Question

Tìm x

\(\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+9\left(x+1\right)^2=15\)

Answer 1

\(\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+9\left(x+1\right)^2=15\)

\(\Leftrightarrow x^3-9x^2+27x-27-\left(x^3-3^3\right)+9\left(x+1\right)^2=15\)

\(\Leftrightarrow x^3-9x^2+27x-27-x^3+27+9.\left(x^2+2x+1\right)=15\)

\(\Leftrightarrow-9x^2+27x+9x^2+18x+9=15\)

\(\Leftrightarrow45x=6\Leftrightarrow x=\frac{2}{15}\)

Vậy: \(S=\left\{\frac{2}{15}\right\}\)

Answer 2

pt <=> \(\left(x-3\right)^3-\left(x^3-27\right)+9\left(x+1\right)^2=15\)

   <=> \(x^3-3x^2.3+3x.3^2-27-x^3+27+9x^2+18x+9=15\)

   <=>  \(45x=6\)

   <=>  \(x=\frac{6}{45}=\frac{2}{15}\)

Answer 3

= \(\frac{2}{15}\) nha bn