AH

Tìm x, biết :
\(\left|3.x-\dfrac{1}{5}\right|=x-5\)

LL
27 tháng 9 2021 lúc 21:38

\(\left|3x-\dfrac{1}{5}\right|=x-5\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-\dfrac{1}{5}=x-5\\3x-\dfrac{1}{5}=5-x\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=-\dfrac{24}{5}\\4x=\dfrac{26}{5}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{12}{5}\\x=\dfrac{13}{10}\end{matrix}\right.\)

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EC
27 tháng 9 2021 lúc 21:40

Nếu \(3x-\dfrac{1}{5}\ge0\Leftrightarrow x\ge\dfrac{1}{15}\)

 \(\Rightarrow\left|3x-\dfrac{1}{5}\right|=x-5\Leftrightarrow3x-\dfrac{1}{5}=x-5\Leftrightarrow2x=\dfrac{24}{5}\Leftrightarrow x=\dfrac{12}{5}\left(tm\right)\)

Nếu \(3x-\dfrac{1}{5}\le0\Leftrightarrow x\le\dfrac{1}{15}\)

\(\Rightarrow\left|3x-\dfrac{1}{5}\right|=x-5\Leftrightarrow-\left(3x-\dfrac{1}{5}\right)=x-5\Leftrightarrow4x=\dfrac{26}{5}\Leftrightarrow x=\dfrac{13}{10}\left(loại\right)\)

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NT
27 tháng 9 2021 lúc 21:40

\(\left|3x-\dfrac{1}{5}\right|=x-5\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-\dfrac{1}{5}=x-5\left(x\ge\dfrac{1}{15}\right)\\3x-\dfrac{1}{5}=5-x\left(x< \dfrac{1}{15}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-\dfrac{24}{5}\\4x=\dfrac{26}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-12}{5}\left(loại\right)\\x=\dfrac{26}{20}=\dfrac{13}{10}\left(loại\right)\end{matrix}\right.\)

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LL
27 tháng 9 2021 lúc 21:41

\(\left|3x-\dfrac{1}{5}\right|=x-5\left(đk:x\ge5\right)\)

\(\Leftrightarrow3x-\dfrac{1}{5}=x-5\Leftrightarrow x=-\dfrac{12}{5}\left(ktm\right)\)

Vậy \(S=\varnothing\)

 

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