1) ( 3+x) ( x-4 ) - x ( x+6 ) = 0
2) 2x ( x - 1 ) = -3x + 3
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4 câu trả lời 296
\[\begin{array}{l}
1){\rm{ }}\left( {{\rm{ }}3 + x} \right){\rm{ }}\left( {{\rm{ }}x - 4{\rm{ }}} \right){\rm{ }} - {\rm{ }}x{\rm{ }}\left( {{\rm{ }}x + 6{\rm{ }}} \right){\rm{ }} = {\rm{ }}0\\
\Leftrightarrow 3{\rm{x}} - 12 + {x^2} - 4{\rm{x}} - {x^2} - 6{\rm{x}} = 0\\
\Leftrightarrow - 7{\rm{x}} - 12 = 0\\
\Leftrightarrow 7{\rm{x}} = - 12\\
\Leftrightarrow x = \frac{{ - 12}}{7}\\
vay:x = \frac{{ - 12}}{7}\\
2){\rm{ }}2x{\rm{ }}\left( {{\rm{ }}x{\rm{ }} - {\rm{ }}1{\rm{ }}} \right){\rm{ }} = {\rm{ }} - 3x{\rm{ }} + {\rm{ }}3\\
\Leftrightarrow 2{\rm{x}}\left( {x - 1} \right) + 3{\rm{x}} - 3 = 0\\
\Leftrightarrow 2{\rm{x}}\left( {x - 1} \right) + 3\left( {x - 1} \right) = 0\\
\Leftrightarrow \left( {x - 1} \right)\left( {2{\rm{x}} + 3} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x - 1 = 0\\
2{\rm{x}} + 3 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x = \frac{{ - 3}}{2}
\end{array} \right.\\
vay:S = \{ 1;\frac{{ - 3}}{2}\}
\end{array}\]
1) ( 3+x) ( x-4 ) - x ( x+6 ) = 0
x^2-x-12-x^2-6x=0
-7x-12=0
-7x=12
x=-12/7
1, (3+x).(x-4)-x(x+6)=0
(=)3x-12+-4x--6x=0
(=)-3x-12=0
(=)x=-4
2,2x(x-1)=-3x+3
(=)-2x+3x-3=0
(=)+x-3=0
=)x=1 hay x=
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