Question

[Bao Cao Su]

tìm x biết

b) \(\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2+3x^2=-33\)

Answer 1

[Thanh Ngân]

\(\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2+3x^2=-33\)

<=> \(x^3-9x^2+27x-27\) \(-\left(x^3-3^3\right)+6\left(x^2+2x+1\right)+3x^2=-33\)

<=> \(x^3-9x^2+27x-27-x^3+27+6x^2+12x+6+3x^2=-33\)

<=> \(-6x^2+39x+6=-33\)

<=> \(6x^2-39x-6=33\)

<=> \(6x^2-39x-39=0\)

<=> \(6\left(x^2-\frac{39}{6}x-\frac{39}{6}\right)=0\)

<=> \(x^2-2.x.\frac{39}{12}+\frac{1521}{144}-\frac{273}{16}=0\)

<=> \(\left(x-\frac{39}{12}\right)^2-\frac{273}{16}=0\)

<=> \(\left(x-\frac{39}{12}-\frac{\sqrt{273}}{4}\right)\left(x-\frac{39}{12}+\frac{\sqrt{273}}{4}\right)=0\)

<=> \(\left(x-\frac{13+\sqrt{273}}{4}\right).\left(x-\frac{13-\sqrt{273}}{4}\right)=0\)

<=> \(x=\frac{13+\sqrt{273}}{4}\) ( h ) \(x=\frac{13-\sqrt{273}}{4}\)

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