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

Tìm x

MH
6 tháng 2 2022 lúc 20:56

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

\(\left|x-\dfrac{1}{3}\right|+\dfrac{4}{5}=-\dfrac{14}{5}\)

\(\left|x-\dfrac{1}{3}\right|=-\dfrac{14}{5}-\dfrac{4}{5}=-\dfrac{18}{5}\)

Vì \(\left|x-\dfrac{1}{3}\right|\ge0\) ∀x

⇒Phương trình vô nghiệm

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H24
6 tháng 2 2022 lúc 21:04

|x-\(\dfrac{1}{3}\)|+\(\dfrac{4}{5}\)=|(\(\dfrac{-16}{5}\))+\(\dfrac{2}{5}\)|

⇒|x-\(\dfrac{1}{3}\)|+\(\dfrac{4}{5}\)=|\(\dfrac{-14}{5}\)|

⇒|x-\(\dfrac{1}{3}\)|+\(\dfrac{4}{5}\)=\(\dfrac{14}{5}\)

⇒|x-\(\dfrac{1}{3}\)|=\(\dfrac{14}{5}\)-\(\dfrac{4}{5}\)

⇒|x-\(\dfrac{1}{3}\)|=2

⇒x-\(\dfrac{1}{3}\)=2⇒x=\(\dfrac{7}{3}\)

hoặc

⇒x-\(\dfrac{1}{3}\)=-2⇒x=\(\dfrac{-5}{3}\)

Vậy x=\(\dfrac{7}{3}\) hoặc x=\(\dfrac{-5}{3}\)

\(\dfrac{-5}{3}\)

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