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)

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