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\(\frac{1}{2}x+\frac{3}{5}.\left(x-2\right)=3\)
\(\frac{1}{2}.x+\frac{3}{5}.x-\frac{6}{5}=3\)
\(\frac{11}{10}x-\frac{6}{5}=3\)
\(\frac{11}{10}x=\frac{21}{5}\)
\(x=\frac{42}{11}\)
a) 4/3 - x = 3/5 + 1/2
=> 4/3 - x= 0,8
=> x = 4/3 + 0/8
=> x = 5/8
\(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0\)
\(\Leftrightarrow\left(2x+\frac{3}{5}\right)^2=\frac{9}{25}\)
\(\Leftrightarrow\left(2x+\frac{3}{5}\right)^2=\left(\frac{3}{5}\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}2x+\frac{3}{5}=\frac{3}{5}\\2x+\frac{3}{5}=-\frac{3}{5}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x=0\\2x=-\frac{6}{5}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-\frac{3}{5}\end{cases}}\)
_Tần vũ_
\(3\left(3x-\frac{1}{2}\right)^3+\frac{1}{9}=0\)
\(\Leftrightarrow3\left(3x-\frac{1}{2}\right)^3=-\frac{1}{9}\)
\(\Leftrightarrow\left(3x-\frac{1}{2}\right)^3=-\frac{1}{27}\)
\(\Leftrightarrow\left(3x-\frac{1}{2}\right)^3=\left(-\frac{1}{3}\right)^3\)
\(\Leftrightarrow3x-\frac{1}{2}=\frac{-1}{3}\)
\(\Leftrightarrow3x=\frac{1}{6}\)
\(\Leftrightarrow x=\frac{1}{18}\)
_Tần Vũ_
\(|x-\frac{1}{3}|=\frac{5}{6}\)
\(\Rightarrow x-\frac{1}{3}=\frac{5}{6}\) hoặc \(x-\frac{1}{3}=-\frac{5}{6}\)
\(\Rightarrow x=\frac{5}{6}+\frac{1}{3}\) hoặc \(x=-\frac{5}{6}+\frac{1}{3}\)
\(\Rightarrow x=\frac{7}{6}\) hoặc \(x=-\frac{3}{6}=-\frac{1}{2}\)
\(\left|x-\frac{1}{3}\right|=\frac{5}{6}\)
=> Các trường hợp
TH1 : \(x-\frac{1}{3}=\frac{5}{6}\)
\(x=\frac{5}{6}+\frac{1}{3}\)
\(x=\frac{7}{6}\)
TH2 : \(x-\frac{1}{3}=\frac{-5}{6}\)
\(x=\frac{-5}{6}+\frac{1}{3}\)
\(x=\frac{-1}{2}\)