Question

Решите уравнение
(9-4x)²=5(4x+1)

Answer 1

$displaystyle tt (9-4x)^2=5cdot(4x+1)\displaystyle tt 81-72x+16x^2=20x+5\displaystyle tt 81-72x+16x^2-20x-5=0\displaystyle tt 16x^2-92x+76=0::::::|div4\displaystyle tt 4x^2-23x+19=0\displaystyle tt 4x^2-4x-19x+19=0\displaystyle tt 4x(x-1)-19(x-1)=0\displaystyle tt (x-1)(4x-19)=0\\ displaystyle tt x-1=0\displaystyle tt bold{x_1=1}\\ displaystyle tt 4x-19=0\displaystyle tt 4x=19\displaystyle tt x=19div4\displaystyle tt bold{x_2=4,75}$