Đáp án:
Giải thích các bước giải:
1,
`(x-7)(x+2)=0<=>`\(\left[ \begin{array}{l}x-7=0\\x+2=0\end{array} \right.\) `<=>`\(\left[ \begin{array}{l}x=7\\x=-2\end{array} \right.\)
Vậy `x \in {-2;7}`
$\\$
2,
`x^2-6x=0<=>x(x-6)=0<=>`\(\left[ \begin{array}{l}x=0\\x-6=0\end{array} \right.\) `<=>`\(\left[ \begin{array}{l}x=0\\x=6\end{array} \right.\)
Vậy `x \in { 0;6}`
$\\$
3,
`x^4-3x^2=x^2(x^2-3)=0<=>`\(\left[ \begin{array}{l}x^2=0\\x^2-3=0\end{array} \right.\) `<=>`\(\left[ \begin{array}{l}x=0\\x= ±\sqrt3\end{array} \right.\)
Vậy `x \in {+-\sqrt{3};0}`
$\\$
4,
`3x^2-2x=0<=>x(3x-2)=0<=>`\(\left[ \begin{array}{l}x=0\\3x-2=0\end{array} \right.\) `<=>`\(\left[ \begin{array}{l}x=0\\x=\dfrac32\end{array} \right.\)
Vậy `x \in {0;3/2}`
$\\$
5,
`4x^2-12x+9=9<=>4x^2-12x=0<=>4x(x-3)=0<=>`\(\left[ \begin{array}{l}4x=0\\x-3=0\end{array} \right.\) `<=>`\(\left[ \begin{array}{l}x=0\\x=3\end{array} \right.\)
Vậy `x \in {0;3}`
$\\$
6,
`(2x-3)^3=64<=>(2x-3)^3=4^3<=>2x-3=4<=>2x=7<=>x=7/2`
Vậy `x=7/2`
