Question

Đã trả lời bởi chuyên gia

Answer 1

\[\begin{array}{l}
A){\rm{ }}2x\left( {x{\rm{ }} - {\rm{ }}5} \right){\rm{ }} - {\rm{ }}7\left( {x{\rm{ }} - {\rm{ }}5} \right){\rm{ }} = {\rm{ }}0\\
\Leftrightarrow \left( {x - 5} \right)\left( {2{\rm{x}} - 7} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x - 5 = 0\\
2{\rm{x}} - 7 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 5\\
x = \frac{7}{2}
\end{array} \right.\\
Vay:S = \{ 5;\frac{7}{2}\} \\
B){\rm{ }}3x\left( {2x{\rm{ }} - {\rm{ }}7} \right){\rm{ }} - {\rm{ }}4\left( {7{\rm{ }} - {\rm{ }}2x} \right){\rm{ }} = {\rm{ }}0\\
\Leftrightarrow \left( {2{\rm{x}} - 7} \right)\left( {3{\rm{x}} + 4} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
2{\rm{x}} - 7 = 0\\
3{\rm{x + 4 = 0}}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \frac{7}{2}\\
x = \frac{{ - 4}}{3}
\end{array} \right.\\
vay:S = \{ \frac{7}{2};\frac{{ - 4}}{3}\} \\
C){\rm{ }}4x{\rm{ }} - {\rm{ }}6{\rm{ }} + {\rm{ }}5x\left( {2x{\rm{ }} - {\rm{ }}3} \right){\rm{ }} = {\rm{ }}0\\
\Leftrightarrow 2\left( {2{\rm{x}} - 3} \right) + 5{\rm{x}}\left( {2{\rm{x}} - 3} \right) = 0\\
\Leftrightarrow \left( {2{\rm{x}} - 3} \right)\left( {2 + 5{\rm{x}}} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
2{\rm{x}} - 3 = 0\\
2 + 5{\rm{x}} = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \frac{3}{2}\\
x = \frac{{ - 2}}{5}
\end{array} \right.\\
vay:S = \{ \frac{3}{2};\frac{{ - 2}}{5}\}
\end{array}\]

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Answer 2

\[\begin{array}{l}
A){\rm{ }}2x\left( {x{\rm{ }} - {\rm{ }}5} \right){\rm{ }} - {\rm{ }}7\left( {x{\rm{ }} - {\rm{ }}5} \right){\rm{ }} = {\rm{ }}0\\
\Leftrightarrow \left( {x - 5} \right)\left( {2{\rm{x}} - 7} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x - 5 = 0\\
2{\rm{x}} - 7 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 5\\
x = \frac{7}{2}
\end{array} \right.\\
Vay:S = \{ 5;\frac{7}{2}\} \\
B){\rm{ }}3x\left( {2x{\rm{ }} - {\rm{ }}7} \right){\rm{ }} - {\rm{ }}4\left( {7{\rm{ }} - {\rm{ }}2x} \right){\rm{ }} = {\rm{ }}0\\
\Leftrightarrow \left( {2{\rm{x}} - 7} \right)\left( {3{\rm{x}} + 4} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
2{\rm{x}} - 7 = 0\\
3{\rm{x + 4 = 0}}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \frac{7}{2}\\
x = \frac{{ - 4}}{3}
\end{array} \right.\\
vay:S = \{ \frac{7}{2};\frac{{ - 4}}{3}\} \\
C){\rm{ }}4x{\rm{ }} - {\rm{ }}6{\rm{ }} + {\rm{ }}5x\left( {2x{\rm{ }} - {\rm{ }}3} \right){\rm{ }} = {\rm{ }}0\\
\Leftrightarrow 2\left( {2{\rm{x}} - 3} \right) + 5{\rm{x}}\left( {2{\rm{x}} - 3} \right) = 0\\
\Leftrightarrow \left( {2{\rm{x}} - 3} \right)\left( {2 + 5{\rm{x}}} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
2{\rm{x}} - 3 = 0\\
2 + 5{\rm{x}} = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \frac{3}{2}\\
x = \frac{{ - 2}}{5}
\end{array} \right.\\
vay:S = \{ \frac{3}{2};\frac{{ - 2}}{5}\}
\end{array}\]

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