mk ko bt 123

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

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

\(\frac{1999^{1999+1}}{1999^{2000+1}}=1-\frac{1}{1999^{2000+1}};\)\(\frac{1999^{1998+1}}{1999^{1999+1}}=1-\frac{1}{1999^{1999+1}}\)

Vì \(1-\frac{1}{1999^{2000+1}}< 1-\frac{1}{1999^{1999+1}}\)nên \(\frac{1999^{1999+1}}{1999^{2000+1}}>\frac{1999^{1998+1}}{1999^{1999+1}}\)

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so sanh ma bạn

Bài này dễ  mà?

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.

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