Question

Đã trả lời bởi chuyên gia

Answer 1

\[\begin{array}{l}
\;a){\rm{ }}x\left( {x + 3} \right){\rm{ }} - {\rm{ }}x{\rm{ }} - {\rm{ }}3 = 0\\
\Leftrightarrow x\left( {x + 3} \right){\rm{ - }}\left( {x + 3} \right){\rm{ = 0}}\\
\Leftrightarrow \left( {x + 3} \right)\left( {x - 1} \right){\rm{ = 0}}\\
\Leftrightarrow \left[ \begin{array}{l}
x + 3 = 0\\
x - 1 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = - 3\\
x = 1
\end{array} \right.\\
{\rm{Vay:S = \{ - 3;1\} }}\\
b){\rm{ }}5\left( {{x^2}{\rm{ }} - {\rm{ }}16} \right){\rm{ }} - {\rm{ }}x + 4 = 0\\
\Leftrightarrow 5\left( {x - 4} \right)\left( {x + 4} \right){\rm{ - }}\left( {x - 4} \right){\rm{ = 0}}\\
\Leftrightarrow \left( {x - 4} \right)\left( {5x + 20 - 1} \right){\rm{ = 0}}\\
\Leftrightarrow \left( {x - 4} \right)\left( {5x + 19} \right){\rm{ = 0}}\\
\Leftrightarrow \left[ \begin{array}{l}
x - 4 = 0\\
5x + 19 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 4\\
x = \frac{{ - 19}}{5}
\end{array} \right.\\
{\rm{vay:S = \{ 4;}}\frac{{ - 19}}{5}{\rm{\} }}\\
c){\rm{ }}9{\left( {x + 2} \right)^2} = {\left( {3{\rm{ }} - {\rm{ }}4x} \right)^2}\\
\Leftrightarrow {[3\left( {x + 2} \right)]^2} = {\left( {3 - 4x} \right)^2}\\
\Leftrightarrow {\left( {3x + 6} \right)^2} = {\left( {3 - 4x} \right)^2}\\
\Leftrightarrow \left[ \begin{array}{l}
3x + 6 = 3 - 4x\\
3x + 6 = 4x - 3
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
3x + 4x = 3 - 6\\
4x - 3x = 6 + 3
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
7x = - 3\\
x = 9
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \frac{{ - 3}}{7}\\
x = 9
\end{array} \right.\\
Vay:S = \{ \frac{{ - 3}}{7};9\}
\end{array}\]

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Answer 2

\[\begin{array}{l}
\;a){\rm{ }}x\left( {x + 3} \right){\rm{ }} - {\rm{ }}x{\rm{ }} - {\rm{ }}3 = 0\\
\Leftrightarrow x\left( {x + 3} \right){\rm{ - }}\left( {x + 3} \right){\rm{ = 0}}\\
\Leftrightarrow \left( {x + 3} \right)\left( {x - 1} \right){\rm{ = 0}}\\
\Leftrightarrow \left[ \begin{array}{l}
x + 3 = 0\\
x - 1 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = - 3\\
x = 1
\end{array} \right.\\
{\rm{Vay:S = \{ - 3;1\} }}\\
b){\rm{ }}5\left( {{x^2}{\rm{ }} - {\rm{ }}16} \right){\rm{ }} - {\rm{ }}x + 4 = 0\\
\Leftrightarrow 5\left( {x - 4} \right)\left( {x + 4} \right){\rm{ - }}\left( {x - 4} \right){\rm{ = 0}}\\
\Leftrightarrow \left( {x - 4} \right)\left( {5x + 20 - 1} \right){\rm{ = 0}}\\
\Leftrightarrow \left( {x - 4} \right)\left( {5x + 19} \right){\rm{ = 0}}\\
\Leftrightarrow \left[ \begin{array}{l}
x - 4 = 0\\
5x + 19 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 4\\
x = \frac{{ - 19}}{5}
\end{array} \right.\\
{\rm{vay:S = \{ 4;}}\frac{{ - 19}}{5}{\rm{\} }}\\
c){\rm{ }}9{\left( {x + 2} \right)^2} = {\left( {3{\rm{ }} - {\rm{ }}4x} \right)^2}\\
\Leftrightarrow {[3\left( {x + 2} \right)]^2} = {\left( {3 - 4x} \right)^2}\\
\Leftrightarrow {\left( {3x + 6} \right)^2} = {\left( {3 - 4x} \right)^2}\\
\Leftrightarrow \left[ \begin{array}{l}
3x + 6 = 3 - 4x\\
3x + 6 = 4x - 3
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
3x + 4x = 3 - 6\\
4x - 3x = 6 + 3
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
7x = - 3\\
x = 9
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \frac{{ - 3}}{7}\\
x = 9
\end{array} \right.\\
Vay:S = \{ \frac{{ - 3}}{7};9\}
\end{array}\]

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