\(=\dfrac{2\left(1+2+2^2+...+2^{2008}\right)-\left(1+2+2^2+...+2^{2008}\right)}{1-2^{2009}}\)

\(=\dfrac{\left(2+2^2+2^3+...+2^{2009}\right)-\left(1+2+2^2+...+2^{2008}\right)}{1-2^{2009}}\)

\(=\dfrac{2^{2009}-1}{1-2^{2009}}=-1\)

ta có: \(A=\dfrac{2008^{2009}+2}{2008^{2009}-1}=\dfrac{2008^{2009}-1+3}{2008^{2009}-1}=1+\dfrac{3}{2008^{2009}-1}\)

B=\(\dfrac{2008^{2009}}{2008^{2009}-3}=\dfrac{2008^{2009}-3+3}{2008^{2009}-3}=1+\dfrac{3}{2008^{2009}-3}\)

ta thấy: \(1+\dfrac{3}{2008^{2009}-1}\)<\(1+\dfrac{3}{2008^{2009}-3}\)

vậy A<B

a) \(A=2^{2010}-2^{2009}-2^{2008}-...-2-1\)

\(A=2^{2010}\left(2^{2009}+2^{2008}+...+2+1\right)\)

Đặt \(\text{A = 1 + 2 + . . . + 2^{2008} + 2^{2009}}\)

\(\text{⇒ 2 A = 2 + 2 2 + . . + 2^{2010}}\)

⇒ \(A=2^{2010}-1\)

⇒ \(A=2^{2010}-\left(2^{2010}-1\right)\)

⇒ \(A=1\)

b) \(B=2072\)

c) \(\dfrac{4949}{19800}\)

Xin lỗi mình không có nhiều thời gian để giải thích trên đây á nên tạm gửi ảnh mình tạo nhé . Học tốt !

\(B=\sqrt{1+2008^2+\dfrac{2008^2}{2009^2}}+\dfrac{2008}{2009}=\sqrt{\dfrac{2009^2+2008^2.2009^2+2008^2}{2009^2}}+\dfrac{2008}{2009}=\dfrac{\sqrt{2009^2+\left(2009-1\right)^2.2009^2+2008^2}}{2009}+\dfrac{2008}{2009}=\dfrac{\sqrt{2009^2+2009^4-2.2009.2009^2+2009^2+2008^2}+2008}{2009}=\dfrac{\sqrt{2009^4+2.2009^2-2.\left(2008+1\right).2009^2+2008^2}+2008}{2009}=\dfrac{\sqrt{2009^4+2.2009^2-2.2008.2009^2-2.2009^2+2008^2}+2008}{2009}=\dfrac{\sqrt{2009^4-2.2008.2009^2+2008^2}+2008}{2009}=\dfrac{\sqrt{\left(2009^2-2008\right)^2}+2008}{2009}=\dfrac{2009^2-2008+2008}{2009}=2009\in N\)

Vậy B có giá trị là một số tự nhiên

Xét các số thực a, b, c thỏa mãn \(a+b+c=0\)

\(\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}=\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2}{ab}+\frac{2}{bc}+\frac{2}{ca}-\frac{2}{ab}-\frac{2}{bc}-\frac{2}{ca}}\)

\(=\sqrt{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2-2.\frac{a+b+c}{abc}}\)

\(=\sqrt{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}=\left|\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right|\)

Ta có:

\(B=\sqrt{1+2008^2+\frac{2008^2}{2009^2}}+\frac{2008}{2009}\)

\(=\sqrt{2008^2}.\sqrt{\frac{1}{2018^2}+\frac{1}{1^2}+\frac{1}{2009^2}}+\frac{2008}{2009}\)

\(=2008.\sqrt{\frac{1}{2018^2}+\frac{1}{1^2}+\frac{1}{\left(-2009\right)^2}}+\frac{2008}{2009}\)

\(=2008.\left|\frac{1}{2008}+1-\frac{1}{2009}\right|+\frac{2008}{2009}\)

\(=2008.\left(\frac{1}{2008}+1-\frac{1}{2009}\right)+\frac{2008}{2009}\)

\(=2008.\left(\frac{1}{2008}+1-\frac{1}{2009}+\frac{1}{2009}\right)\)

\(=2008.\frac{2009}{2008}=2009\in\text{N}\)

`A=\sqrt{1+2008^2+2008^2/2009^2}+2008/2009`

`=\sqrt{1+2008^2+2.2008+2008^2/2009^2-2.2008}+2008/2009`

`=\sqrt{(2008+1)^2-2.2008+2008^2/2009^2}+2008/2009`

`=\sqrt{2009-2.2008/2009*2009+2008^2/2009^2}+2008/2009`

`=\sqrt{(2009-2008/2009)^2}+2008/2009`

`=|2009-2008/2009|+2008/2009`

`=2009-2008/2009+2008/2009`

`=2009` là 1 số tự nhiên

Đặt \(2008=a\)

\(\Leftrightarrow A=\sqrt{1+a^2+\dfrac{a^2}{\left(a+1\right)^2}}+\dfrac{a}{a+1}\\ A=\sqrt{\left(a+1\right)^2-\dfrac{2a\left(a+1\right)}{a+1}+\dfrac{a^2}{\left(a+1\right)^2}}+\dfrac{a}{a+1}\\ A=\sqrt{\left(a+1-\dfrac{a}{a+1}\right)^2}+\dfrac{a}{a+1}\\ A=a+1-\dfrac{a}{a+1}+\dfrac{a}{a+1}=a+1=2009\left(đpcm\right)\)

Bài 1: 

Ta có: \(a+b\ge2\sqrt{ab}\)

\(b+c\ge2\sqrt{bc}\)

\(a+c\ge2\sqrt{ac}\)

Do đó: \(2\left(a+b+c\right)\ge2\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\right)\)

hay \(a+b+c\ge\sqrt{ab}+\sqrt{cb}+\sqrt{ac}\)

1.

\(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}+\frac{1}{2^{100}}+\frac{1}{2^{100}}\)

\(=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}+\left(\frac{1}{2^{100}}+\frac{1}{2^{100}}\right)\)

\(=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}+\frac{1}{2^{99}}\)

cứ làm như vậy ta được :

\(=1+1=2\)

2. Ta có :

\(\frac{2008+2009}{2009+2010}=\frac{2008}{2009+2010}+\frac{2009}{2009+2010}\)

vì \(\frac{2008}{2009}>\frac{2008}{2009+2010}\); \(\frac{2009}{2010}>\frac{2009}{2009+2010}\)

\(\Rightarrow\frac{2008}{2009}+\frac{2009}{2010}>\frac{2008+2009}{2009+2010}\)

Đặt \(C=1+2+2^2+...+2^{2007}+2^{2008}\)

\(\Rightarrow2C=2+2^2+2^3+...+2^{2008}+2^{2009}\)

\(\Rightarrow2C-C=2^{2009}-1\)

\(\Rightarrow C=2^{2009}-1\)

\(\Rightarrow B=\dfrac{2^{2009}-1}{1-2^{2009}}=\dfrac{-1\left(1-2^{2009}\right)}{1-2^{2009}}=-1\)

Giải:

B=1+2+2²+2³+...+2²⁰⁰⁸/1-2²⁰⁰⁹

Ta gọi phần tử là A, ta có:

A=1+2+2²+2³+...+2²⁰⁰⁸

2A=2+2²+2³+2⁴+...+2²⁰⁰⁹

2A-A=(2+2²+2³+2⁴+...+2²⁰⁰⁹)-(1+2+2²+2³+...+2²⁰⁰⁸)

A=2²⁰⁰⁹-1

Vậy B=2²⁰⁰⁹-1/1-2²⁰⁰⁹

Chúc bạn học tốt!