Ta có: (`x` `- 15)` `. x` `= 0`
` ⇔` \(\left[ \begin{array}{l}x - 15=0\\x=0\end{array} \right.\)
` ⇔` \(\left[ \begin{array}{l}x =15\\x=0\end{array} \right.\)
Vậy `x ∈ {0; 15}`
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Cách 2:
Ta có: (`x` `- 15)` `. x` `= 0`
* TH1: `x - 15 = 0`
`⇔ x = 15`
* TH2: `x = 0`
Vậy `x ∈ {0; 15}`
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Công thức: `A . B = 0 ⇔ `\(\left[ \begin{array}{l}TH1: A=0\\TH2: B =0\end{array} \right.\)
