Đáp án:
`x\in{17/6;-15/6}`
Giải thích các bước giải:
`d)`
`67/9-(x-1/6)^2=1/3`
`=>(x-1/6)^2=67/9-1/3`
`=>(x-1/6)^2=67/9-3/9`
`=>(x-1/6)^2=64/9`
`=>`\(\left[ \begin{array}{l}(x-\dfrac{1}{6})^2=(\dfrac{8^2}{3^2})\\(x-\dfrac{1}{6})^2=(\dfrac{-8^2}{3^2})\end{array} \right.\)
`=>`\(\left[ \begin{array}{l}(x-\dfrac{1}{6})^2=(\dfrac{8}{3})^2\\(x-\dfrac{1}{6})^2=(\dfrac{-8}{3})^2\end{array} \right.\)
`=>`\(\left[ \begin{array}{l}x-\dfrac{1}{6}=\dfrac{8}{3}\\x-\dfrac{1}{6}=\dfrac{-8}{3}\end{array} \right.\)
`=>`\(\left[ \begin{array}{l}x=\dfrac{8}{3}+\dfrac{1}{6}\\x=\dfrac{-8}{3}+\dfrac{1}{6}\end{array} \right.\)
`=>`\(\left[ \begin{array}{l}x=\dfrac{16}{6}+\dfrac{1}{6}\\x=x=\dfrac{-16}{3}+\dfrac{1}{6}\end{array} \right.\)
`=>`\(\left[ \begin{array}{l}x=\dfrac{17}{6}\\x=\dfrac{-15}{6}\end{array} \right.\)
`text{Vậy}\quadx\in{17/6;-15/6}`
