定义新运算:(a,b)⊗(c,d)=(ac,bd),(a,b)⊕(c,d)=(a+c,b+d),(a,b)*(c,d)=a2+c2-bd(1)求(1,2)*(3,-4)的值;(2)已知(1,2)⊗(p,q)=(2,-4),分别求出p与q的值;(3)在(2
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![定义新运算:(a,b)⊗(c,d)=(ac,bd),(a,b)⊕(c,d)=(a+c,b+d),(a,b)*(c,d)=a2+c2-bd(1)求(1,2)*(3,-4)的值;(2)已知(1,2)⊗(p,q)=(2,-4),分别求出p与q的值;(3)在(2](/uploads/image/z/2428891-43-1.jpg?t=%E5%AE%9A%E4%B9%89%E6%96%B0%E8%BF%90%E7%AE%97%EF%BC%9A%EF%BC%88a%2Cb%EF%BC%89%26%238855%3B%EF%BC%88c%2Cd%EF%BC%89%3D%EF%BC%88ac%2Cbd%EF%BC%89%2C%EF%BC%88a%2Cb%EF%BC%89%E2%8A%95%EF%BC%88c%2Cd%EF%BC%89%3D%EF%BC%88a%2Bc%2Cb%2Bd%EF%BC%89%2C%EF%BC%88a%2Cb%EF%BC%89%2A%EF%BC%88c%2Cd%EF%BC%89%3Da2%2Bc2-bd%EF%BC%881%EF%BC%89%E6%B1%82%EF%BC%881%2C2%EF%BC%89%2A%EF%BC%883%2C-4%EF%BC%89%E7%9A%84%E5%80%BC%EF%BC%9B%EF%BC%882%EF%BC%89%E5%B7%B2%E7%9F%A5%EF%BC%881%2C2%EF%BC%89%26%238855%3B%EF%BC%88p%2Cq%EF%BC%89%3D%EF%BC%882%2C-4%EF%BC%89%2C%E5%88%86%E5%88%AB%E6%B1%82%E5%87%BAp%E4%B8%8Eq%E7%9A%84%E5%80%BC%EF%BC%9B%EF%BC%883%EF%BC%89%E5%9C%A8%EF%BC%882)
定义新运算:(a,b)⊗(c,d)=(ac,bd),(a,b)⊕(c,d)=(a+c,b+d),(a,b)*(c,d)=a2+c2-bd(1)求(1,2)*(3,-4)的值;(2)已知(1,2)⊗(p,q)=(2,-4),分别求出p与q的值;(3)在(2
定义新运算:(a,b)⊗(c,d)=(ac,bd),(a,b)⊕(c,d)=(a+c,b+d),
(a,b)*(c,d)=a2+c2-bd
(1)求(1,2)*(3,-4)的值;
(2)已知(1,2)⊗(p,q)=(2,-4),分别求出p与q的值;
(3)在(2)的条件下,求(1,2)⊕(p,q)的结果;
(4)已知x2+2xy+y2=5,x2-2xy+y2=1,求(x,5)*(y,xy)的值.
定义新运算:(a,b)⊗(c,d)=(ac,bd),(a,b)⊕(c,d)=(a+c,b+d),(a,b)*(c,d)=a2+c2-bd(1)求(1,2)*(3,-4)的值;(2)已知(1,2)⊗(p,q)=(2,-4),分别求出p与q的值;(3)在(2
由题意可得:
(1)(1,2)*(3,-4)=1²+3²-2·(-4)=1+9+8=18;
.
(2)因为:(1,2)⊗(p,q)=(1·p,2·q)=(p,2q)=(2,-4)
所以解得:p=2,q=-2
.
(3)在(2)的条件下:
(1,2)♁(p,q)=(1,2)♁(2,-4)=(1+2,2-4)=(3,-2)
.
(4)已知x²+2xy+y²=5,x²-2xy+y²=1,那么:
4xy=(x²+2xy+y²)-(x²-2xy+y²)=4,2(x²+y²)=(x²+2xy+y²)+(x²-2xy+y²)=6
即得:xy=1,x²+y²=3
所以:(x,5)*(y,xy)=x²+y²-5xy=3-5=-2