[Toán 9] Phương trình chứa căn thức

T

tathivanchung

$\sqrt{x + 2\sqrt{x + ... + 2\sqrt{x + 2\sqrt{3x}}}} = x$
\Leftrightarrow $x + 2\sqrt{x + ... + 2\sqrt{x + 2\sqrt{3x}}} = x^2$
\Leftrightarrow $x+2x=x^2$
\Leftrightarrow $x^2-3x=0$
\Leftrightarrow $\begin{bmatrix}x=0&\\x=3&\end{bmatrix}$
Vậy.....
 
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