(2x-4)x=0
x(x+2)(x+6)=0
(16-8x)x=0 tìm x thuộc z
help me thanks ^^
(2x-4)x=0
x(x+2)(x+6)=0
(16-8x)x=0 tìm x thuộc z
help me thanks ^^
`a, (2x – 4)x = 0`
`⇒` \(\left[ \begin{array}{l}2x – 4=0\\x=0\end{array} \right.\)
`⇒` \(\left[ \begin{array}{l}2x = 4\\x=0\end{array} \right.\)
`⇒` \(\left[ \begin{array}{l}x = 2\\x = 0 \end{array} \right.\)
`b, x(x + 2)(x + 6) = 0`
`⇒` \(\left[ \begin{array}{l}x = 0\\x + 2=0\\x + 6 = 0 \end{array} \right.\)
`⇒` \(\left[ \begin{array}{l}x = 0\\x = -2\\x = -6 \end{array} \right.\)
`c, (16 – 8x)x = 0`
`⇒` \(\left[ \begin{array}{l}16 – 8x =0\\x = 0 \end{array} \right.\)
`⇒` \(\left[ \begin{array}{l}8x =16\\x = 0 \end{array} \right.\)
`⇒` \(\left[ \begin{array}{l}x = 2\\x = 0 \end{array} \right.\)
Đáp án + Giải thích các bước giải:
`a//(2x-4)x=0`
`=>` \(\left[ \begin{array}{l}2x-4=0\\x=0\end{array} \right.\)
`=>` \(\left[ \begin{array}{l}x=2\\x=0\end{array} \right.\)
Vậy `x∈{2;0}`
`b//x(x+2)(x+6)=0`
`⇒` \(\left[ \begin{array}{l}x=0\\x+2=0\\x+6=0\end{array} \right.\)
`⇒` \(\left[ \begin{array}{l}x=0\\x=-2\\x=-6\end{array} \right.\)
Vậy `x∈{0;-2;-6}`
`c//(16-8x)x=0`
`⇒` \(\left[ \begin{array}{l}16-8x=0\\x=0\end{array} \right.\)
`⇒` \(\left[ \begin{array}{l}x=2\\x=0\end{array} \right.\)
Vậy `x∈{2;0}`