x²y+xy²-3+x+y+-3xy=(x²y+xy²-3xy)+(x+y-3)=
=xy*(x+y-3)+(x+y-3)=
=(x+y-3)*(xy+1)
См. вложение
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<span>|tgx|-x^2tgx=0
1. tgx</span>≥0 tgx(1-x²)=0 tgx=0 x=2πn (с учетом tgx≥0) n∈Z х=+-1
2. tgx<0 -tgx-x²tgx=0 tgx+x²tgx=0 tgx(1+x²)=0
x²≠-1 tgx=0 x=π+2πn (с учетом tgx<0)