a) |x^2 + 2 | x-1/2 | | = x^2 + 2 b) | | 2x-1| + 1/2 | = 4/5 04/12/2021 Bởi Madelyn a) |x^2 + 2 | x-1/2 | | = x^2 + 2 b) | | 2x-1| + 1/2 | = 4/5
$a) 2|x-\dfrac{1}{2}|=2$ $⇔ |x-\dfrac{1}{2}|=1$ $⇔$ \(\left[ \begin{array}{l}x-\dfrac{1}{2}=1\\x-\dfrac{1}{2}=-1\end{array} \right.\) $⇔$ \(\left[ \begin{array}{l}x=1+\dfrac{1}{2}\\x=-1+\dfrac{1}{2}\end{array} \right.\) $⇔$ \(\left[ \begin{array}{l}x=\dfrac{3}{2}\\x=\dfrac{-1}{2}\end{array} \right.\) $b) |2x-1|+\dfrac{1}{2}=\dfrac{4}{5}$ $⇔ |2x-1|=\dfrac{4}{5}-\dfrac{1}{2}$ $⇔ |2x-1|=\dfrac{3}{10}$ $⇔$ \(\left[ \begin{array}{l}2x-1=\dfrac{3}{10}\\2x-1=\dfrac{-3}{10}\end{array} \right.\) $⇔$ \(\left[ \begin{array}{l}2x=\dfrac{3}{10}+1\\2x=\dfrac{3}{10}+1\end{array} \right.\) $⇔$ \(\left[ \begin{array}{l}2x=\dfrac{13}{10}\\2x=\dfrac{7}{10}\end{array} \right.\) $⇔$ \(\left[ \begin{array}{l}x=\dfrac{13}{20}\\x=\dfrac{7}{20}\end{array} \right.\) Chúc Bạn Học Tốt ^.^ #NoCopy Bình luận
$a) 2|x-\dfrac{1}{2}|=2$
$⇔ |x-\dfrac{1}{2}|=1$
$⇔$ \(\left[ \begin{array}{l}x-\dfrac{1}{2}=1\\x-\dfrac{1}{2}=-1\end{array} \right.\)
$⇔$ \(\left[ \begin{array}{l}x=1+\dfrac{1}{2}\\x=-1+\dfrac{1}{2}\end{array} \right.\)
$⇔$ \(\left[ \begin{array}{l}x=\dfrac{3}{2}\\x=\dfrac{-1}{2}\end{array} \right.\)
$b) |2x-1|+\dfrac{1}{2}=\dfrac{4}{5}$
$⇔ |2x-1|=\dfrac{4}{5}-\dfrac{1}{2}$
$⇔ |2x-1|=\dfrac{3}{10}$
$⇔$ \(\left[ \begin{array}{l}2x-1=\dfrac{3}{10}\\2x-1=\dfrac{-3}{10}\end{array} \right.\)
$⇔$ \(\left[ \begin{array}{l}2x=\dfrac{3}{10}+1\\2x=\dfrac{3}{10}+1\end{array} \right.\)
$⇔$ \(\left[ \begin{array}{l}2x=\dfrac{13}{10}\\2x=\dfrac{7}{10}\end{array} \right.\)
$⇔$ \(\left[ \begin{array}{l}x=\dfrac{13}{20}\\x=\dfrac{7}{20}\end{array} \right.\)
Chúc Bạn Học Tốt ^.^
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