Question

Answer 1

a: \(\left(4x+3\right)^2-16x\left(x+6\right)+7\left(10x+1\right)=0\)

=>\(16x^2+24x+9-16x^2-96x+70x+7=0\)

=>-2x+16=0

=>-2x=-16

=>x=8

c: \(\left(9-2x\right)^2-4\left(x+3\right)^2=81-24x\)

=>\(4x^2-36x+81-4\left(x^2+6x+9\right)-81+24x=0\)

=>\(4x^2-12x-4x^2-24x-36=0\)

=>-36x-36=0

=>36x=-36

=>x=-1

d: \(\left(x^2-1\right)^2-6\left(x-1\right)\left(x+1\right)=-9\)

=>\(\left(x^2-1\right)^2-2\cdot\left(x^2-1\right)\cdot3+3^2=0\)

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

=>\(x^2-4=0\)

=>\(x^2=4\)

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