z=x^2+xy^2+xy+3x-5 la ecuación
El profesor se sorprenderá mucho al ver tu solución correcta😉
Solución
Solución detallada
Tenemos una ecuación lineal:
z = x^2+x*y^2+x*y+3*x-5
Obtenemos la respuesta: z = -5 + x^2 + 3*x + x*y + x*y^2
Suma y producto de raíces
[src]
2 2 / / 2\\ / 2\
-5 + re (x) - im (x) + 3*re(x) + I*\3*im(x) + 2*im(x)*re(x) + im(x*y) + im\x*y // + re(x*y) + re\x*y /
i(2re(x)im(x)+3im(x)+im(xy)+im(xy2))+(re(x))2+3re(x)+re(xy)+re(xy2)−(im(x))2−5
2 2 / / 2\\ / 2\
-5 + re (x) - im (x) + 3*re(x) + I*\3*im(x) + 2*im(x)*re(x) + im(x*y) + im\x*y // + re(x*y) + re\x*y /
i(2re(x)im(x)+3im(x)+im(xy)+im(xy2))+(re(x))2+3re(x)+re(xy)+re(xy2)−(im(x))2−5
2 2 / / 2\\ / 2\
-5 + re (x) - im (x) + 3*re(x) + I*\3*im(x) + 2*im(x)*re(x) + im(x*y) + im\x*y // + re(x*y) + re\x*y /
i(2re(x)im(x)+3im(x)+im(xy)+im(xy2))+(re(x))2+3re(x)+re(xy)+re(xy2)−(im(x))2−5
2 2 / / 2\\ / 2\
-5 + re (x) - im (x) + 3*re(x) + I*\3*im(x) + 2*im(x)*re(x) + im(x*y) + im\x*y // + re(x*y) + re\x*y /
i(2re(x)im(x)+3im(x)+im(xy)+im(xy2))+(re(x))2+3re(x)+re(xy)+re(xy2)−(im(x))2−5
-5 + re(x)^2 - im(x)^2 + 3*re(x) + i*(3*im(x) + 2*im(x)*re(x) + im(x*y) + im(x*y^2)) + re(x*y) + re(x*y^2)
2 2 / / 2\\ / 2\
z1 = -5 + re (x) - im (x) + 3*re(x) + I*\3*im(x) + 2*im(x)*re(x) + im(x*y) + im\x*y // + re(x*y) + re\x*y /
z1=i(2re(x)im(x)+3im(x)+im(xy)+im(xy2))+(re(x))2+3re(x)+re(xy)+re(xy2)−(im(x))2−5
z1 = i*(2*re(x)*im(x) + 3*im(x) + im(x*y) + im(x*y^2)) + re(x)^2 + 3*re(x) + re(x*y) + re(x*y^2) - im(x)^2 - 5