Question

Answer 1

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

\(\left(x+4\right)\left(3x+2\right)=0\)

\(\left[{}\begin{matrix}x+4=0\\3x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-4\\x=\dfrac{-2}{3}\end{matrix}\right.\)

 

Answer 2

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

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

\(\Leftrightarrow\left(x+4\right)\left(3x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=-\dfrac{2}{3}\end{matrix}\right.\)

b) \(x^2\left(3x-2\right)-8+12x=0\)

\(\Leftrightarrow x^2\left(3x-2\right)+4\left(3x-2\right)=0\)

\(\Leftrightarrow\left(3x-2\right)\left(x^2+4\right)=0\)

\(\Leftrightarrow x=\dfrac{2}{3}\)( do \(x^2+4\ge4>0\))

Answer 3

\(a,\Leftrightarrow\left(2x+3-x+1\right)\left(2x+3+x-1\right)=0\\ \Leftrightarrow\left(x+4\right)\left(3x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-4\\x=-\dfrac{2}{3}\end{matrix}\right.\\ b,\Leftrightarrow x^2\left(3x-2\right)-4\left(3x-2\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x+2\right)\left(3x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\\x=\dfrac{2}{3}\end{matrix}\right.\)

Answer 4

-x^2(2-3x)-4(2-3x)=0

(-x^2-4)(2-3x)=0

=>x=2/3

Answer 5

b: Ta có: \(x^2\left(3x-2\right)+12x-8=0\)

\(\Leftrightarrow3x-2=0\)

hay \(x=\dfrac{2}{3}\)