\(\left(3x-5\right)^2-\left(3x+1\right)^2=8\)
\(\Leftrightarrow\left(3x-5-3x-1\right)\left(3x-5+3x+1\right)=8\)
\(\Leftrightarrow-6\left(6x-4\right)=8\)
\(\Leftrightarrow6x-4=-\dfrac{4}{3}\)
\(\Leftrightarrow6x=\dfrac{8}{3}\)
\(\Leftrightarrow x=\dfrac{4}{9}\)
Vậy \(x=\dfrac{4}{9}\)
:D
Ta có:
\(\left(3x-5\right)^2-\left(3x+1\right)^2=8\)
\(\Rightarrow\left(3x-5+3x+1\right)\left(3x-5-3x-1\right)=8\)
\(\Rightarrow\left(6x-4\right)\left(-6\right)=8\)
\(\Rightarrow2\left(3x-2\right).\left(-6\right)=8\)
\(\Rightarrow-12\left(3x-2\right)=8\)
\(\Rightarrow3x-2=-\dfrac{8}{12}\)
\(\Rightarrow3x-2=-\dfrac{2}{3}\)
\(\Rightarrow3x=-\dfrac{2}{3}+2\)
\(\Rightarrow3x=\dfrac{4}{3}\)
\(\Rightarrow x=\dfrac{4}{3}:3\)
\(\Rightarrow x=\dfrac{4}{9}\)
