a) |x^2 + 2 | x-1/2 | | = x^2 + 2 b) | | 2x-1| + 1/2 | = 4/5

Question

a) |x^2 + 2 | x-1/2 | | = x^2 + 2
b) | | 2x-1| + 1/2 | = 4/5

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Madelyn 6 ngày 2021-12-04T08:03:19+00:00 1 Answers 1 views 0

Answers ( )

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    2021-12-04T08:04:35+00:00

    $a) 2|x-\dfrac{1}{2}|=2$

    $⇔ |x-\dfrac{1}{2}|=1$

    $⇔$ \(\left[ \begin{array}{l}x-\dfrac{1}{2}=1\\x-\dfrac{1}{2}=-1\end{array} \right.\) 

    $⇔$ \(\left[ \begin{array}{l}x=1+\dfrac{1}{2}\\x=-1+\dfrac{1}{2}\end{array} \right.\) 

    $⇔$ \(\left[ \begin{array}{l}x=\dfrac{3}{2}\\x=\dfrac{-1}{2}\end{array} \right.\) 

    $b) |2x-1|+\dfrac{1}{2}=\dfrac{4}{5}$

    $⇔ |2x-1|=\dfrac{4}{5}-\dfrac{1}{2}$

    $⇔ |2x-1|=\dfrac{3}{10}$

    $⇔$ \(\left[ \begin{array}{l}2x-1=\dfrac{3}{10}\\2x-1=\dfrac{-3}{10}\end{array} \right.\) 

    $⇔$ \(\left[ \begin{array}{l}2x=\dfrac{3}{10}+1\\2x=\dfrac{3}{10}+1\end{array} \right.\) 

    $⇔$ \(\left[ \begin{array}{l}2x=\dfrac{13}{10}\\2x=\dfrac{7}{10}\end{array} \right.\) 

    $⇔$ \(\left[ \begin{array}{l}x=\dfrac{13}{20}\\x=\dfrac{7}{20}\end{array} \right.\) 

    Chúc Bạn Học Tốt ^.^

    #NoCopy

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