phương trình chứa căn khó?

L

linh123658

Viết lại đề:$x+\sqrt{4-x^2}=2+3x\sqrt{4-x^2}$
Đặt $\sqrt{4-x^2}=y$\Leftrightarrow$x^2+y^2=4$($1$)($y$\geq$0$)
Ta có:$x+y=3xy+2$
\Leftrightarrow$x=\dfrac{y-2}{3y-1}$($2$)
Thay ($2$) vào ($1$) Ta được:
$\dfrac{(y-2)^2}{(3y-1)^2}+y^2-4=0$
\Leftrightarrow$\dfrac{(y-2)^2}{(3y-1)^2}+(y+2)(y-2)=0$
\Leftrightarrow$(y-2)(\dfrac{y-2}{(3y-1)^2}+y+2)=0$
\Leftrightarrow$(y-2)y(10y^2+24y+14)$
$..........................................................$
 
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