Х+у=4 3х+у=6 розвязать систему уравнений

Question

Алгебра: Х+у=4 3х+у=6 розвязать систему уравнений

Nikitos228456 · Accepted Answer

Метод неопределенных коэффициентов (2x+1)/[(x-1)^2*(x^2+2x+3)] = A1/(x-1) + A2/(x-1)^2 + (A3*x+A4)/(x^2+2x+3) = = [A1*(x-1)(x^2+2x+3) + A2*(x^2+2x+3) + (A3*x+A4)(x-1)^2] / [(x-1)^2*(x^2+2x+3)] = = [A1(x^3+x^2+x-3)+A2(x^2+2x+3)+A3*x(x^2-2x+1)+A4(x^2-2x+1)] / / [(x-1)^2*(x^2+2x+3)] = = [x^3(A1+A3)+x^2(A1+A2-2A3+A4)+x(A1+2A2+A3-2A4)+(-3A1+3A2+A4)] / / [(x-1)^2*(x^2+2x+3)] = (2x+1)/[(x-1)^2*(x^2+2x+3)] Система { A1 + A3 = 0 { A1 + A2 - 2A3 + A4 = 0 { A1 + 2A2 + A3 - 2A4 = 2 { -3A1 + 3A2 + A4 = 1 { A3...

MOGZ ответил