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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=
,
(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=
,
(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)=___.
,
(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 a+3b 2 2
2222
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 |
2222
▼优质解答
答案和解析
(1)∵2⊕5=
=
,
∴4(2⊕5)=4×
=34.
故答案为34;
(2)4⊕x=
,
解方程
=5,得x=2,
故答案为x=2;
(3)∵A=x2+2xy+y2,B=x2-2xy+y2,
∴(A⊕B)=
=2x2-2xy+2y2,
(B⊕A)=
=2x2+2xy+2y2,
∴(A⊕B)+(B⊕A)=4x2+4y2.
故答案为4x2+4y2.
2+3×5 2 2+3×5 2+3×5 2+3×52 2 2=
,
∴4(2⊕5)=4×
=34.
故答案为34;
(2)4⊕x=
,
解方程
=5,得x=2,
故答案为x=2;
(3)∵A=x2+2xy+y2,B=x2-2xy+y2,
∴(A⊕B)=
=2x2-2xy+2y2,
(B⊕A)=
=2x2+2xy+2y2,
∴(A⊕B)+(B⊕A)=4x2+4y2.
故答案为4x2+4y2.
17 2 17 17 172 2 2,
∴4(2⊕5)=4×
=34.
故答案为34;
(2)4⊕x=
,
解方程
=5,得x=2,
故答案为x=2;
(3)∵A=x2+2xy+y2,B=x2-2xy+y2,
∴(A⊕B)=
=2x2-2xy+2y2,
(B⊕A)=
=2x2+2xy+2y2,
∴(A⊕B)+(B⊕A)=4x2+4y2.
故答案为4x2+4y2.
17 2 17 17 172 2 2=34.
故答案为34;
(2)4⊕x=
,
解方程
=5,得x=2,
故答案为x=2;
(3)∵A=x2+2xy+y2,B=x2-2xy+y2,
∴(A⊕B)=
=2x2-2xy+2y2,
(B⊕A)=
=2x2+2xy+2y2,
∴(A⊕B)+(B⊕A)=4x2+4y2.
故答案为4x2+4y2.
4+3x 2 4+3x 4+3x 4+3x2 2 2,
解方程
=5,得x=2,
故答案为x=2;
(3)∵A=x2+2xy+y2,B=x2-2xy+y2,
∴(A⊕B)=
=2x2-2xy+2y2,
(B⊕A)=
=2x2+2xy+2y2,
∴(A⊕B)+(B⊕A)=4x2+4y2.
故答案为4x2+4y2.
4+3x 2 4+3x 4+3x 4+3x2 2 2=5,得x=2,
故答案为x=2;
(3)∵A=x22+2xy+y22,B=x22-2xy+y22,
∴(A⊕B)=
=2x2-2xy+2y2,
(B⊕A)=
=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) 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)2 2 2=2x22-2xy+2y22,
(B⊕A)=
=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) 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)2 2 2=2x22+2xy+2y22,
∴(A⊕B)+(B⊕A)=4x22+4y22.
故答案为4x22+4y22.
2+3×5 |
2 |
17 |
2 |
∴4(2⊕5)=4×
17 |
2 |
故答案为34;
(2)4⊕x=
4+3x |
2 |
解方程
4+3x |
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 |
(B⊕A)=
x2-2xy+y2+3(x2+2xy+y2) |
2 |
∴(A⊕B)+(B⊕A)=4x2+4y2.
故答案为4x2+4y2.
2+3×5 |
2 |
17 |
2 |
∴4(2⊕5)=4×
17 |
2 |
故答案为34;
(2)4⊕x=
4+3x |
2 |
解方程
4+3x |
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 |
(B⊕A)=
x2-2xy+y2+3(x2+2xy+y2) |
2 |
∴(A⊕B)+(B⊕A)=4x2+4y2.
故答案为4x2+4y2.
17 |
2 |
∴4(2⊕5)=4×
17 |
2 |
故答案为34;
(2)4⊕x=
4+3x |
2 |
解方程
4+3x |
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 |
(B⊕A)=
x2-2xy+y2+3(x2+2xy+y2) |
2 |
∴(A⊕B)+(B⊕A)=4x2+4y2.
故答案为4x2+4y2.
17 |
2 |
故答案为34;
(2)4⊕x=
4+3x |
2 |
解方程
4+3x |
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 |
(B⊕A)=
x2-2xy+y2+3(x2+2xy+y2) |
2 |
∴(A⊕B)+(B⊕A)=4x2+4y2.
故答案为4x2+4y2.
4+3x |
2 |
解方程
4+3x |
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 |
(B⊕A)=
x2-2xy+y2+3(x2+2xy+y2) |
2 |
∴(A⊕B)+(B⊕A)=4x2+4y2.
故答案为4x2+4y2.
4+3x |
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 |
(B⊕A)=
x2-2xy+y2+3(x2+2xy+y2) |
2 |
∴(A⊕B)+(B⊕A)=4x2+4y2.
故答案为4x2+4y2.
x2+2xy+y2+3(x2-2xy+y2) |
2 |
(B⊕A)=
x2-2xy+y2+3(x2+2xy+y2) |
2 |
∴(A⊕B)+(B⊕A)=4x2+4y2.
故答案为4x2+4y2.
x2-2xy+y2+3(x2+2xy+y2) |
2 |
∴(A⊕B)+(B⊕A)=4x22+4y22.
故答案为4x22+4y22.
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