Question

\(\left(3x-1\right).\left(\frac{-1}{2}x+5\right)=0\)

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

\(\left(2x+\frac{3}{5}\right)^{^2}-\frac{9}{25}=0\)

\(3\left(3x-\frac{1}{2}\right)+\frac{1}{9}=0\)

Answer 1

\(\left(3x-1\right)\left(\frac{-1}{2}x+5\right)=0\)

\(\orbr{\begin{cases}3x-1=0\\\frac{-1}{2}x+5=0\end{cases}}\)

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

Answer 2

\(\frac{1}{4}+\frac{1}{3}:(2x-1)=-5\)

\(\Rightarrow\frac{1}{3}:(2x-1)=-5-\frac{1}{4}\)

\(\Rightarrow\frac{1}{3}:(2x-1)=\frac{-21}{4}\)

\(\Rightarrow2x-1=\frac{1}{3}:-\frac{21}{4}\)

\(\Rightarrow2x-1=\frac{1}{3}\cdot-\frac{4}{21}\)

\(\Rightarrow2x-1=\frac{-4}{63}\)

\(\Rightarrow2x=-\frac{4}{63}+1\)

\(\Rightarrow2x=\frac{59}{63}\Leftrightarrow x=\frac{59}{126}\)

Answer 3

\(\left[2x+\frac{3}{5}\right]^2-\frac{9}{25}=0\)

\(\Rightarrow\left[2x+\frac{3}{5}\right]^2=\frac{9}{25}\)

\(\Rightarrow\left[2x+\frac{3}{5}\right]^2=\left[\frac{9}{25}\right]^2\)

\(\Rightarrow2x+\frac{3}{5}=\pm\frac{9}{25}\)

\(\Rightarrow\orbr{\begin{cases}2x+\frac{3}{5}=\frac{9}{25}\\2x+\frac{3}{5}=-\frac{9}{25}\end{cases}}\Rightarrow\orbr{\begin{cases}x=-\frac{3}{25}\\x=-\frac{12}{25}\end{cases}}\)