Question

(а+2)(а-5)-3а(1-2а)

(х+3)(х-2)-(х+4)(х-1)=3х​

Answer 1

ответ:

$\displaystyle 1)7a^2-6a-10\\\\2)-\frac{2}{5}=-0,4$

решение:

$\displaystyle 1)(a+2)(a-5)-3a(1-2a)=a^2-5a+2a-10-3a(1-2a)=\\a^2-5a+2a-10-3a+6a^2=7a^2-5a+2a-10-3a=7a^2-6a-10\\\\2)(x+3)(x-2)-(x+4)(x-1)=3x\\x^2-2x+3x-6-(x+4)(x-1)=3x\\x^2-2x+3x-6-(x^2-x+4x-4)=3x\\x^2-2x+3x-6-(x^2+3x-4)=3x\\x^2-2x-6-(x^2+3x-4)=0\\x^2-2x-6-x^2-3x+4=0\\-2x-6-3x+4=0\\-5x-6+4=0\\-5x-2=0\\-5x=2\\\\x=-\frac{2}5}=-0,4$