Question

Giai ptr

\(\dfrac{1}{2x}+\dfrac{1}{2\left(25-x\right)}=\dfrac{1}{12}\)

Answer 1

ĐKXĐ: \(\left\{{}\begin{matrix}2x\ne0\\2\left(25-x\right)\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\x\ne25\end{matrix}\right.\)

\(\dfrac{1}{2x}+\dfrac{1}{2\left(25-x\right)}=\dfrac{1}{12}\\ \Leftrightarrow\dfrac{25-x+x}{2x\left(25-x\right)}=\dfrac{1}{12}\\ \Leftrightarrow\dfrac{25}{-2x^2+50x}=\dfrac{1}{12}\\ \Leftrightarrow-2x^2+50x=300\\ \Leftrightarrow-2x^2+50x-300=0\\ \Leftrightarrow\left[{}\begin{matrix}x=15\left(tm\right)\\x=10\left(tm\right)\end{matrix}\right.\)

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