Question

tìm x

(9x² - 4)(x + 1) = (3x +2)(x²-1)

Answer 1

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

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}3x+2=0\\x+1=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-2}{3}\\x=-1\\x=\dfrac{1}{2}\end{matrix}\right.\)

Vậy phương trình có tập nghiệm S = \(\left\{\dfrac{-2}{3};-1;\dfrac{1}{2}\right\}\)

Answer 2

\(\Leftrightarrow\left(3x+2\right)\left(3x-2\right)\left(x+1\right)-\left(3x+2\right)\left(x-1\right)\left(x+1\right)=0\\ \Leftrightarrow\left(3x+2\right)\left(x+1\right)\left(3x-2-\left(x-1\right)\right)\\ \Leftrightarrow\left(3x+2\right)\left(x+1\right)\left(2x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}\\x=-1\\x=\dfrac{1}{2}\end{matrix}\right.\)