Question

Tìm x: a) ( x - 2018) ^ 2014 = 1

b) (\(\dfrac{1}{3}\) x ) : \(\dfrac{2}{3}\) = \(^7_4:\dfrac{2}{5}\)

Answer 1

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Answer 2

a) \(\left(x-2018\right)^{2014}=1\)

\(\Rightarrow\left(x-2018\right)^{2014}=1^{2014}\)

\(\Rightarrow\left[{}\begin{matrix}x-2018=1\\x-2018=-1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=1+2018\\x=-1+2018\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=2019\\x=2017\end{matrix}\right.\)

b) \(\left(\dfrac{1}{3}x\right):\dfrac{2}{3}=\dfrac{7}{4}:\dfrac{2}{5}\)

\(\Rightarrow\left(\dfrac{1}{3}x\right):\dfrac{2}{3}=\dfrac{7}{4}.\dfrac{2}{5}\)

\(\Rightarrow\left(\dfrac{1}{3}x\right):\dfrac{2}{3}=\dfrac{7}{10}\)

\(\Rightarrow\dfrac{1}{3}x=\dfrac{7}{10}.\dfrac{2}{3}\)

\(\Rightarrow\dfrac{1}{3}x=\dfrac{7}{15}\)

\(\Rightarrow x=\dfrac{7}{15}:\dfrac{1}{3}\)

\(\Rightarrow x=\dfrac{7}{15}.3\)

\(\Rightarrow x=\dfrac{7}{5}\)

Answer 3

a) \(\left(x-2018\right)^{2014}=1\Leftrightarrow\)\(\left[{}\begin{matrix}x-2018=\sqrt[2014]{1}\\x-2018=-\sqrt[2014]{1}\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}x-2018=1\\x-2018=-1\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=2019\\x=2017\end{matrix}\right.\)

Vậy S={2017;2019}

b) \(\left(\dfrac{1}{3}x\right):\dfrac{2}{3}=\dfrac{7}{4}:\dfrac{2}{5}\Leftrightarrow\left(\dfrac{1}{3}x\right):\dfrac{2}{3}=\dfrac{7}{4}.\dfrac{5}{2}\Leftrightarrow\left(\dfrac{1}{3}x\right):\dfrac{2}{3}=\dfrac{35}{8}\Leftrightarrow\dfrac{1}{3}x=\dfrac{35}{8}.\dfrac{2}{3}\Leftrightarrow\dfrac{1}{3}x=\dfrac{35}{12}\Leftrightarrow x=\dfrac{35}{12}:\dfrac{1}{3}\Leftrightarrow x=\dfrac{35}{4}\)

Vậy S={\(\dfrac{35}{4}\)}