Sửa đề: \(\left(x-3\right)^3-\left(x-3\right)\left(x^2-3x+9\right)+9\left(x+1\right)^2=15\)
Ta có: \(\left(x-3\right)^3-\left(x-3\right)\left(x^2-3x+9\right)+9\left(x+1\right)^2=15\)
\(\Leftrightarrow x^3-3x^2+3x-1-\left(x^3-27\right)+9\left(x^2+2x+1\right)-15=0\)
\(\Leftrightarrow x^3-3x^2+3x-1-x^3+27+9x^2+18x+9-15=0\)
\(\Leftrightarrow6x^2+21x+20=0\)
\(\Leftrightarrow6\left(x^2+\frac{7}{2}x+\frac{10}{3}\right)=0\)
\(\Leftrightarrow x^2+2\cdot x\cdot\frac{7}{4}+\frac{49}{16}+\frac{13}{48}=0\)
\(\Leftrightarrow\left(x+\frac{7}{4}\right)^2+\frac{13}{48}=0\)(Vô lý)
Vậy: x∈∅
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