若p{x竖线x=n，n属于z}，Q{x竖线x=n/2，n属于z}，s{x竖线x=n+1/2}，则Q=P并S，为什么？_百度知道
若p{x竖线x=n，n属于z}，Q{x竖线x=n/2，n属于z}，s{x竖线x=n+1/2}，则Q=P并S，为什么？
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若P={x|x=nn∈Z}Q={x|x=n/2n∈Z}S={x|x=n+1/2n∈Z}则Q=P∪S解：于Q={x|x=n/2n∈Z}n要奇数要偶数即
n=2k或2k+1
Q={x|x=2k/2或(2k+1)/2k∈Z}即
Q={x|x=k或k+1/2k∈Z}=P∪S
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出门在外也不愁s、t属于{x|x=m^2+n^2,m、n属于N} 求证 s/t=p^2+q^2 t不等于0 其中p、q是有理数_百度知道
s、t属于{x|x=m^2+n^2,m、n属于N} 求证 s/t=p^2+q^2 t不等于0 其中p、q是有理数
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只要知道s= m1 ^2 + n1 ^2 = (m1+ i *n1)(m1-i*n1), t = m2 ^2 + n2 ^2 = (m2+ i *n2)(m2-i*n2), 因此s/t=(m1+ i *n1)(m1-i*n1)/((m2+ i *n2)(m2-i*n2)) = [(m1+ i *n1)/(m2-i*n2)]* [(m2+ i *n2)/(m1-i*n1)] 展开即得：(a+i b)(a- ib)=a^2+b^2 其中a,b为m1,n1,m2,n2的有理函数。i为虚数单位。
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谢谢 虚数我们还没有学到 不过我会做了
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出门在外也不愁已知M={x|x^2-sx+s-1=0},N={x|x^2-6x+5=0},P={x|x^2-tx+6=0},且M并U=N,N交P=P,求s和t的值_百度知道
已知M={x|x^2-sx+s-1=0},N={x|x^2-6x+5=0},P={x|x^2-tx+6=0},且M并U=N,N交P=P,求s和t的值
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x^2-sx+s-1=0(x-1)[x-(s-1)]=0所以x=1或x=s-1故M={x|x^2-sx+s-1=0}={1,s-1}x^2-6x+5=0(x-1)(x-5)=0x=1或x=5故N={x|x^2-6x+5=0}={1,5}M∪N=N那么s-1=1或s-1=5故s=2或s=6N∩P=PP={x|x^2-tx+6=0}P可以是空集或P={1}或P={5}或P={1,5}P是空集时Δ=t^2-24&0,-2√6&t&2√6P={1}时,1^2-t+6=0,t=7,但此时P={1,6},不符,舍同理x^2-tx+6=0均不符合所以s=2或s=6,-2√6&t&2√6如果不懂,请Hi我,祝学习愉快
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出门在外也不愁C O M P A N Y, P.C.
C O M P A N Y, P.C.
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