Question

даю 20б y+6 - y+2 + 5
__________
4y+8 4y-8 y^2-4 ​

Answer 1

$\dfrac{y+6}{4y+8}-\dfrac{y+2}{4y-8}+\dfrac5{y^2-4}=\dfrac{y+6}{4(y+2)}-\dfrac{y+2}{4(y-2)}+\dfrac5{(y-2)(y+2)}=$

$=\dfrac{(y+6)(y-2)-(y+2)(y+2)+5\cdot4}{4(y-2)(y+2)}=\dfrac{y^2-2y+6y-12-(y^2+4y+4)+20}{4(y-2)(y+2)}=$$=\dfrac{y^2+4y+8-y^2-4y-4}{4(y-2)(y+2)}=\dfrac{4}{4(y-2)(y+2)}=\dfrac1{y^2-4}$