a.x-\(\dfrac{3}{5}\)=\(\left(\dfrac{-1}{2}\right)^2\)
=>x-\(\dfrac{3}{5}\)=\(\dfrac{1}{4}\)
=>x=\(\dfrac{1}{4}\)+\(\dfrac{3}{5}\)=\(\dfrac{17}{20}\)
vậy x=\(\dfrac{17}{20}\)
b.0.75-/\(\dfrac{1}{3}\)x+1/=/-0.5/
=>\(\dfrac{3}{4}\)-/\(\dfrac{1}{3}\)x+1/=\(\dfrac{1}{2}\)
=>/\(\dfrac{1}{3}\)x+1/=\(\dfrac{3}{4}\)-\(\dfrac{1}{2}\)=\(\dfrac{1}{4}\)
=>\(\left\{{}\begin{matrix}\dfrac{1}{3}x+1=\dfrac{-1}{4}=>\dfrac{1}{3}x=\dfrac{-1}{4}-1=\dfrac{-5}{4}=>x=\dfrac{-5}{4}:\dfrac{1}{3}=\dfrac{-15}{4}\\\dfrac{1}{3}x+1=\dfrac{1}{4}=>\dfrac{1}{3}x=\dfrac{1}{4}-1=\dfrac{-3}{4}=>x=\dfrac{-3}{4}:\dfrac{1}{3}=\dfrac{-9}{4}\end{matrix}\right.\)
a) x- \(\dfrac{3}{5}\) = \(\left(-\dfrac{1}{2}\right)^2\)
x- \(\dfrac{3}{5}\) = \(\dfrac{1}{4}\)
x = \(\dfrac{1}{4}+\dfrac{3}{5}\)
x= \(\dfrac{5}{20}\)+\(\dfrac{12}{20}\)
x=\(\dfrac{17}{20}\)
1. \(x-\dfrac{3}{5}=\left(-\dfrac{1}{2}\right)^2\Rightarrow x-\dfrac{3}{5}=\dfrac{1}{4}\)
\(\Rightarrow x=\dfrac{1}{4}+\dfrac{3}{5}=\dfrac{17}{20}\)
Vậy \(x=\dfrac{17}{20}\)
2. \(0.75-\left|\dfrac{1}{3}x+1\right|=\left|-0.5\right|\)
\(\Rightarrow0.75-\left|\dfrac{1}{3}x+1\right|=0.5\)
\(\Rightarrow\left|\dfrac{1}{3}x+1\right|=0,25=\dfrac{1}{4}\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{1}{3}x+1=\dfrac{1}{4}\\\dfrac{1}{3}x+1=-\dfrac{1}{4}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\dfrac{1}{3}x=-\dfrac{3}{4}\\\dfrac{1}{3}x=-\dfrac{5}{4}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{9}{4}\\x=-\dfrac{15}{4}\end{matrix}\right.\)
