Найдите наименьший корень уравнения x/(x+3)-4/(x+1)=2 - вопрос №34125145745 от zlatochkaluninа 14.04.2023 15:34

Question

Алгебра: Найдите наименьший корень уравнения x/(x+3)-4/(x+1)=2

zlatochkaluninа · Answer

x/(x+3) - 4/(x+1) = 2x(x+1)/(x+3)(x+1) - 4(x+3) /(x+1)(x+3) = 2(x(x+1)-4(x+3)) /(x+1)(x+3) = 2(x²+x-4x-12)/(x+1)(x+3) = 2(x²-3x-12)/(x²+3x+x+3) = 2(x²-3x-12)/(x²+4x+3) = 2(x²-3x-12)/(x²+4x+3) - 2 = 0(x²-3x-12)/(x²+4x+3) - 2*(x²+4x+3)/(x²+4x+3) = 0(x²-3x-12)/(x²+4x+3) - (2x²+8x+6)/(x²+4x+3) = 0(x²-3x-12)-(2x²+8x+6) /(x²+4x+3) = 0(x²-3x-12-2x²-8x-6)/(x²+4x+3) = 0(-x²-11x-18)/(x²+4x+3) = 0 |*(x²+4x+3) ОДЗ: (x²+4x+3)≠0(-x²-11x-18)*(x²+4x+3) = 0-x²-11x-18=0 |*(-1)x²+11x+18=0D=121-72= 49x1,2 = (-11±7)/2x1=...