\(9\left(x+1\right)^2-\left(3x-2\right)\left(3x+2\right)=10\)
\(9\left(x^2+2x+1\right)-\left(9x^2-4\right)=10\)
\(9x^2+18x+9-9x^2+4-10=0\)
\(18x+3=0\)
\(18x=-3\)
\(x=\frac{-3}{18}\)
\(x=\frac{-1}{6}\)
vậy \(x=\frac{-1}{6}\)
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Ta có:
\(9\left(x+1\right)^2-\left(3x-2\right)\left(3x+2\right)=10\)
\(\Rightarrow9\left(x^2+2x+1\right)-\left(9x^2-4\right)=10\)
\(\Rightarrow9x^2+18x+9-9x^2+4=10\)
\(\Rightarrow9x^2-9x^2+18x+13=10\)
\(\Rightarrow18x=10-13\)
\(\Rightarrow18x=-3\)
\(\Rightarrow18x=-\frac{1}{6}\)
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