Question

已知x²+xy=3,xy+y²=-2,求下列各式的值（1）x²－y²；（2）x²+4xy+3y²；

已知x²+xy=3,xy+y²=-2,求下列各式的值（1）x²－y²；（2）x²+4xy+3y²；（3）2x²－xy－3y²需要过程。

Answer 1

（1）x²－y²
=(x²+xy)-(xy+y²)
=3-(-2)
=5
（2）x²+4xy+3y²
=(x²+xy)+3(xy+y²)
=3+3×(-2)
=3-6
=-3

（3）2x²－xy－3y²
=2(x²+xy)-3(xy+y²)
=2×3-3×(-3)
=6+9
=15

追答

（1）x²－y²
=(x²+xy)-(xy+y²)
=3-(-2)
=5
（2）x²+4xy+3y²
=(x²+xy)+3(xy+y²)
=3+3×(-2)
=3-6
=-3

（3）2x²－xy－3y²
=2(x²+xy)-3(xy+y²)
=2×3-3×(-2)
=6+6
=12
(前面看错了一个数字，呵呵)

Answer 2

x²－y²=x² + xy - ( xy + y² ) = 3 + 2 = 5x²+4xy+3y² = x²+xy+3(xy+y²)=3-6=-3

Answer 3

x+y+x-y。

Answer 4

Answer 5

追答

好评哦亲