1999.00075

Đặt 2000 = a thì ta có

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

\(=\sqrt{\frac{a^4-2a^3+3a^2-2a+1}{a^2}}+\frac{a-1}{a}\)

\(=\frac{a^2-a+1}{a}+\frac{a-1}{a}=a=2000\)

Cảm ơn bạn rất nhiều . 

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

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

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

(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:

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

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

\(=2000\)

Với số nguyên dương n, ta có: 

\(1+n^2+\left(\frac{n}{n+1}\right)^2=\frac{\left(n+1\right)^2+n^2\left(n+1\right)^2+n^2}{\left(n+1\right)^2}=\frac{n^2+2n+1+n^2+n^2\left(n+1\right)^2}{\left(n+1\right)^2}\)

\(=\frac{n^2\left(n+1\right)^2+2n\left(n+1\right)+1}{\left(n+1\right)^2}=\frac{\left[n\left(n+1\right)+1\right]^2}{\left(n+1\right)^2}=\left(\frac{n^2+n+1}{n+1}\right)^2\)

\(\Rightarrow\sqrt{1+n^2+\left(\frac{n}{n+1}\right)^2}=\frac{n^2+n+1}{n+1}=n+\frac{1}{n+1}\)

\(\Rightarrow P=\left(1999+\frac{1}{2000}\right)+\frac{1999}{2000}=1999+1=2000\)

Cách ez hđt lp 8 nhé 

\(P=\sqrt{\left(1+2.1999+1999^2\right)-2.1999+\frac{1999^2}{2000^2}}+\frac{1999}{2000}\)

\(P=\sqrt{\left(1+1999\right)^2-2.1999+\frac{1999^2}{2000^2}}+\frac{1999}{2000}\)

\(P=\sqrt{2000^2-2.1999+\frac{1999^2}{2000^2}}+\frac{1999}{2000}\)

\(P=\sqrt{\left(2000-\frac{1999}{2000}\right)^2}+\frac{1999}{2000}\)

\(P=\left|2000-\frac{1999}{2000}\right|+\frac{1999}{2000}=2000-\frac{1999}{2000}+\frac{1999}{2000}=2000\)

... 

\(\frac{A}{B}=\frac{\frac{2000}{1}+\frac{1999}{2}+...+\frac{1}{2000}+2000}{1+\frac{1999}{2}+\frac{1998}{3}+...+\frac{1}{2000}}\)

\(=\frac{\left[\frac{2001}{1}+1\right]+\left[\frac{2001}{2}+1\right]+...+\left[\frac{2001}{2000}+1\right]+2001}{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2000}}\)

\(=\frac{2001\left[1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2000}\right]}{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2000}}=2001\)

$\ge $ 

là sao?

cho mình hỏi và đề đâu

(1999 + 1999^2 + 1999^3 +...+ 1999^1998)

=1999(1+1999)+1999^3(1+1999)+...+1999^1997(1+1999)

=2000(1999+1999^3+...+1999^19997) 

Do 2000 chia hết cho 2000

=>2000(1999+1999^3+...+1999^19997) chia hết cho 2000

Vậy (1999 + 1999^2 + 1999^3 +...+ 1999^1998) chia hết cho 2000

S= (1999+1999^2+1999^3 +....+1999^1998)

=(1999+1999^2)+(1999^3+1999^4)+...+(1999^1997+1999^1998)

=1999(1+1999)+1999^3(1+1999)+...+1999^1997(1+1999)

=1999.2000+1999^3.2000+...+1999^1997.2000

=2000(1999+1999^3+...+1999^1997) CHIA HET CHO 2000

Vậy S chia het cho 2000(đpcm)

S = 1999 + 1999² + … + 1999¹⁹⁹⁸

S = 1999 ( 1 + 1999 + 1999² + … + 1999¹⁹⁹⁷ )

S = 1999 [ ( 1 + 1999 )( 1 + 1999² + 1999⁴ + … + 1999¹⁹⁹⁶ ) ]

S = 1999 [ 2000 ( 1 + 1999² + 1999⁴ + … + 1999¹⁹⁹⁶ ) ] chia hết cho 2000.

Vậy ta có điều phải chứng minh.