Question

\(\left(\frac{1}{3}-\frac{3}{2}x\right)^2=\frac{9}{4}\)

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Answer 1

bạn nào giúp mik vs

Answer 2

\(\left(\frac{1}{3}-\frac{3}{2}x\right)^2=\frac{9}{4}\)

\(=>\left(\frac{1}{3}-\frac{3}{2}x\right)^2=\left(\frac{3}{2}\right)^2\)

\(=>\frac{1}{3}-\frac{3}{2}x=\frac{3}{2}\)

\(=>\frac{3}{2}x=\frac{1}{3}-\frac{3}{2}=-\frac{7}{6}\)

\(=>x=-\frac{7}{6}:\frac{3}{2}=-\frac{7}{9}\)

Answer 3

\(\left(\frac{1}{3}-\frac{3}{2}x\right)^2=\frac{9}{4}\)

\(\Leftrightarrow\orbr{\begin{cases}\frac{1}{3}-\frac{3}{2}x=\frac{3}{2}\\\frac{1}{3}-\frac{3}{2}x=\frac{-3}{2}\end{cases}\Leftrightarrow\orbr{\begin{cases}-\frac{3}{2}x=\frac{7}{6}\\-\frac{3}{2}x=\frac{-11}{6}\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{7}{9}\\x=\frac{11}{9}\end{cases}}}\)

Answer 4

\(\left(\frac{1}{3}-\frac{3}{2}x\right)^2=\frac{9}{4}\)

<=>\(\left(\frac{1}{3}-\frac{3}{2}x\right)^2=\left(\frac{3}{2}\right)^2\)

<=>\(\frac{1}{3}-\frac{3}{2}x=\frac{3}{2}\)

<=>\(\frac{3}{2}x=\frac{-7}{6}\)

<=>\(x=\frac{-7}{9}\)

Answer 5

\(\left(\frac{1}{3}-\frac{3}{2}x\right)^2=\frac{9}{4}\)

\(\left(\frac{1}{3}-\frac{3}{2}x\right)^2=\sqrt{\frac{9}{4}}=\frac{3}{2}\)

\(Th1:\frac{1}{3}-\frac{3}{2}x=\frac{3}{2}\Leftrightarrow\frac{3}{2}x=-\frac{7}{6}\Leftrightarrow\frac{-7}{9}\)

\(Th2:\frac{1}{3}-\frac{3}{2}x=-\frac{3}{2}\Leftrightarrow\frac{3}{2}x=\frac{11}{6}\Leftrightarrow\frac{11}{9}\)

Answer 6

\(\left(\frac{1}{3}-\frac{3}{2}x\right)^2=\frac{9}{4}\)

\(\left(\frac{1}{3}-\frac{3}{2}x\right)^2=\left(\frac{3}{2}\right)^2\)

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

\(\Rightarrow\orbr{\begin{cases}x=-\frac{7}{9}\\x=-\frac{6}{9}\end{cases}}\)

vậy.......................................

Answer 7

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