(1)已知a+b=6,ab=7,求ab2+a2b的值.
(2)若x+y=2,且(x+2)(y+2)=5,求x2+xy+y2的值.
解:(1)原式=ab(a+b),
又∵a+b=6,ab=7,
∴ab(a+b)=7×6=42;
(2)(x+2)(y+2)=5,x+y=2,
∴xy+2(x+y)+4=5,
∴xy=﹣3,
∴x2+xy+y2=(x+y)2﹣xy=4﹣(﹣3)=7.
(1)已知a+b=6,ab=7,求ab2+a2b的值.
(2)若x+y=2,且(x+2)(y+2)=5,求x2+xy+y2的值.
解:(1)原式=ab(a+b),
又∵a+b=6,ab=7,
∴ab(a+b)=7×6=42;
(2)(x+2)(y+2)=5,x+y=2,
∴xy+2(x+y)+4=5,
∴xy=﹣3,
∴x2+xy+y2=(x+y)2﹣xy=4﹣(﹣3)=7.