Una lección y una práctica en matemáticas de primer grado

La primera pregunta (x+2)/1+(x+7)/1=(x+3)/1+(x+6)/1

(x +2) (x+7)/(x+7)+(x+7)(x+2)/(x+2)=(x+3)(x+6)/(x+6)+( x+6)(x+3)/(x+3)

(x+2)(x+7)/(2x+9)=(x+3)(x+6)/ (2x+9)

Cuando 2x+9=0, hay x=-2/9

Cuando 2x+9≠0, hay

(x+2) (x+7) = (x+3) (x+6)

x^2+9x+14=x^2+9x+18

No hay solución

Entonces x=-2/9

(x+2)/(x+3)+(x+3)/(x+4)=(x+ 4 )/(x+5)+(x+1)/(x+2)

(x+2)/(x+3)-(x+1)/(x+2) = (x+4)/(x+5)-(x+3)/(x+4)

(x+2)(x+1)/(x+1)(x+ 3) -(x+2)(x+1)/(x+2)(x+2)=(x+4)(x+3)/(x+5)(x+3)-(x+ 4)( x+3)/(x+4)(x+4)

(x+2)(x+1)/[(x^2+4x+3)-(x^ 2+4x +4)]=(x+4)(x+3)/[(x^2+8x+15)-(x^2+8x+16)]

(x+ 2)(x +1)/(-1)=(x+4)(x+3)/(-1)

(x+2)(x+1)=(x+4 )(x+ 3)

x^2+3x+2=x^2+7x+12

4x=-10

x=-2 /5