Đáp án:
Giải thích các bước giải:
d)$(x+\dfrac{1}{2}).(\dfrac{2}{3}-2x)=0$
\(\left[ \begin{array}{l}x+\dfrac{1}{2}=0\\\dfrac{2}{3}-2x=0\end{array} \right.\)
\(\left[ \begin{array}{l}x=-\dfrac{1}{2}\\\dfrac{2}{3}=2x\end{array} \right.\)
\(\left[ \begin{array}{l}x=-\dfrac{1}{2}\\\dfrac{1}{3}=x\end{array} \right.\)
Vậy $x=\dfrac{-1}{2}$ hoặc $x=\dfrac{1}{3}$
h)$x^2=3x$
$x^2-3x=0$
$x(x-3)=0$
\(\left[ \begin{array}{l}x=0\\x-3=0\end{array} \right.\)
\(\left[ \begin{array}{l}x=0\\x=3\end{array} \right.\)
Vậy $x=0$ hoặc $x=3$
k)$(x-1)^5=(x-1)^7$
$(x-1)^5-(x-1)^7=0$
$(x-1)^5.[1-(x-1)^2]=0$
\(\left[ \begin{array}{l}(x-1)^5=0\\1-(x-1)^2=0\end{array} \right.\)
\(\left[ \begin{array}{l}x-1=0\\(x-1)^2=1\end{array} \right.\)
\(\left[ \begin{array}{l}x=1\\x=2\end{array} \right.\)
Vậy $x=2$ hoặc $x=1$
