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

1999/1990 và 2000/1991

1999/1990-1=9/1990

2000/1991-1=9/1991

Vì 9/1990>9/1991 nên 1999/1990 < 2000/1991

2780/2770 và 2555/2550

2780/2770 -1=1/277

2555/2550-1=1/510

Vì 1/277>1/510 nên 2780/2770 < 2555/2550

8089/8080 và 9879/9870

8089/8080-1=9/8080

9879/9870-1=3/3290=9/9870

Vì 9/8080> 9/9870 nên 8089/8080< 9879/9870

\(B=\frac{1999+2000}{2000+2001}\)

\(B=\frac{1999}{2000+2001}+\frac{2000}{2000+2001}\)

Vì \(\frac{1999}{2000+2001}< \frac{1999}{2000}\) ; \(\frac{2000}{2000+2001}< \frac{2000}{2001}\)

\(\Rightarrow\)\(B=\frac{1999}{2000+2001}+\frac{2000}{2000+2001}\)<  \(A=\frac{1999}{2000}+\frac{2000}{2001}\)

\(\Rightarrow\)B < A

Vậy B < A

vì 2 phan số = 1 nên khi cộng với 1 thì = 2 mà 2= 2 nên 2 phân số bằng nhau

\(P=\sqrt{1+1999^2+\dfrac{1999^2}{2000^2}}+\dfrac{1999}{2000}=\sqrt{\dfrac{2000^2+1999^2.2000^2+1999^2}{2000^2}}+\dfrac{1999}{2000}=\dfrac{\sqrt{2000^2+\left(2000-1\right)^2.2000^2+1999^2}}{2000}+\dfrac{1999}{2000}=\dfrac{\sqrt{2000^2+\left(2000^2-2.2000+1\right).2000^2+1999^2}+1999}{2000}=\dfrac{\sqrt{2000^2+2000^4-2.2000.2000^2+2000^2+1999^2}+1999}{2000}=\dfrac{\sqrt{2000^4+2.2000^2-2.\left(1999+1\right).2000^2+1999^2}+1999}{2000}=\dfrac{\sqrt{2000^4+2.2000^2-2.1999.2000^2-2.2000^2+1999^2}+1999}{2000}=\dfrac{\sqrt{2000^4-2.1999.2000^2+1999^2}+1999}{2000}=\dfrac{\sqrt{\left(2000^2-1999\right)^2}+1999}{2000}=\dfrac{2000^2-1999+1999}{2000}=\dfrac{2000^2}{2000}=2000\)

\(\sqrt{1+a^2+\dfrac{a^2}{\left(a+1\right)^2}}\)

\(=\sqrt{1^2+a^2+\left(\dfrac{a}{a+1}\right)^2+2a-\dfrac{2a}{a+1}-\dfrac{2a^2}{a+1}}\)

(vì \(2a-\dfrac{2a}{a+1}-\dfrac{2a^2}{a+1}=\dfrac{2a^2+2a-2a-2a^2}{a+1}=0\))

\(=\sqrt{\left(1+a-\dfrac{a}{a+1}\right)^2}\)

\(=\left|1+a-\dfrac{a}{a+1}\right|\)

Áp dụng vào P, ta có:

\(P=\sqrt{1+1999^2+\dfrac{1999^2}{2000^2}}+\dfrac{1999}{2000}\)

\(=\left|1+1999-\dfrac{1999}{2000}\right|+\dfrac{1999}{2000}\)

\(=2000\)

bằng nhau                    

Ta có:

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

\(=\dfrac{\left(1999^{1999}+1\right)^2-\left(1999^{1998}+1\right)\left(1999^{2000}+1\right)}{\left(1999^{1998}+1\right)\left(1999^{1999}+1\right)}\)

\(=\dfrac{1999^{3998}+2\cdot1999^{1999}+1-\left(1999^{3998}+1999^{1998}+1999^{2000}+1\right)}{1999^{3997}+1999^{1998}+1999^{1999}+1}\)

\(=\dfrac{2\cdot1999^{1999}-1999^{1998}-1999^{2000}}{1999^{3997}+1999^{1998}+1999^{1999}+1}\)

Mà \(2\cdot1999^{1999}-1999^{1998}-1999^{2000}=-\left[\left(1999^{999}\right)^2-2\cdot1999^{999}\cdot1999^{1000}+\left(1999^{1000}\right)^2\right]\)

\(=-\left(1999^{999}-1999^{1000}\right)^2< 0\)

Mà mẫu số > 0

\(\Rightarrow A-B< 0\Leftrightarrow A< B\)

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

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

Vì 1999¹⁹⁹⁸+1<1999¹⁹⁹⁹+1 nên

\(\dfrac{1}{1999^{1998}+1}\)>\(\dfrac{1}{1999^{1999}+1}\) nên \(\dfrac{-1}{1999^{1998}+1}< \dfrac{-1}{1999^{1999}+1}\)

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

A<B