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6 tháng 9 2015

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

26 tháng 10 2019

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

23 tháng 8 2019

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

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

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

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

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

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

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

=> \(\sqrt{a}-\sqrt{b}=0\)(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

23 tháng 8 2019

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

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

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

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

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

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

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

=> 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

29 tháng 9 2015

Câu a:

Có dạng tổng quát:\(\frac{1}{\left(k+1\right)\sqrt{k}+k\sqrt{x+1}}=\frac{1}{\sqrt{\left(k+1\right)k}\left(\sqrt{k+1}+\sqrt{k}\right)}=\frac{\sqrt{k+1}-\sqrt{k}}{\sqrt{\left(k+1\right)k}}=\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k-1}}\)

Áp dụng kết quả trên suy ra câu a

`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

26 tháng 9 2021

Đặt \(2008=a\)

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

9 tháng 1 2017

Tổng quát \(n\in N\text{*};n\ge2\) ta có \(\sqrt{1+\frac{1}{n^2}+\frac{1}{\left(n+1\right)^2}}=\sqrt{1+\frac{1}{n^2}+\frac{1}{\left(n+1\right)^2}+\frac{2\left(n+1-n-1\right)}{n\left(n+1\right)}}\)

\(=\sqrt{1+\frac{1}{n^2}+\frac{1}{\left(n+1\right)^2}+2\cdot1\cdot\frac{1}{n}-2\cdot1\cdot\frac{1}{n+1}-2\cdot\frac{1}{n}\cdot\frac{1}{n+1}}\)

\(=\sqrt{\left(1+\frac{1}{n}-\frac{1}{n-1}\right)^2}=1+\frac{1}{n}-\frac{1}{n-1}\).Áp dụng vào ta có:

\(\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+...+\sqrt{1+\frac{1}{2008^2}+\frac{1}{2009^2}}=1+\frac{1}{2}-\frac{1}{3}+1+\frac{1}{3}-\frac{1}{4}+...+1+\frac{1}{2008}-\frac{1}{2009}\)

\(=\left(1+1+...+1\right)+\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2008}-\frac{1}{2009}\right)\)

Super dễ nhé !! Cho bn xử nốt