Tìm x a/ x( $x^{2}$ + 4) = 0 b/ (3x – 2)( 4 + 7x) = 0 c/ $x^{2}$ – 8x = 0

Đáp án:

 a, `x(x^2+4)=0`

`⇒`\(\left[ \begin{array}{l}x=0\\x^2+4=0\end{array} \right.\) 

`⇒`\(\left[ \begin{array}{l}x=0\\x^2=0-4\end{array} \right.\) 

`⇒`\(\left[ \begin{array}{l}x=0\\x^2=-4 (loại)\end{array} \right.\) 

Vậy `x= 0`

b, `(3x – 2)( 4 + 7x) = 0`

`⇒`\(\left[ \begin{array}{l}3x – 2=0\\4 + 7x = 0\end{array} \right.\)

`⇒`\(\left[ \begin{array}{l}3x=2\\7x=-4\end{array} \right.\)

`⇒`\(\left[ \begin{array}{l}x=\dfrac{2}{3}\\x=\dfrac{-4}{7}\end{array} \right.\)

Vậy `x∈{2/3 ; -4/7}`

c, `x^2-8x=0`

`⇒x(x-8)=0`

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

`⇒`\(\left[ \begin{array}{l}x=0\\x=0+8\end{array} \right.\) 

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

Vậy `x∈{0;8}`

`#Hien`

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