Question

Đã trả lời bởi chuyên gia

Answer 1

\[\begin{array}{l}
1)2.(x + 3) - x.(x + 3) = 0\\
\Leftrightarrow \left( {x + 3} \right)\left( {2 - x} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x + 3 = 0\\
2 - x = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = - 3\\
x = 2
\end{array} \right.\\
vay:S = \{ - 3;2\} \\
2)5x.(x - 1) = x - 1\\
\Leftrightarrow 5x\left( {x - 1} \right) - \left( {x - 1} \right) = 0\\
\Leftrightarrow \left( {x - 1} \right)\left( {5x - 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x - 1 = 0\\
5x - 1 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x = \frac{1}{5}
\end{array} \right.\\
vay:S = \{ 1;\frac{1}{5}\} \\
3)2.(x + 5) - {x^2} - 5x = 0\\
\Leftrightarrow 2\left( {x + 5} \right) - x\left( {x + 5} \right) = 0\\
\Leftrightarrow \left( {x + 5} \right)\left( {2 - x} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x + 5 = 0\\
2 - x = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = - 5\\
x = 2
\end{array} \right.\\
vay:S = \{ - 5;2\} \\
4){x^2} - 10x = - 25\\
\Leftrightarrow {x^2} - 10x + 25 = 0\\
\Leftrightarrow {\left( {x - 5} \right)^2} = 0\\
\Leftrightarrow x - 5 = 0\\
\Leftrightarrow x = 5\\
vay:x = 5
\end{array}\]

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Answer 2

\[\begin{array}{l}
1)2.(x + 3) - x.(x + 3) = 0\\
\Leftrightarrow \left( {x + 3} \right)\left( {2 - x} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x + 3 = 0\\
2 - x = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = - 3\\
x = 2
\end{array} \right.\\
vay:S = \{ - 3;2\} \\
2)5x.(x - 1) = x - 1\\
\Leftrightarrow 5x\left( {x - 1} \right) - \left( {x - 1} \right) = 0\\
\Leftrightarrow \left( {x - 1} \right)\left( {5x - 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x - 1 = 0\\
5x - 1 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x = \frac{1}{5}
\end{array} \right.\\
vay:S = \{ 1;\frac{1}{5}\} \\
3)2.(x + 5) - {x^2} - 5x = 0\\
\Leftrightarrow 2\left( {x + 5} \right) - x\left( {x + 5} \right) = 0\\
\Leftrightarrow \left( {x + 5} \right)\left( {2 - x} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x + 5 = 0\\
2 - x = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = - 5\\
x = 2
\end{array} \right.\\
vay:S = \{ - 5;2\} \\
4){x^2} - 10x = - 25\\
\Leftrightarrow {x^2} - 10x + 25 = 0\\
\Leftrightarrow {\left( {x - 5} \right)^2} = 0\\
\Leftrightarrow x - 5 = 0\\
\Leftrightarrow x = 5\\
vay:x = 5
\end{array}\]

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