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6 tháng 2 2022

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

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

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

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

⇒Phương trình vô nghiệm

6 tháng 2 2022

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

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

21 tháng 5 2015

=> |x-1/3| +4/5 = 0,7

=> |x-1/3|  = -3,8

mà |x-1/3|  \(\ge\)0

=> ko tồn tại x

11 tháng 8 2021

mà[x-1/3] > 0 nha

25 tháng 10 2019

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

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

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

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

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

\(\left|x-\frac{1}{3}\right|=\frac{10}{5}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{1}{3}=\frac{-10}{5}\\x-\frac{1}{3}=\frac{10}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-5}{3}\\x=\frac{7}{3}\end{matrix}\right.\)

Vậy: \(x\in\left\{\frac{-5}{3};\frac{7}{3}\right\}\)

Chúc bạn học tốt nhé !!!

22 tháng 9 2020

∣ ∣ ∣ x − 1 3 ∣ ∣ ∣ + 4 5 = ∣ ∣ ∣ − 3.2 + 2 5 ∣ ∣ ∣ ⇒ ∣ ∣ ∣ x − 1 3 ∣ ∣ ∣ + 4 5 = ∣ ∣ ∣ − 28 5 ∣ ∣ ∣ ⇒ ∣ ∣ ∣ x − 1 3 ∣ ∣ ∣ + 4 5 = 28 5 ⇒ ∣ ∣ ∣ x − 1 3 ∣ ∣ ∣ = 28 5 − 4 5 = 24 5 ⇒ x − 1 3 = ( ± 24 5 ) ⇒ ⎡ ⎢ ⎢ ⎣ x − 1 3 = 24 5 x − 1 3 = − 24 5 ⇒ ⎡ ⎢ ⎢ ⎣ x = 24 5 + 1 3 x = − 24 5 + 1 3 ⇒ ⎡ ⎢ ⎢ ⎣ x = 77 15 x = − 67 15

28 tháng 3 2018

\(\left|x-\dfrac{1}{3}\right|+\dfrac{4}{5}=\left|\left(-3,2\right)+\dfrac{2}{5}\right|\\ \Rightarrow\left|x-\dfrac{1}{3}\right|+\dfrac{4}{5}=\left|\dfrac{-14}{5}\right|\\ \Rightarrow\left|x-\dfrac{1}{3}\right|+\dfrac{4}{5}=\dfrac{14}{5}\\ \Rightarrow\left|x-\dfrac{1}{3}\right|=2\\ \Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{3}=-2\\x-\dfrac{1}{3}=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-5}{3}\\x=\dfrac{7}{3}\end{matrix}\right.\)

14 tháng 7 2019

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

\(\Rightarrow\left|x-\frac{1}{3}\right|+\frac{4}{5}=\left[-\frac{3}{2}+\frac{2}{5}\right]\)

\(\Rightarrow\left|x-\frac{1}{3}\right|+\frac{4}{5}=-\frac{11}{10}\)

\(\Rightarrow\left|x-\frac{1}{3}\right|=-\frac{11}{10}-\frac{4}{5}\)

\(\Rightarrow\left|x-\frac{1}{3}\right|=-\frac{19}{10}\)

\(\Rightarrow\orbr{\begin{cases}x-\frac{1}{3}=\frac{19}{10}\\x-\frac{1}{3}=-\frac{19}{10}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{67}{30}\\x=-\frac{47}{30}\end{cases}}\)

14 tháng 7 2019

Bạn ơi còn b,c nữa 

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

=>\(\left|x-\dfrac{1}{3}\right|+0,8=\left|-3,2+0,4\right|=2,8\)

=>\(\left|x-\dfrac{1}{3}\right|=2,8-0,8=2\)

=>\(\left[{}\begin{matrix}x-\dfrac{1}{3}=2\\x-\dfrac{1}{3}=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=-\dfrac{5}{3}\end{matrix}\right.\)

14 tháng 11 2023

a: \(\left|7-2x\right|+7=2x\)

=>\(\left|2x-7\right|+7=2x\)

=>\(\left|2x-7\right|=2x-7\)

=>2x-7>=0

=>\(x>=\dfrac{7}{2}\)

b: \(\left|1-x\right|=4x+1\)

=>\(\left|x-1\right|=4x+1\)

=>\(\left\{{}\begin{matrix}4x+1>=0\\\left(4x+1\right)^2=\left(x-1\right)^2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>=-\dfrac{1}{4}\\\left(4x+1\right)^2-\left(x-1\right)^2=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>=-\dfrac{1}{4}\\\left(4x+1-x+1\right)\left(4x+1+x-1\right)=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>=-\dfrac{1}{4}\\5x\left(3x+2\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{1}{4}\\\left[{}\begin{matrix}x=0\left(nhận\right)\\x=-\dfrac{2}{3}\left(loại\right)\end{matrix}\right.\end{matrix}\right.\)

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

=>\(\left|x-\dfrac{1}{3}\right|=\dfrac{16}{5}+\dfrac{2}{5}-\dfrac{4}{5}=\dfrac{14}{5}\)

=>\(\left[{}\begin{matrix}x-\dfrac{1}{3}=\dfrac{14}{5}\\x-\dfrac{1}{3}=-\dfrac{14}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{14}{5}+\dfrac{1}{3}=\dfrac{42+5}{15}=\dfrac{47}{15}\\x=-\dfrac{14}{5}+\dfrac{1}{3}=\dfrac{-42+5}{15}=-\dfrac{37}{15}\end{matrix}\right.\)

d: \(\left|x-7\right|+2x+5=6\)

=>\(\left|x-7\right|=6-2x-5=-2x+1\)

=>\(\left\{{}\begin{matrix}-2x+1>=0\\\left(-2x+1\right)^2=\left(x-7\right)^2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x< =\dfrac{1}{2}\\\left(2x-1\right)^2-\left(x-7\right)^2=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x< =\dfrac{1}{2}\\\left(2x-1+x-7\right)\left(2x-1-x+7\right)=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x< =\dfrac{1}{2}\\\left(3x-8\right)\left(x+6\right)=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x< =\dfrac{1}{2}\\\left[{}\begin{matrix}x=\dfrac{8}{3}\left(loại\right)\\x=-6\left(nhận\right)\end{matrix}\right.\end{matrix}\right.\)

e: 3x-|2x-1|=2

=>|2x-1|=3x-2

=>\(\left\{{}\begin{matrix}3x-2>=0\\\left(3x-2\right)^2=\left(2x-1\right)^2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>=\dfrac{2}{3}\\\left(3x-2\right)^2-\left(2x-1\right)^2=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>=\dfrac{2}{3}\\\left(3x-2-2x+1\right)\left(3x-2+2x-1\right)=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>=\dfrac{2}{3}\\\left(x-1\right)\left(5x-3\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{2}{3}\\\left[{}\begin{matrix}x-1=0\\5x-3=0\end{matrix}\right.\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>=\dfrac{2}{3}\\\left[{}\begin{matrix}x=1\left(nhận\right)\\x=\dfrac{3}{5}\left(loại\right)\end{matrix}\right.\end{matrix}\right.\)

16 tháng 3 2018

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

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

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

\(\Rightarrow x-\frac{1}{3}=\hept{\begin{cases}2\\-2\end{cases}}\)

\(\Rightarrow x=\hept{\begin{cases}\frac{7}{3}\\\frac{-5}{3}\end{cases}}\)