\(\frac{2008}{\sqrt{2009}}+\frac{2009}{\sqrt{2008}}\) và \(\sqrt{2008}+\sqrt{2009}\)
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20 tháng 7 2017
Ta có : \(\frac{2008}{\sqrt{2009}}+\frac{2009}{\sqrt{2008}}=\frac{2009-1}{\sqrt{2009}}+\frac{2008+1}{\sqrt{2008}}=\sqrt{2009}+\sqrt{2008}+\left(\frac{1}{\sqrt{2008}}-\frac{1}{\sqrt{2009}}\right)\)
Vì \(\frac{1}{\sqrt{2008}}>\frac{1}{\sqrt{2009}}\) nên \(\frac{1}{\sqrt{2008}}-\frac{1}{\sqrt{2009}}>0\)
\(\Rightarrow\sqrt{2009}+\sqrt{2008}+\left(\frac{1}{\sqrt{2008}}-\frac{1}{\sqrt{2009}}\right)>\sqrt{2009}+\sqrt{2008}\)
Hay \(\frac{2008}{\sqrt{2009}}+\frac{2009}{\sqrt{2008}}>\sqrt{2008}+\sqrt{2009}\)
26 tháng 2 2020
Ta có :
\(\frac{2008}{\sqrt{2009}}+\frac{2009}{\sqrt{2008}}=\frac{2009}{\sqrt{2009}}-\frac{1}{\sqrt{2009}}+\frac{2008}{\sqrt{2008}}+\frac{1}{\sqrt{2008}}\)
\(=\sqrt{2008}+\sqrt{2009}+\frac{1}{\sqrt{2008}}-\frac{1}{\sqrt{2009}}\)
Mà \(\sqrt{2008}< \sqrt{2009}\Rightarrow\frac{1}{\sqrt{2008}}>\frac{1}{\sqrt{2009}}\Leftrightarrow\frac{1}{\sqrt{2008}}-\frac{1}{\sqrt{2009}}\)
\(\Leftrightarrow\sqrt{2008}+\sqrt{2009}+\frac{1}{\sqrt{2008}}-\frac{1}{\sqrt{2009}}>\sqrt{2008}+\sqrt{2009}\)
⇒ đpcm
R
so sánh \(\frac{2008}{\sqrt[]{2009}}+\frac{2009}{\sqrt[]{2008}}\) và \(\sqrt[]{2008}+\sqrt[]{2009}\)
1
5 tháng 2 2020
Ta có : \(\frac{2008}{\sqrt{2009}}+\frac{2009}{\sqrt{2008}}\) = \(\frac{2009-1}{\sqrt{2009}}+\frac{2008+1}{\sqrt{2008}}\)
= \(\frac{2009}{\sqrt{2009}}-\frac{1}{\sqrt{2009}}+\frac{2008}{\sqrt{2008}}+\frac{1}{\sqrt{2008}}\)
= \(\frac{\left(\sqrt{2009}\right)^2}{\sqrt{2009}}-\frac{1}{\sqrt{2009}}+\frac{\left(\sqrt{2008}\right)^2}{\sqrt{2008}}+\frac{1}{\sqrt{2008}}\)
= \(\sqrt{2009}-\frac{1}{\sqrt{2009}}+\sqrt{2008}+\frac{1}{\sqrt{2008}}\)
Mà \(\frac{1}{\sqrt{2008}}>\frac{1}{\sqrt{2009}}\)
=> \(\frac{1}{\sqrt{2008}}-\frac{1}{\sqrt{2009}}>0\)
=> \(\frac{1}{\sqrt{2008}}-\frac{1}{\sqrt{2009}}+\sqrt{2008}+\sqrt{2009}>\sqrt{2008}+\sqrt{2009}\)
Vậy \(\frac{1}{\sqrt{2008}}-\frac{1}{\sqrt{2009}}+\sqrt{2008}+\sqrt{2009}>\sqrt{2008}+\sqrt{2009}\) .
DM
1
I
1
6 tháng 9 2015
Mời bạn đi nối này http://olm.vn/hoi-dap/question/189394.html
23 tháng 8 2019
1,\(\sqrt{4x+1}-\sqrt{3x-2}=\frac{x+3}{5}\)(đk :\(x\ge\frac{2}{3}\)) (1)
Đặt \(4x+1=a\left(a\ge0\right)\) , \(3x-2=b\left(b\ge0\right)\)
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
23 tháng 8 2019
1,\(\sqrt{4x+1}-\sqrt{3x-2}=\frac{x+3}{5}\) (1) (đk: \(x\ge\frac{2}{3}\))
Đặt \(4x+1=a\left(a\ge0\right)\) ,\(3x-2=b\left(b\ge0\right)\)
=> \(a-b=4x+1-3x+2=x+3\)
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
VT
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
so sánh nha mn