Đáp án:
Giải thích các bước giải:
a) $27x^2(x+3)-12(x^2+3x)=0\\
\Leftrightarrow 27x^2(x+3)-12x(x+3)=0\\
\Leftrightarrow x(x+3)[27x-12]=0\\
\Leftrightarrow x=0, x+3=0, 27x-12=0\\
\Leftrightarrow x=0, x=-3, x=\frac{12}{27}=\frac{4}{9}\\\\
b) (x^2+x+1)(6-2x)=0\\
\Leftrightarrow x^2+x+1=0(VN)),6-2x=0\\
\Leftrightarrow 2x=6\Leftrightarrow x=3
c) (8x-4)(x^2+2x+2)=0\\
\Leftrightarrow 8x-4=0,x^2+2x+2=0(VN)\\
\Leftrightarrow 8x=4 \Leftrightarrow x=\frac{1}{2}\\
d)x-\frac{4}{5}+3x-\frac{2}{10}-x=2x-\frac{5}{3} -7x+\frac{2}{6}\\
\Leftrightarrow x+3x-x-2x+7x=-\frac{5}{3}+\frac{2}{6}+\frac{2}{10}+\frac{4}{5}\\
\Leftrightarrow 8x=\frac{-1}{3}\\
\Leftrightarrow x=\frac{-1}{24}$
