Question

помогите пожалуйста с 13 14 17 номерами

Answer 1

13
1)p²/q-q²/p=(p³-q³)/pq
2)1/(p+q)*(p³-q³)/hq=(p³-q³)/pq(p+q)
3)(p³-q³)/pq-(p³-q³)/(p+q)(pq=(p³+p²q-pq²-q³-p³+q³)/(p+q)pq=
=pq(p-q)/(p+q)pq=(p-q)/(p+q)
4)(p-q)/(p+q):(p-q)/p=(p-q)/(p+q)*p/(p-q)=p/(p+q)
14
1)1/b²-1/c²=(c²-b²)/b²c²
2)(b²+c²)/b²c²*(c²-b²)/b²c²=(c^4-b^4)/b^4c^4
3)1/a²-1/c²=(c²-a²)/a²c²
4)(c²-a²)/a²c²*(a²+c²)/a²c²=(c^4-a^4)/a^4c^4
5)(c^4-b^4)/b^4c^4-(c^4-a^4)/a^4c^4=(a^4c^4-a^4b^4-b^4c^4+a^4b^4)/a^4b^4c^4=c^4(a^4-b^4)/a^4b^4c^4(a^4-b^4)/a^4b^4
6)(a^4-b^4)/a^4b^4 :(a²+b²)/a²b²=(a²-b²)(a²+b²)/a^4b^4c^4 *a²b^2/(a²+b²)=
=(a²-b²)/a²b²
17
1)(x³+y³)/(x+y):(x²-y²)=(x+y)(x³-xy+y³)/(x+y) *1/(x³-y³)=(x³-xy+y³)/(x²-y²)
2)(x²-xy+y²)/(x-y)(x+y)+2y/(x+y)=(x²-xy+y²+2xy-2y²)/(x+y)(x-y)=
=(x²+xy-y²)/(x-y)(x+y)
3)(x²+xy-y²)/(x²-y²)-xy/(x²-y²)=(x²+xy-y²-xy)/(x²-y²)=(x²-y²)/(x²-y²)=1

Answer 2

Решение на фотографиях