要求,p,q均为两位数(p,q为50到80之间的素数)求出公钥和私钥(d和n)
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=2(p+q)(3p-2q)再问:步骤?再答:就一步,是提取公因式2(p+q)
(p+q)^3=p^3+q^3+3p^2q+3pq^2=(p^3+q^3)+3pq(p+q) 所以p^3+q3=(p+q)^3-3pq(p+q)-------------(1) 又因为(p+q)/
若P+q>2,则p>2-q,由于x^3是R上的增函数,∴p^3>(2-q)^3=8-12q+6q^2-q^3,∴p^3+q^3>6(q-1)^2+2>=2,矛盾.∴p+q
p∧q为假命题┐(p∧q)=(┑p)∨(┐q)为真命题(┑p)真或(┐q)真p假或q假 因此由“命题p∧q为假”得到“p,q均为假命题”的结论是错的.再问:抱歉哈~题中说的是若p并q~您说的是交集.
pV(p->q)pV(非pVq)1P->(Q->P)非pV(q->p)非pV(非qVp)1非p->(p->q)p->(q->p)
6p(p+q)-4p(p+q)=(p+q)(6p-4p)=(p+q)2p=2p(p+q)
[(p-q)³-2(q-p)²-2/3(q-p)]/[(p-q)/3]=3(p-q)²-6(p-q)+2(p-q)
=(2p²-pq+4pq-2q²)-(p²-q²)=p²+3pq-q²
previousquestion先决问题
m²(p-q)-p+q=m²(p-q)-(p-q)=(m²-1)(p-q)=(m+1)(m-1)(p-q)
(1)原式=2(p+q)(3p+3q-1)(2)原式=(x-y)(2x-2y-x)=(x-y)(x-2y)(3)原式=(a-b)(m+n)(4)原式=12.1×(1.3+0.9-1.2)=12.1(5
由p^2*q+12p-12≤3p^2+4pq-4q?p^2q+12p-12-(3p^2+4pq-4q)≤0?p^2*(q-3)+4p(3-q)-4(3-q)≤0?(p-2)^2*(q-3)≤0?.(1
6p(p+q)-4q(p+q)=(6p-4q)(p+q)=2(3p-2q)(p+q)
6p+(p+q)-4q(p+q)=6p+p+q-4qp-4q^2=7p+q-4pq-4q^2
6p(p+q)-4q(p+q)=(6P-4q)(p+q)m(a-3)+2(3-a)=(m-2)(a-3)
6p(p+q)-4q(p+q)=(p+q)(6p-4q)=2(p+q)(3p-2q)
再答:我的回答满意吗?望采纳
6p(p+q)²-4q(p+q)²=2(p+q)²(3p-2q)
6p{(p+q)(p+q)}-4q(p+q)=2(p+q)[3p(p+q)-2q]=2(p+q)(3p²++3pq-2q)