Question

da dayy ạ

Answer 1

a: \(\left(x-3\right)\left(x+3\right)-x\left(x+6\right)=0\)

=>\(x^2-9-x^2-6x=0\)

=>-6x-9=0

=>2x+3=0

=>2x=-3

=>\(x=-\dfrac{3}{2}\)

b: \(x\left(x-5\right)+2x-10=0\)

=>\(x\left(x-5\right)+2\left(x-5\right)=0\)

=>(x-5)(x+2)=0

=>\(\left[{}\begin{matrix}x-5=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)

c: \(x^3-4x=\left(2x+3\right)\left(2-x\right)\)

=>\(x\left(x-2\right)\left(x+2\right)=\left(-2x-3\right)\left(x-2\right)\)

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

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

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

=>\(\left[{}\begin{matrix}x-2=0\\x^2+4x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x^2+4x+4=7\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=2\\\left(x+2\right)^2=7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x+2=\sqrt{7}\\x+2=-\sqrt{7}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\sqrt{7}-2\\x=-\sqrt{7}-2\end{matrix}\right.\)

Answer 2

a) \(\left(x-3\right)\left(x+3\right)-x\left(x+6\right)=0\)

\(\Leftrightarrow x^2-9-x^2-6x=0\)

\(\Leftrightarrow-6x-9=0\)

\(\Leftrightarrow-6x=9\Leftrightarrow x=-\dfrac{3}{2}\)

b) \(x\left(x-5\right)+2x-10=0\)

\(\Leftrightarrow x\left(x-5\right)+2\left(x-5\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)

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

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

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

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

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

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