Violympic toán 6

CK

a) \(\dfrac{-3}{4}x-\dfrac{4}{10}=\dfrac{1}{5}\)

b) \(\left|x-\dfrac{3}{4}\right|:\dfrac{1}{2}+\dfrac{3}{2}=2\dfrac{1}{2}\)

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SP
4 tháng 5 2018 lúc 22:40

a) \(\dfrac{-3}{4}x-\dfrac{4}{10}=\dfrac{1}{5}\)

\(\dfrac{-3}{4}x=\dfrac{1}{5}+\dfrac{4}{10}=\dfrac{1}{5}+\dfrac{2}{5}=\dfrac{3}{5}\)

\(x=\dfrac{3}{5}:\dfrac{-3}{4}=\dfrac{3}{5}.\dfrac{-4}{3}=\dfrac{3.\left(-4\right)}{5.3}=\dfrac{1.\left(-4\right)}{5.1}=\dfrac{-4}{5}\)

b) \(\left|x-\dfrac{3}{4}\right|:\dfrac{1}{2}+\dfrac{3}{2}=2\dfrac{1}{2}\)

\(\left|x-\dfrac{3}{4}\right|:\dfrac{1}{2}=2\dfrac{1}{2}-\dfrac{3}{2}=\dfrac{5}{2}-\dfrac{3}{2}=\dfrac{2}{2}=1\)

\(\left|x-\dfrac{3}{4}\right|=1.\dfrac{1}{2}=\dfrac{1}{2}\)

\(\Rightarrow x-\dfrac{3}{4}=\dfrac{1}{2}\) hoặc \(x-\dfrac{3}{4}=\dfrac{-1}{2}\)

\(+x-\dfrac{3}{4}=\dfrac{1}{2}\)

\(x=\dfrac{1}{2}+\dfrac{3}{4}=\dfrac{2}{4}+\dfrac{3}{4}=\dfrac{5}{4}\)

\(+x-\dfrac{3}{4}=\dfrac{-1}{2}\)

\(x=\dfrac{-1}{2}+\dfrac{3}{4}=\dfrac{-2}{4}+\dfrac{3}{4}=\dfrac{1}{4}\)

\(\Rightarrow x\in\left\{\dfrac{5}{4};\dfrac{1}{4}\right\}\)

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H24
5 tháng 5 2018 lúc 5:15

\(\dfrac{-3}{4}x-\dfrac{4}{10}=\dfrac{1}{5}\)

\(\Rightarrow\dfrac{-3}{4}x=\dfrac{1}{5}+\dfrac{4}{10}=\dfrac{3}{5}\)

\(\Rightarrow x=\dfrac{3}{5}.\dfrac{-4}{3}=\dfrac{-4}{5}\)

b, \(\left|x-\dfrac{3}{4}\right|:\dfrac{1}{2}+\dfrac{3}{2}=2\dfrac{1}{2}\)

\(\Rightarrow\left|x-\dfrac{3}{4}\right|:\dfrac{1}{2}=1\Rightarrow\left|x-\dfrac{3}{4}\right|=\dfrac{1}{2}\)

\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{3}{4}=\dfrac{1}{2}\\x-\dfrac{3}{4}=\dfrac{-1}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{4}\\x=\dfrac{1}{4}\end{matrix}\right.\)

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