Question

ПОМОГИТЕ ПОЖАЛУЙСТА. С РЕШЕНИЕМ!!!​

Answer 1

Ответ:

$\boxed{\ \ (a\pm b)^2=a^2\pm 2ab+b^2\ \ }\\\\1)\ \ 4x^2-(2x-1)^2=15\\\\4x^2-(4x^2-4x+1)=15\\\\4x^2-4x^2+4x-1=15\\\\4x=16\\\\x=4\\\\\\2)\ \ 9x^2-1=(3x-2)^2\\\\9x^2-1=9x^2-12x+4\\\\12x=5\\\\x=\dfrac{5}{12}$

$3)\ \ (3x+1)^2-(3x-1)^2=11x+1,2\\\\9x^2+6x+1-(9x^2-6x+1)=11x+1,2\\\\9x^2+6x+1-9x^2+6x-1=11x+1,2\\\\12x=11x+1,2\\\\x=1,2\\\\\\4)\ \ (5+2y)(y-3)-2(y-1)^2=0\\\\5y-15+2y^2-6y-2(y^2-2y+1)=0\\\\-y-15+2y^2-2y^2+4y-2=0\\\\3y=17\\\\y=\dfrac{17}{3}\ \ ,\ \ \ y=5\dfrac{2}{3}$