`d)|x^2-x+1|-|1-3x|=3`
Điều kiện:`x≤1/3`
` ⇔)|x^2-x+1|-|1-3x|=3`
Ta có:`x^2-x+1`
`=x^2-2.1/2.x+(1/2)^2+3/4`
`=(x-1/2)^2+3/4`
Mà:
`(x-1/2)^2≥0`
`=>(x-1/2)^2+3/4≥3/4`
`⇔x^2-x+1-|1-3x|=3`
`⇔x^2-x-2-|1-3x|=0`
`⇔|1-3x|=x^2-x-2`
`⇔`\(\left[ \begin{array}{l}1-3x=x^2-x-2\\1-3x=-(x^2-x+2)+3\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}1-3x=x^2-x-2\\1-3x=-x^2+x+1\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}-x^2+x+2+1-3x=0\\x^2-x-2+1-3x=0\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x-x^2-3x+3=0\\x^2-4x-1=0\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x(x-1)-3(x-1)=0\\x^2-4x+2-3=0\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}(x-1)(x-3)=0\\(x-2)^2-(\sqrt{3})^2=0\end{array} \right.\)
Phần còn lại mình làm trong hình nhé.
