Question

tim x

(3x+1)(3x-1)-9(x-1)^2=1

Answer 1

\(\left(3x+1\right)\left(3x-1\right)-9\left(x-1\right)^2=1\\ \Leftrightarrow\left(3x\right)^2-1^2-9\left(x^2-2x+1\right)=1\\ \Leftrightarrow9x^2-1-9x^2+18x-9=1\\ \Leftrightarrow18x-10=1\\ \Leftrightarrow18x=1+10\\ \Leftrightarrow18x=11\\ \Leftrightarrow x=\dfrac{11}{18}\)

Answer 2

    \(\left(3x+1\right)+\left(3x-1\right)-9\left(x-1\right)^2=1\)

⇔  \(9x^2-1-9\left(x-1\right)^2=1\)

⇔ \(9x^2-9\left(x^2-2x+1\right)=2\)

⇔\(18x=11\)

⇔\(x=\dfrac{11}{18}\)

Vậy ...