Respuestas

2012-07-02T22:08:47+02:00

9-x²=x+3

x²+x-6=0

x²+3x-2x-6=0

x(x+3)-2(x+3)=0

(x-2)(x+3)=0

x=2 ∨ x=-3

 

 \\\int \limits_{-3}^2 (9-x^2-(x+3))\, dx=\\ \int \limits_{-3}^2 (-x^2-x+6)\, dx=\\ \Big[-\frac{x^3}{3}-\frac{x^2}{2}+6x\Big]_{-3}^2=\\ -\frac{2^3}{3}-\frac{2^2}{2}+6\cdot2-(-\frac{(-3)^3}{3}-\frac{(-3)^2}{2}+6\cdot(-3))=\\ -\frac{8}{3}-2+12-(\frac{27}{3}-\frac{9}{2}-18)=\\ -\frac{35}{3}+\frac{9}{2}+28=\\ -\frac{70}{6}+\frac{27}{6}+\frac{168}{6}=\\ \frac{125}{6}

2012-07-02T22:20:02+02:00

9-x²=x+3

x²+x-6=0

x²+3x-2x-6=0

x(x+3)-2(x+3)=0

(x-2)(x+3)=0

x=2 ∨ x=-3

  espero haberte ayudado mua♥