Tìm x, biết:
a/ [tex]x^{2}[/tex]+x = 6
b/ 6[tex]x^{3}[/tex] +[tex]x^{2}[/tex] = 2x
a) $x^2+x = 6\\\Leftrightarrow x^2+x-6=0\\\Leftrightarrow x^2+3x-2x-6=0\\\Leftrightarrow x(x+3)-2(x+3)=0\\\Leftrightarrow (x-2)(x+3)=0 \\\Leftrightarrow x=2 hoac x=-3$
Vậy........
b)$6x^3 +x^2 = 2x\\\Leftrightarrow 6x^3+x^2-2x=0\\\Leftrightarrow 6x^3-3x^2+4x^2-2x=0\\\Leftrightarrow 3x^2(2x-1)+2x(2x-1)=0\\\Leftrightarrow (3x^2+2x)(2x-1)=0\\\Leftrightarrow x(3x+2)(2x-1)=0\\\Leftrightarrow x=0 hoac x=\frac{-2}{3} hoac x=\frac{1}{2}$
Vậy.......