已知xy+y²=3,x²+xy=二分之一,求2x²+3xy+y²的值

问题描述:

已知xy+y²=3,x²+xy=二分之一,求2x²+3xy+y²的值

解2x²+3xy+y²
=2x²+2xy+xy+y²
=2(x²+xy)+xy+y²
=2×1/2+3
=1+3
=4

等于4
求式=(2x²+2xy)+(xy+y²)=2X二分之一+3=4

2x²+3xy+y²
=2x²+2xy+xy+y²
=2(x²+xy)+xy+y²
=2*1/2+3
=1+3
=4