2 câu trả lời
Đáp án:
$x\in\left\{\dfrac12;1\right\}$.
Giải thích các bước giải:
$\begin{array}{l}\left(2x-1\right)^3=\left(2x-1\right)^2\\\Rightarrow \left(2x-1\right)^3-\left(2x-1\right)^2=0\\\Rightarrow\left(2x-1\right)^2.\![(2x-1)-1]=0\\\Rightarrow \left(2x-1\right)^2=0\quad {\rm{or}}\quad 2x-2=0\\\Rightarrow 2x-1=0\qquad\ {\rm{or}}\quad 2x=2\\\Rightarrow 2x=1\qquad\qquad{\rm{or}}\quad x=1\end{array}\\\Rightarrow x=\dfrac12\qquad\qquad\ {\rm{or}}\quad x=1$
Vậy $x\in\left\{\dfrac12;1\right\}$.
`(2.x-1)^3=(2.x-1)^2`
`<=>(2.x-1)^2.(2.x-1)=(2.x-1)^2 `
`<=>(2.x-1)^2.(2.x-1)-(2.x-1)^2=0 `
`<=>(2.x-1)^2.(2.x-1-1)=0 `
`<=>(2.x-1)^2.(2.x-2)=0 `
`<=>`\(\left[ \begin{array}{l}2x-1=0\\2x-2=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}2x=1\\2x=2\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=\frac{1}{2}\\x=1\end{array} \right.\)
