Verified answer

Ответ:

[tex]1)(2x-3)(2x+3)-x(4x-18)=0\\ 4x^{2} -9-4x^{2} +18x=0\\-9+18x=0\\18x=9\\x=9:18\\x=\frac{1}{2} \\2) x(5-x)+(x-9)(x+9)=(x+3)(x+1)-x^{2} \\5x-x^{2} +x^{2} -81=x^{2} +x+3x+3-x^{2} \\5x-81=4x+3\\5x-4x=3+81\\x=84\\3) (2x-3)(2x+3)-(4x^{2} +12x)=3x+18\\4x^{2} -9-4x^{2} -12x=3x+18\\-9-12x=3x+18\\-12x-3x=18+9\\-15x=27\\x=-\frac{27}{15} =-\frac{9}{5} =-1\frac{4}{5}[/tex]

Пошаговое объяснение: