Question

Tìm x :

a ) \(\left(3x-1\right)^2-3x\left(3x+2\right)=0\)

Answer 1

\(\left(3x-1\right)^2-3x\left(3x+2\right)=0\)

<=> \(9x^2-6x+1-9x^2-6x=0\)

<=>\(-12x+1=0\)

<=> \(x=\frac{1}{12}\)

Answer 2

\(\left(3x-1\right)^2-3x\left(3x+2\right)=0\)

\(\Leftrightarrow9x^2-6x+1-9x^2-6x=0\)

\(\Leftrightarrow-12x+1=0\)

\(\Leftrightarrow-12x=0-1\)

\(\Leftrightarrow-12x=-1\)

\(\Leftrightarrow x=-1:-12\)

\(\Leftrightarrow x=\frac{1}{12}\)

Answer 3

\(\left(3x-1\right)^2-3x\left(3x+2\right)=0\)

\(\Leftrightarrow9x^2-6x+1-9x^2-6x=0\)

\(\Leftrightarrow9x^2-9x^2-6x-6x+1=0\)

\(\Leftrightarrow-12+1=0\)

\(\Leftrightarrow-12x=-1\)

\(\Leftrightarrow x=\frac{-1}{-12}\)

\(\Rightarrow x=\frac{1}{12}\)

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