已知集合A={x|x=a+b根号2,a,b∈Q},若X1∈A,X2∈A(1)试问X1X2,X1/X2是否属于A(2)若B=={x|x=a+b根号2,a,b∈Z},试问X1X2,X1/X2是否属于B,为什么?
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已知集合A={x|x=a+b根号2,a,b∈Q},若X1∈A,X2∈A(1)试问X1X2,X1/X2是否属于A(2)若B=={x|x=a+b根号2,a,b∈Z},试问X1X2,X1/X2是否属于B,为什么?
已知集合A={x|x=a+b根号2,a,b∈Q},若X1∈A,X2∈A
(1)试问X1X2,X1/X2是否属于A
(2)若B=={x|x=a+b根号2,a,b∈Z},试问X1X2,X1/X2是否属于B,为什么?
已知集合A={x|x=a+b根号2,a,b∈Q},若X1∈A,X2∈A(1)试问X1X2,X1/X2是否属于A(2)若B=={x|x=a+b根号2,a,b∈Z},试问X1X2,X1/X2是否属于B,为什么?
X1=a1+b1根号2
X2=a2+b2根号2
X1×X2=(a1+b1根号2)(a2+b2根号2)=a1a2+2b1b2+(a1b2+a2b1)根号2
因为a1a2+2b1b2,a1b2+a2b1都属于Q,所以X1X2属于A
X1/X2=(a1+b1根号2)/(a2+b2根号2)=(a1a2-2b1b2)/(a2^2-2b2^2)+(a2b1-a1b2)/(a2^2-2b2^2)根号2
因为(a1a2-2b1b2)/(a2^2-2b2^2),(a2b1-a1b2)/(a2^2-2b2^2)不属于Q
所以X1/X2不属于A
第二题也一样,最后求得X1X2属于B , X1/X2不属于B
X1=a1+b1根号2
X2=a2+b2根号2
X1×X2=(a1+b1根号2)(a2+b2根号2)=a1a2+2b1b2+(a1b2+a2b1)根号2
因为a1a2+2b1b2,a1b2+a2b1都属于Q,所以X1X2属于A
X1/X2=(a1+b1根号2)/(a2+b2根号2)=(a1a2-2b1b2)/(a2^2-2b2^2)+(a2b1-a1b2)/(a2^...
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X1=a1+b1根号2
X2=a2+b2根号2
X1×X2=(a1+b1根号2)(a2+b2根号2)=a1a2+2b1b2+(a1b2+a2b1)根号2
因为a1a2+2b1b2,a1b2+a2b1都属于Q,所以X1X2属于A
X1/X2=(a1+b1根号2)/(a2+b2根号2)=(a1a2-2b1b2)/(a2^2-2b2^2)+(a2b1-a1b2)/(a2^2-2b2^2)根号2
因为(a1a2-2b1b2)/(a2^2-2b2^2),(a2b1-a1b2)/(a2^2-2b2^2)不属于Q
所以X1/X2不属于A
第二题也一样,最后求得X1X2属于B , X1/X2不属于B
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(1)X1=a1+b1根号2
X2=a2+b2根号2
X1×X2=(a1+b1根号2)(a2+b2根号2)=a1a2+2b1b2+(a1b2+a2b1)根号2
因为a1a2+2b1b2,a1b2+a2b1都属于Q,所以X1X2属于A
X1/X2=(a1+b1根号2)/(a2+b2根号2)=(a1a2-2b1b2)/(a2^2-2b2^2)+(a2b1-a1b2)/(...
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(1)X1=a1+b1根号2
X2=a2+b2根号2
X1×X2=(a1+b1根号2)(a2+b2根号2)=a1a2+2b1b2+(a1b2+a2b1)根号2
因为a1a2+2b1b2,a1b2+a2b1都属于Q,所以X1X2属于A
X1/X2=(a1+b1根号2)/(a2+b2根号2)=(a1a2-2b1b2)/(a2^2-2b2^2)+(a2b1-a1b2)/(a2^2-2b2^2)根号2
因为(a1a2-2b1b2)/(a2^2-2b2^2),(a2b1-a1b2)/(a2^2-2b2^2)不属于Q
所以X1/X2不属于A
(2)方法和第一问一样,在这里我就不详细说了。
X1X2一定属于B,x1/x2不属于B。
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