Question

найдите корень уравнения
(x-5)^2-x^2=3
^2 -это во второй степени

(2y+1)^2-4y^2=5

Answer 1

$(x-5)^2 - x^2 = 3 \ x^2 -10x+25 -x^2 -3 = 0 \ 10x-22 = 0 \ 10x = 22 \ x = 2.2$

$(2y+1)^2 -4y^2 = 5\ 4y^2 + 4y +1 -4y^2 -5 = 0 \ 4y - 4 = 0\ 4y = 4\ y = 1$