Question

решите пожалуйста четвёртое задание!!!!

Answer 1

$\frac{2x^2-5x-3}{3x^2-8x-3}=\\\\2x^2-5x-3=2(x-3)(x+1/2)\\D=(-5)^2-4*2*(-3)=25+24=49=7^2\\x_1=(5+7)/4=3;\; \; x_2=(5-7)/4=-1/2 \\\\3x^3-8x-3=3(x-3)(x+1/3)\\D=(-8)^2-4*3*(-3)=64+36=100=10^2\\x_1=(8+10)/6=3;\; \; x_2=(8-10)/6=-1/3\\\\\frac{2x^2-5x-3}{3x^2-8x-3}= \frac{2(x-3)(x+1/2)}{3(x-3)(x+1/3)} = \frac{2(x+1/2)}{3(x+1/3)}= \frac{2x+1}{3x+1}$