早教吧作业答案频道 -->数学-->
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
题目详情
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.
 看了 填空题:(请将结果直接写在横...的网友还看了以下:
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 2020-04-09 …
(2008•温州模拟)如图,在平面直角坐标系xOy中,A(1,0),B(1,1),C(0,1),映 2020-06-14 …
确定常数λ,使在右半平面x>0上的向量A(x,y)=2xy(x4+y2)λi-x2(x4+y2)λ 2020-06-27 …
已知对任意的x∈(-∞,0)∪(0,+∞),y∈[-1,1],不等式x2+16x2-2xy-8x1 2020-07-21 …
已知一切x,y∈R,不等式x2+81x2-2xy+18x2-y2-a≥0恒成立,则实数a的取值范围 2020-07-29 …
已知对满足x+y+4=2xy的任意正实数x,y,都有x2+2xy+y2-ax-ay+1≥0,则实数 2020-07-30 …
已知两个二次函数Y1和Y2.当x=a(a>0)时,Y1取得最大值5,且Y2=25,又Y2的最小值为- 2020-10-31 …
二次函数题目已知a小于-1,点(a-1,y1),(a,y2),(a+1,y3)都在函数y=2x^2的 2020-10-31 …
如何求解带参数的四元三次方程组symsy1y2y3an(n是参数)9*(n-4)*y2*y3+9*y 2020-10-31 …
(2013•梅州一模)若不等式x2+2xy≤a(x2+y2)对于一切正数x,y恒成立,则实数a的最小 2020-11-12 …