Violympic toán 7

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1 tìm x
\(\dfrac{15-x}{8}=\dfrac{x-23}{10}\)
\(5.3^x=8.3^{^{ }9}\)
\(\dfrac{1}{2}\left|2x-1\right|-3\dfrac{2}{5}=\left(\dfrac{-1}{2}\right).\left(2015\right)^0\)

Trần Nguyễn Bảo Quyên
8 tháng 10 2017 lúc 16:31

*** \(\dfrac{15-x}{8}=\dfrac{x-23}{10}\)

\(\Rightarrow10\left(15-x\right)=8\left(x-23\right)\)

\(\Rightarrow150-10x=8x-184\)

\(\Rightarrow150+184=10x+8x\)

\(\Rightarrow18x=334\)

\(\Rightarrow x=\dfrac{167}{9}\)

*** \(\dfrac{1}{2}\left|2x-1\right|-3\dfrac{2}{5}=\left(-\dfrac{1}{2}\right).\left(2015\right)^0\)

\(\Rightarrow\dfrac{1}{2}\left|2x-1\right|-3\dfrac{2}{5}=\left(-\dfrac{1}{2}\right).1\)

\(\Rightarrow\dfrac{1}{2}\left|2x-1\right|-3\dfrac{2}{5}=\left(-\dfrac{1}{2}\right)\)

\(\Rightarrow\dfrac{1}{2}\left|2x-1\right|-3\dfrac{2}{5}=\left(-\dfrac{1}{2}\right)\)

\(\Rightarrow\dfrac{1}{2}\left|2x-1\right|=\left(-\dfrac{1}{2}\right)+3\dfrac{2}{5}\)

\(\Rightarrow\dfrac{1}{2}\left|2x-1\right|=\dfrac{29}{10}\)

\(\Rightarrow\left|2x-1\right|=\dfrac{29}{10}:\dfrac{1}{2}\)

\(\Rightarrow\left|2x-1\right|=\dfrac{29}{5}\)

\(\Rightarrow\left[{}\begin{matrix}2x-1=\dfrac{29}{5}\\2x-1=-\dfrac{29}{5}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x=\dfrac{34}{5}\\2x=-\dfrac{24}{5}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{17}{5}\\x=-\dfrac{12}{5}\end{matrix}\right.\)