Question

Giải các phương trình :

\(a,\left(4x+2\right)\left(x^2+1\right)=0\)

\(b,\left(3x+2\right)\left(x^2-1\right)=\left(9x^2-4\right)\left(x+1\right)\)

Answer 1

a. \(\left(4x+2\right)\left(x^2+1\right)=0\)

\(\Leftrightarrow4x+2=0\left(x^2+1\ne0\right)\)

\(\Leftrightarrow x=-\frac{1}{2}\)

b. \(\left(3x+2\right)\left(x^2-1\right)=\left(9x^2-4\right)\left(x+1\right)\)

\(\Leftrightarrow\left(3x+2\right)\left(x-1\right)\left(x+1\right)=\left(3x+2\right)\left(3x-2\right)\left(x+1\right)\)

\(\Leftrightarrow\left(3x+2\right)\left(x+1\right)\left(x-1-3x+2\right)=0\)

\(\Leftrightarrow-\left(3x+2\right)\left(x+1\right)\left(2x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{2}{3}\\x=-1\\x=\frac{1}{2}\end{matrix}\right.\)