bạn nào giúp mik vs
\(\left(\frac{1}{3}-\frac{3}{2}x\right)^2=\frac{9}{4}\)
\(=>\left(\frac{1}{3}-\frac{3}{2}x\right)^2=\left(\frac{3}{2}\right)^2\)
\(=>\frac{1}{3}-\frac{3}{2}x=\frac{3}{2}\)
\(=>\frac{3}{2}x=\frac{1}{3}-\frac{3}{2}=-\frac{7}{6}\)
\(=>x=-\frac{7}{6}:\frac{3}{2}=-\frac{7}{9}\)
\(\left(\frac{1}{3}-\frac{3}{2}x\right)^2=\frac{9}{4}\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{1}{3}-\frac{3}{2}x=\frac{3}{2}\\\frac{1}{3}-\frac{3}{2}x=\frac{-3}{2}\end{cases}\Leftrightarrow\orbr{\begin{cases}-\frac{3}{2}x=\frac{7}{6}\\-\frac{3}{2}x=\frac{-11}{6}\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{7}{9}\\x=\frac{11}{9}\end{cases}}}\)
\(\left(\frac{1}{3}-\frac{3}{2}x\right)^2=\frac{9}{4}\)
<=>\(\left(\frac{1}{3}-\frac{3}{2}x\right)^2=\left(\frac{3}{2}\right)^2\)
<=>\(\frac{1}{3}-\frac{3}{2}x=\frac{3}{2}\)
<=>\(\frac{3}{2}x=\frac{-7}{6}\)
<=>\(x=\frac{-7}{9}\)
\(\left(\frac{1}{3}-\frac{3}{2}x\right)^2=\frac{9}{4}\)
\(\left(\frac{1}{3}-\frac{3}{2}x\right)^2=\sqrt{\frac{9}{4}}=\frac{3}{2}\)
\(Th1:\frac{1}{3}-\frac{3}{2}x=\frac{3}{2}\Leftrightarrow\frac{3}{2}x=-\frac{7}{6}\Leftrightarrow\frac{-7}{9}\)
\(Th2:\frac{1}{3}-\frac{3}{2}x=-\frac{3}{2}\Leftrightarrow\frac{3}{2}x=\frac{11}{6}\Leftrightarrow\frac{11}{9}\)
\(\left(\frac{1}{3}-\frac{3}{2}x\right)^2=\frac{9}{4}\)
\(\left(\frac{1}{3}-\frac{3}{2}x\right)^2=\left(\frac{3}{2}\right)^2\)
\(\Rightarrow\orbr{\begin{cases}\frac{1}{3}-\frac{3}{2}x=\frac{3}{2}\\\frac{1}{3}-\frac{3}{2}x=-\frac{3}{2}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-\frac{7}{9}\\x=-\frac{6}{9}\end{cases}}\)
vậy.......................................
Từ biểu thức, suy ra:
\(\left(\frac{1}{3}-\frac{3}{2}x\right)^2=\left(\frac{3}{2}\right)^2\)
\(\Leftrightarrow\frac{1}{3}-\frac{3}{2}x=\frac{3}{2}\)
\(\Leftrightarrow\frac{3}{2}x=\frac{1}{3}-\frac{3}{2}=\frac{-7}{6}\)
\(\Leftrightarrow x=\frac{-7}{9}\)
Vậy x=-7/9