Question

Помогите пожалуйста с 17 18 19 номерами

Answer 1

Решение на фотографиях

Answer 2

17
1)(x³+y³)/(x+y):(x³-y³)=(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
18
1)1/a²+1/b²=(b²+a²)a²b²
2)1/(a+b)²*(b²+c²)/a²b²=(b²+c²)/(a+b)²a²b²
3)1/a+1/b=(b+a)/ab
4)2/(a+b)³+(b+a)/ab=2/(a+b)²ab
5)(b²+a²)/(a+b)²a²b²+2/(a+b)²ab=(b²+a²+2ab)/(a+b)²a²b²=(a+b)²/(a+b)²a²b²=1/a²b²
6)1/a²b²*a²b²=1
19
1)[(x+y)/x+(x-y)/y]²=[(xy+y²+x²-xy)/xy]²=(x²+y²)²/x²y²
2)[(x+y)/x-(x-y)/y]²=[(xy+y²-x²+xy)/xy]²=(y²-x²)²/x²y²
3)(x²+y²)²/x²y²-(y²-x²)²/x²y²=(x²+y²-y²+x²)(x²+y²+y²-x²)/x²y²=
=2x²*2y²/x²y²=4