[Toán 9] $x + \sqrt{x+\frac{1}{2}+\sqrt{x+\frac{1}{4}}}=2$

C

coberacroi_kt

uhm

em tách cái trong dấu căn kia kia hằng đẳg thức
x+1/2 +căn[x+1/4]= x+1/4 +căn[x+1/4] +1/4 =(căn[x+1/4] +1/2)^2
..................
 
L

l4s.smiledonghae

[tex]x + \sqrt {x + \frac{1}{2} + \sqrt {x + \frac{1}{4}} } = 2[/tex]
[tex]{t^2} = x + \frac{1}{4} \Rightarrow x = {t^2} - \frac{1}{4}[/tex]
[tex] \Leftrightarrow {t^2} - \frac{1}{4} + \sqrt {{t^2} + t + \frac{1}{4}} = 2[/tex]
[tex] \Leftrightarrow \sqrt {{t^2} + t + \frac{1}{4}} = \frac{9}{4} - {t^2}[/tex]
[tex] \Rightarrow {t^2} + t + \frac{1}{4} = {\left( {\frac{9}{4} - {t^2}} \right)^2}[/tex]
[tex] \Leftrightarrow {t^4} - \frac{{11}}{2}{t^2} - t + \frac{{77}}{{16}} = 0[/tex]
[tex] \Leftrightarrow 16{t^4} - 88{t^2} - 16t + 77 = 0[/tex]
\Rightarrow tới đây mình bí rồi :-\"
 
L

lovelybones311

$x + \sqrt[]{x+\frac{1}{2}+\sqrt[]{x+\frac{1}{4}}}=2$
<=>[tex]$x+\sqrt[]{(\sqrt[]{x+\frac{1}{4}})^2+2.(\sqrt[]{x+\frac{1}{4}}).\frac{1}{2} + \frac{1}{4}=2$[/tex]
<=>$x+\frac{1}{2}+\sqrt[]{x+\frac{1}{4}}$=2
<=>$(\sqrt{x+\dfrac{1}{4}}+\dfrac{1}{2})^2-2=0$
Tới đây tự làm nhé
 
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