VD1: $3{x^2} + 42x + 83 =
\left( {6x + 42} \right)\sqrt {x + 3} +
\left( {6x + 10} \right)\sqrt {3x + 1} $
Dùng CASIO ta biết pt có nghiệm kép duy nhất x=1
Doán pt có dạng $n{\left(
{\sqrt {x + 3} + ax + b} \right)^2} +
m{\left( {\sqrt {3x + 1} + cx + d}
\right)^2} = 0$
Ta phân tích $\left( {6x +
42} \right)\sqrt {x + 3} =
\frac{1}{{12}}.2\left( {3x + 21} \right)12\sqrt {x + 3} $
\[\left( {6x + 10}
\right)\sqrt {3x + 1} =
\frac{1}{4}2\left( {3x + 5} \right)4\sqrt {3x + 1} \]
Vậy pt:
$\begin{array}{l}
\Leftrightarrow \frac{{{{\left( {3x + 21} \right)}^2} + 144\left( {x + 3}
\right)}}{{12}} + \frac{{{{\left( {3x + 5} \right)}^2} + 16\left( {3x + 1}
\right)}}{4} - \left( {3{x^2} + 42x + 83} \right)\\ = \frac{1}{{12}}{\left( {3x
+ 21 - 12\sqrt {x + 3} } \right)^2} + \frac{1}{4}{\left( {3x + 5 - 4\sqrt {3x +
1} } \right)^2}\\ \Leftrightarrow 0 = \frac{1}{{12}}{\left( {3x + 21 - 12\sqrt
{x + 3} } \right)^2} + \frac{1}{4}{\left( {3x + 5 - 4\sqrt {3x + 1} }
\right)^2}\\ \Leftrightarrow \left\{ \begin{array}{l}3x + 21 - 12\sqrt {x +
3} = 0\\3x + 5 - 4\sqrt {3x + 1} = 0\end{array} \right. \Leftrightarrow x = 1\end{array}$
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