分析:(1)先根據(jù)去括號(hào)、合并同類項(xiàng)化簡(jiǎn),然后再把x、y的值代入求解;
(2)先根據(jù)非負(fù)數(shù)的性質(zhì)“兩個(gè)非負(fù)數(shù)相加,和為0,這兩個(gè)非負(fù)數(shù)的值都為0.”解出a、b的值,再把(a-b)2+4ab利用完全平方公式化簡(jiǎn)后代入數(shù)據(jù)計(jì)算即可.
解答:解:(1)x
2+(2xy-3y
2)-2(x
2+yx-2y
2),
=x
2+2xy-3y
2-2x
2-2yx+4y
2,
=-x
2+y
2,
當(dāng)x=-1,y=2時(shí),原式=-(-1)
2+2
2=-1+4=3.
(2)∵|a-
|+(b+
)
2=0,
∴a-
=0,b+
=0,
∴
a=,b=-,
(a-b)
2+4ab
=a
2-2ab+b
2+4ab
=a
2+2ab+b
2,
=(a+b)
2,
=(
-
)
2,
=
.
故答案為:
.
點(diǎn)評(píng):本題考查了非負(fù)數(shù)的性質(zhì),完全平方公式,整式的化簡(jiǎn),兩個(gè)非負(fù)數(shù)相加,和為0,這兩個(gè)非負(fù)數(shù)的值都為0.