Toán Lớp 6: Bài 9: Tìm x Z biết:
a) x (x – 7) = 0; b) x (x + 11) = 0;
c) (x + 8) (x – 12) = 0; d) (x – 3) (x2 + 3) = 0;
Làm hộ mk với MN ơi khó quá :(((
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Toán Lớp 6: Bài 9: Tìm x Z biết:
a) x (x – 7) = 0; b) x (x + 11) = 0;
c) (x + 8) (x – 12) = 0; d) (x – 3) (x2 + 3) = 0;
Làm hộ mk với MN ơi khó quá :(((
Comments ( 2 )
=>\(\left[ \begin{array}{l}x=0\\x-7=0\end{array} \right.\)
=>\(\left[ \begin{array}{l}x=0\\x=0+7\end{array} \right.\)
=>\(\left[ \begin{array}{l}x=0\\x=7\end{array} \right.\)
Vậy x in {0;7}
=>\(\left[ \begin{array}{l}x=0\\x+11=0\end{array} \right.\)
=>\(\left[ \begin{array}{l}x=0\\x=0-11\end{array} \right.\)
=>\(\left[ \begin{array}{l}x=0\\x=-11\end{array} \right.\)
Vậy x in {0;-11}
=>\(\left[ \begin{array}{l}x+8=0\\x-12=0\end{array} \right.\)
=>\(\left[ \begin{array}{l}x=0-8\\x=0+12\end{array} \right.\)
=>\(\left[ \begin{array}{l}x=-8\\x=12\end{array} \right.\)
Vậy x in {12;-8}
(x – 3) (2x + 3) = 0
=>\(\left[ \begin{array}{l}x-3=0\\2x+3=0\end{array} \right.\)
=>\(\left[ \begin{array}{l}x=0+3\\2x=0-3\end{array} \right.\)
=>\(\left[ \begin{array}{l}x=3\\2x=-3\end{array} \right.\)
=>\(\left[ \begin{array}{l}x=3\\x=-3:2\end{array} \right.\)
=>\(\left[ \begin{array}{l}x=3\\x=\frac{-3}{2}∉ Z(loại)\end{array} \right.\)
Vậy x=3