\(a,PT\Leftrightarrow x^3-6x^2+12x-8-x^3+x+6x^2-18x-10=0\)
\(\Leftrightarrow-5x-18=0\)
\(\Leftrightarrow x=-\dfrac{18}{5}\)
Vậy ...
\(b,PT\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6x^2+12x-6+10=0\)
\(\Leftrightarrow12x+6=0\)
\(\Leftrightarrow x=-\dfrac{1}{2}\)
Vậy ...
\(c,PT\Leftrightarrow\left(x+1\right)^3+3^3=0\)
\(\Leftrightarrow\left(x+1+3\right)\left(x^2+2x+1-3x-3+9\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(x^2-x+7\right)=0\)
Thấy : \(x^2-\dfrac{2.x.1}{2}+\dfrac{1}{4}+\dfrac{27}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{27}{4}\ge\dfrac{27}{4}>0\)
\(\Rightarrow x+4=0\)
\(\Leftrightarrow x=-4\)
Vậy ...
\(d,PT\Leftrightarrow\left(x-2\right)^3+1^3=0\)
\(\Leftrightarrow\left(x-2+1\right)\left(x^2-4x+4-x+2+1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2-5x+7\right)=0\)
Thấy : \(x^2-5x+7=x^2-\dfrac{5.x.2}{2}+\dfrac{25}{4}+\dfrac{3}{4}=\left(x-\dfrac{5}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\)
\(\Rightarrow x-1=0\)
\(\Leftrightarrow x=1\)
Vậy ...