1.a) x ² -10x =0 b) 8x ²-72x=0 c)x ²+9x =0

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1. `a, x^2-10x=0`

   `<=>x.(x-10)=0`

   `<=>` \(\left[ \begin{array}{l}x=0\\x-10=0\end{array} \right.\) 

   `<=>` \(\left[ \begin{array}{l}x=0\\x=10\end{array} \right.\) 

    Vậy `x ∈ {0;10}`

   `b,8x^2-72x=0`

   `<=> 8x(x-9)=0`

   `<=>` \(\left[ \begin{array}{l}8x=0\\x-9=0\end{array} \right.\) 

   `<=>` \(\left[ \begin{array}{l}x=0\\x=9\end{array} \right.\) 

   `c,x^2+9x=0`

   `<=>x(x+9)=0`

   `<=>` \(\left[ \begin{array}{l}x=0\\x+9=0\end{array} \right.\) 

   `<=>` \(\left[ \begin{array}{l}x=0\\x=-9\end{array} \right.\) 

   xin hay nhất +5*+cảm ơn

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2. Đáp án:

    

   Giải thích các bước giải:

   $a)x^2-10x=0$

   $⇔x(x-10)=0$

   \(\left[ \begin{array}{l}x=0\\x-10=0\end{array} \right.\) 

   \(\left[ \begin{array}{l}x=0\\x=10\end{array} \right.\) 

    $b)8x^2-72x=0$

    $⇔8x(x-9)=0$

   \(\left[ \begin{array}{l}8x=0\\x-9=0\end{array} \right.\) 

   \(\left[ \begin{array}{l}x=0\\x=9\end{array} \right.\) 

   $c)x^2+9x =0$

   $⇔x(x+9) =0$

   \(\left[ \begin{array}{l}x=0\\x+9=0\end{array} \right.\) 

   \(\left[ \begin{array}{l}x=0\\x=-9\end{array} \right.\) 

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