Toán Lớp 9: x^2-6(x+3) .căn(x+1)+14x+3 .căn (x+1)+13=0
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Toán Lớp 9: x^2-6(x+3) .căn(x+1)+14x+3 .căn (x+1)+13=0
Comments ( 1 )
{x^2} – 6\left( {x + 3} \right)\sqrt {x + 1} + 14x + 3\sqrt {x + 1} + 13 = 0\\
\Leftrightarrow {x^2} – \left( {6x + 15} \right)\sqrt {x + 1} + 14x + 13 = 0\\
\Leftrightarrow {x^2} + 14x + 13 = \left( {6x + 15} \right)\sqrt {x + 1} \\
\Leftrightarrow {x^2} + 14x + 13 = \left( {6x + 15} \right)\left( {\sqrt {x + 1} – 3} \right) + 3\left( {6x + 15} \right)\\
\Leftrightarrow {x^2} – 4x – 32 = \left( {6x + 15} \right)\left( {\sqrt {x + 1} – 3} \right)\\
\Leftrightarrow \left( {x – 8} \right)\left( {x + 4} \right) = \left( {6x + 15} \right).\dfrac{{x – 8}}{{\sqrt {x + 1} + 3}}\\
\Leftrightarrow \left( {x – 8} \right)\left( {x + 4 – \dfrac{{6x + 15}}{{\sqrt {x + 1} + 3}}} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 8\\
x + 4 – \dfrac{{6x + 15}}{{\sqrt {x + 1} + 3}} = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 8\\
\left( {x + 4} \right)\left( {\sqrt {x + 1} + 3} \right) = 6x + 15
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 8\\
\left( {x + 1} \right)\sqrt {x + 1} + 3\left( {x + 4} \right) = 6x + 15
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 8\\
\left( {x + 1} \right)\sqrt {x + 1} = 3x + 3
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 8\\
\left( {x + 1} \right)\sqrt {x + 1} = 3\left( {x + 1} \right)
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 8\\
\left( {x + 1} \right)\left( {\sqrt {x + 1} – 3} \right) = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 8\\
x = – 1\\
\sqrt {x + 1} = 3
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 8\\
x = – 1
\end{array} \right.\\
\Rightarrow S = \left\{ {8; – 1} \right\}
\end{array}$