có nhầm đề không vậy phải là 2010-

\(C=\frac{\frac{1}{2008}-\frac{1}{2009}-\frac{1}{2010}}{\frac{5}{2008}-\frac{5}{2009}-\frac{5}{2010}}+\frac{\frac{2}{2007}-\frac{2}{2008}-\frac{2}{2009}}{\frac{3}{2007}-\frac{3}{2008}-\frac{3}{2009}}\)

\(=\frac{\frac{1}{2008}-\frac{1}{2009}-\frac{1}{2010}}{5.\left(\frac{1}{2008}-\frac{1}{2009}-\frac{1}{2010}\right)}+\frac{2.\left(\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}\right)}{3.\left(\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}\right)}\)

\(=\frac{1}{5}+\frac{2}{3}\)

\(=\frac{13}{15}\)

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

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

Mời bạn đi nối này http://olm.vn/hoi-dap/question/189394.html

đặt \(2008=a\)

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

=> A= 2008+1 = 2009

1,\(\sqrt{4x+1}-\sqrt{3x-2}=\frac{x+3}{5}\)(đk :\(x\ge\frac{2}{3}\)) (1)

Đặt \(4x+1=a\left(a\ge0\right)\) , \(3x-2=b\left(b\ge0\right)\)

Có \(a-b=4x+1-3x+2=x+3\)

=> \(\sqrt{a}-\sqrt{b}=\frac{a-b}{5}\)

<=> \(5\left(\sqrt{a}-\sqrt{b}\right)=\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)\)

<=> \(5\left(\sqrt{a}-\sqrt{b}\right)-\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)=0\)

<=> \(\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}+5\right)=0\)

=> \(\sqrt{a}-\sqrt{b}=0\)(vì \(\sqrt{a}+\sqrt{b}+5\ge5\) do a,b\(\ge0\))

<=> \(\sqrt{a}=\sqrt{b}\) <=>\(4x+1=3x-2\) <=> \(x=-3\)(k tm đk)

Vậy pt (1) vô nghiệm

1,\(\sqrt{4x+1}-\sqrt{3x-2}=\frac{x+3}{5}\) (1) (đk: \(x\ge\frac{2}{3}\))

Đặt \(4x+1=a\left(a\ge0\right)\) ,\(3x-2=b\left(b\ge0\right)\)

=> \(a-b=4x+1-3x+2=x+3\)

Có \(\sqrt{a}-\sqrt{b}=\frac{a-b}{5}\)

<=> \(5\left(\sqrt{a}-\sqrt{b}\right)-\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)=0\)

<=> \(\left(\sqrt{a}-\sqrt{b}\right)\left(5-\sqrt{a}-\sqrt{b}\right)=0\)

=> \(\left[{}\begin{matrix}\sqrt{a}=\sqrt{b}\\5=\sqrt{a}+\sqrt{b}\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}4x+1=3x-2\\25=a+b+2\sqrt{ab}\end{matrix}\right.\)<=>\(\left[{}\begin{matrix}x=-3\left(ktm\right)\\25=a+b+2\sqrt{ab}\end{matrix}\right.\)

=> 25=4x+1+3x-2+\(2\sqrt{\left(4x+1\right)\left(3x-2\right)}\)

<=> 26-7x=2\(\sqrt{12x^2-5x-2}\)

<=> \(676-364x+49x^2=48x^2-20x-8\)

<=> \(676-364x+49x^2-48x^2+20x+8=0\)

<=> \(x^2-344x+684=0\)

<=> \(x^2-342x-2x+684=0\)

<=> \(x\left(x-342\right)-2\left(x-342\right)=0\)

<=> (x-2)(x-342)=0

=> \(\left[{}\begin{matrix}x=2\left(tm\right)\\x=342\left(ktm\right)\end{matrix}\right.\)

Vậy pt (1) có nghiệm x=2

Gọi a là tử số, b là mẫu số của phân số A

a = \(\frac{2008}{1}\)+ \(\frac{2007}{2}\)+ \(\frac{2006}{3}\)+ ... + \(\frac{1}{2008}\)

Dãy số a có (2008 - 1)  : 1 + 1 = 2008 số. Và a = ( \(\frac{2008}{1}\)+ \(\frac{1}{2008}\)) x (2008 : 2) 

b = \(\frac{1}{2}\)+ \(\frac{1}{3}\)+ \(\frac{1}{4}\)+ ... + \(\frac{1}{2009}\)

Dãy số b có (2009 - 2) : 1 + 1 = 2008 số. Và b = (\(\frac{1}{2}\)+ \(\frac{1}{2009}\)) x (2008 : 2)

A = [ ( \(\frac{2008}{1}\)+ \(\frac{1}{2008}\)) x (2008 : 2)] : [ (\(\frac{1}{2}\)+ \(\frac{1}{2009}\)) x (2008 : 2)] = ( \(\frac{2008}{1}\)+ \(\frac{1}{2008}\)) :  (\(\frac{1}{2}\)+ \(\frac{1}{2009}\)) 

A = \(\frac{\text{2008 x2008 + 1}}{2008}\)x \(\frac{2x2009+2}{2x2009}\)

A = 2008