\(\left|x+\dfrac{4}{5}\right|-\dfrac{3}{5}=\dfrac{6}{5}\)
\(\left|x+\dfrac{4}{5}\right|=\dfrac{6}{5}+\dfrac{3}{5}\)
\(\left|x+\dfrac{4}{5}\right|=\dfrac{9}{5}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{4}{5}=\dfrac{9}{5}\\x+\dfrac{4}{5}=-\dfrac{9}{5}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{5}-\dfrac{4}{5}\\x=-\dfrac{9}{5}-\dfrac{4}{5}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{13}{5}\end{matrix}\right.\)
∣x+4/5∣−3/5=65
<=>∣x+4/5∣=6/5+3/5
<=>∣x+4/5∣=9/5
=> Có hai trường hợp: 1, x+4/5=9/5
<=>x=9/5-4/5
<=>x=5/5=1
2,x+4/5=-9/5
<=>x=-9/5-4/5
<=>x=-13/5
Vậy tập nghiệm của phương trình trên là:
S={1;-13/5}
\(\left|x+\dfrac{4}{5}\right|-\dfrac{3}{5}=\dfrac{6}{5}\)
\(\left|x+\dfrac{4}{5}\right|=\dfrac{6}{5}+\dfrac{3}{5}\)
\(\left|x+\dfrac{4}{5}\right|=\dfrac{9}{5}\)
TH1: \(x+\dfrac{4}{5}=\dfrac{9}{5}\)
\(\Leftrightarrow x=\dfrac{9}{5}-\dfrac{4}{5}\)
\(\Leftrightarrow x=1\)
TH2: \(x+\dfrac{4}{5}=-\dfrac{9}{5}\)
\(\Leftrightarrow x=-\dfrac{9}{5}-\dfrac{4}{5}\)
\(\Leftrightarrow x=-\dfrac{13}{5}\)
Vậy \(x=1\) hoặc \(x=-\dfrac{13}{5}\)
\(\Leftrightarrow|x+\dfrac{4}{5}|=\dfrac{6}{5}+\dfrac{3}{5}\)
\(\Leftrightarrow|x+\dfrac{4}{5}|=\dfrac{9}{5}\)
*TH1: \(x+\dfrac{4}{5}\le0\)(1)
(1) \(\Leftrightarrow-\left(x+\dfrac{4}{5}\right)=\dfrac{9}{5}\)
\(\Leftrightarrow-x-\dfrac{4}{5}=\dfrac{9}{5}\)
\(\Leftrightarrow x=-\dfrac{9}{5}-\dfrac{4}{5}\)
\(\Leftrightarrow x=-\dfrac{13}{5}\)
*TH2: \(x+\dfrac{4}{5}\ge0\left(2\right)\)
(2)\(\Leftrightarrow x+\dfrac{4}{5}=\dfrac{9}{5}\)
\(\Leftrightarrow x=\dfrac{9}{5}-\dfrac{4}{5}\)
\(\Leftrightarrow x=1\)
\(\left|x+\dfrac{4}{5}\right|-\dfrac{3}{5}=\dfrac{6}{5}\)
\(\left|x+\dfrac{4}{5}\right|=\dfrac{9}{5}\)
Trường hợp 1: \(x+\dfrac{4}{5}\ge0\) \(\Rightarrow x\ge\dfrac{-4}{5}\)
Thì \(x+\dfrac{4}{5}=\dfrac{9}{5}\)\(\Leftrightarrow x=1\) (thỏa mãn)
Trường hợp 2: \(x+\dfrac{4}{5}< 0\) \(\Rightarrow x< \dfrac{-4}{5}\)
Thì \(x+\dfrac{4}{5}=-\dfrac{9}{5}\)\(\Leftrightarrow x=-\dfrac{13}{5}\) (thỏa mãn)
Vậy \(x=1;x=-\dfrac{13}{5}\)
\(\left|x+\dfrac{4}{5}\right|-\dfrac{3}{5}=\dfrac{6}{5}\\ \Rightarrow\left|x+\dfrac{4}{5}\right|=\dfrac{6}{5}+\dfrac{3}{5}=\dfrac{9}{5}\\ \Rightarrow\left[{}\begin{matrix}x+\dfrac{4}{5}=\dfrac{9}{5}\\x+\dfrac{4}{5}=-\dfrac{9}{5}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{13}{5}\end{matrix}\right.\)
