Question

Помогите пожалуйста решить

Answer 1

а)

$\displaystyle \tt (3x+5)(4x-1)=(6x-3)(2x+7)\\\displaystyle \tt 12x^2-3x+20x-5=12x^2+42x-6x-21\\\displaystyle \tt -3x+20x-5=42x-6x-21\\\displaystyle \tt 17x-5=36x-21\\\displaystyle \tt 17x-36x=-21+5\\\displaystyle \tt -19x=-16\\\\\displaystyle \tt \bold{x=\frac{16}{19}}$

б)

$\displaystyle \tt (5x-1)(2-x)=(x-3)(2-5x)\\\displaystyle \tt 10x-5x^2-2+x=2x-5x^2-6+15x\\\displaystyle \tt 10x-2+x=2x-6+15x\\\displaystyle \tt 11x-2=17x-6\\\displaystyle \tt 11x-17x=-6+2\\\displaystyle \tt -6x=-4\\\\ \displaystyle \tt \bold{x=\frac{2}{3}}$