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填空题:(请将结果直接写在横线上)定义新运算“⊕”,对于任意有理数a,b有a⊕b=a+3b2,(1)4(2⊕5)=.(2)方程4⊕x=5的解是.(3)若A=x2+2xy+y2,B=x2-2xy+y2,则

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填空题:(请将结果直接写在横线上)
定义新运算“⊕”,对于任意有理数a,b有a⊕b=
a+3b
2

(1)4(2⊕5)=___.
(2)方程4⊕x=5的解是___.
(3)若A=x2+2xy+y2,B=x2-2xy+y2,则(A⊕B)+(B⊕A)=___.
填空题:(请将结果直接写在横线上)
定义新运算“⊕”,对于任意有理数a,b有a⊕b=
a+3b
2

(1)4(2⊕5)=___.
(2)方程4⊕x=5的解是___.
(3)若A=x2+2xy+y2,B=x2-2xy+y2,则(A⊕B)+(B⊕A)=___.

a+3b
2

(1)4(2⊕5)=___.
(2)方程4⊕x=5的解是___.
(3)若A=x2+2xy+y2,B=x2-2xy+y2,则(A⊕B)+(B⊕A)=___.
a+3b
2
a+3b 2 a+3b a+3b 2 2


2222
▼优质解答
答案和解析
(1)∵2⊕5=
2+3×5
2
=
17
2

∴4(2⊕5)=4×
17
2
=34.
故答案为34;

(2)4⊕x=
4+3x
2

解方程
4+3x
2
=5,得x=2,
故答案为x=2;

(3)∵A=x2+2xy+y2,B=x2-2xy+y2
∴(A⊕B)=
x2+2xy+y2+3(x2-2xy+y2)
2
=2x2-2xy+2y2

(B⊕A)=
x2-2xy+y2+3(x2+2xy+y2)
2
=2x2+2xy+2y2
∴(A⊕B)+(B⊕A)=4x2+4y2
故答案为4x2+4y2
2+3×5
2
2+3×522+3×52+3×52+3×5222=
17
2

∴4(2⊕5)=4×
17
2
=34.
故答案为34;

(2)4⊕x=
4+3x
2

解方程
4+3x
2
=5,得x=2,
故答案为x=2;

(3)∵A=x2+2xy+y2,B=x2-2xy+y2
∴(A⊕B)=
x2+2xy+y2+3(x2-2xy+y2)
2
=2x2-2xy+2y2

(B⊕A)=
x2-2xy+y2+3(x2+2xy+y2)
2
=2x2+2xy+2y2
∴(A⊕B)+(B⊕A)=4x2+4y2
故答案为4x2+4y2
17
2
172171717222,
∴4(2⊕5)=4×
17
2
=34.
故答案为34;

(2)4⊕x=
4+3x
2

解方程
4+3x
2
=5,得x=2,
故答案为x=2;

(3)∵A=x2+2xy+y2,B=x2-2xy+y2
∴(A⊕B)=
x2+2xy+y2+3(x2-2xy+y2)
2
=2x2-2xy+2y2

(B⊕A)=
x2-2xy+y2+3(x2+2xy+y2)
2
=2x2+2xy+2y2
∴(A⊕B)+(B⊕A)=4x2+4y2
故答案为4x2+4y2
17
2
172171717222=34.
故答案为34;

(2)4⊕x=
4+3x
2

解方程
4+3x
2
=5,得x=2,
故答案为x=2;

(3)∵A=x2+2xy+y2,B=x2-2xy+y2
∴(A⊕B)=
x2+2xy+y2+3(x2-2xy+y2)
2
=2x2-2xy+2y2

(B⊕A)=
x2-2xy+y2+3(x2+2xy+y2)
2
=2x2+2xy+2y2
∴(A⊕B)+(B⊕A)=4x2+4y2
故答案为4x2+4y2
4+3x
2
4+3x24+3x4+3x4+3x222,
解方程
4+3x
2
=5,得x=2,
故答案为x=2;

(3)∵A=x2+2xy+y2,B=x2-2xy+y2
∴(A⊕B)=
x2+2xy+y2+3(x2-2xy+y2)
2
=2x2-2xy+2y2

(B⊕A)=
x2-2xy+y2+3(x2+2xy+y2)
2
=2x2+2xy+2y2
∴(A⊕B)+(B⊕A)=4x2+4y2
故答案为4x2+4y2
4+3x
2
4+3x24+3x4+3x4+3x222=5,得x=2,
故答案为x=2;

(3)∵A=x22+2xy+y22,B=x22-2xy+y22,
∴(A⊕B)=
x2+2xy+y2+3(x2-2xy+y2)
2
=2x2-2xy+2y2

(B⊕A)=
x2-2xy+y2+3(x2+2xy+y2)
2
=2x2+2xy+2y2
∴(A⊕B)+(B⊕A)=4x2+4y2
故答案为4x2+4y2
x2+2xy+y2+3(x2-2xy+y2)
2
x2+2xy+y2+3(x2-2xy+y2)2x2+2xy+y2+3(x2-2xy+y2)x2+2xy+y2+3(x2-2xy+y2)x2+2xy+y2+3(x2-2xy+y2)2+2xy+y2+3(x2-2xy+y2)2+3(x2-2xy+y2)2-2xy+y2)2)222=2x22-2xy+2y22,

(B⊕A)=
x2-2xy+y2+3(x2+2xy+y2)
2
=2x2+2xy+2y2
∴(A⊕B)+(B⊕A)=4x2+4y2
故答案为4x2+4y2
x2-2xy+y2+3(x2+2xy+y2)
2
x2-2xy+y2+3(x2+2xy+y2)2x2-2xy+y2+3(x2+2xy+y2)x2-2xy+y2+3(x2+2xy+y2)x2-2xy+y2+3(x2+2xy+y2)2-2xy+y2+3(x2+2xy+y2)2+3(x2+2xy+y2)2+2xy+y2)2)222=2x22+2xy+2y22,
∴(A⊕B)+(B⊕A)=4x22+4y22.
故答案为4x22+4y22.