rightarrow A:B=\left(\dfrac{2000.2001.2002...2999}{1.2.3...1000}\right):\left(\dfrac{1001.1002.1003...2999}{1.2.3...1999}\right)\)

\(=\dfrac{2000.2001.2002...2999}{1.2.3...1000}.\dfrac{1.2.3...1999}{1001.1002.1003...2999}\)

\(=\dfrac{2000.2001.2002...2999}{1.2.3...1000}.\dfrac{1.2.3...1000.\left(1001.1002...1999\right)}{1001.1002.1003....1999.\left(2000.2001.2002.2999\right)}\)\(=\dfrac{1.2.3...1000}{1.2.3...1000}=1\)

Vậy \(\dfrac{A}{B}=1\)

22222222222222222222

so sanh ma bạn