Question

Решите уравнение
(9-4x)²=5(4x+1)

Answer 1

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