一條簡單的中學數

p+qs
———— –t ＝ pq
ps-q


p+qs
———— ＝ pq + t
ps-q


p + qs = (pq + t) (ps - q)
p + qs = (pq)(ps) + (pq)(-q) + t (ps) + t(-q)
p + qs = p^2qs - pq^2 + tps - tq
qs - p^2qs - tps = -pq^2 - tq - p
s(q-p^2q - tp) = -(pq^2 + tq - p)
s = -(pq^2 + tq - p)
-----------------------
(q-p^2q - tp)
= pq^2 + tq - p
-----------------------
p^2q + tp - q

p+qs
———— –t ＝ pq 【s】
ps-q


p+qs tps-tq
-----–--------＝ pq
ps-q ps-q

p+qs-tps+tq = pqxps-q

p+qs-tps+tq = pqps-pqq

p+tq+pqq =pqps-qs+tps

p+tq+pqq = s(pqp-q+tp)

p+tq+pqq
s = -------------
pqp-q+tp

p + qs - tps + tq = p^2qs - pq^2

p + tq + pq^2 = p^2qs - qs + tps

p + tq + pq^2
------------------ = s
p^2q - q + tp

[p+qs / ps-q ]-t = pq [s]

[p+qs / ps-q ]=pq+t

p+qs-ps+q=pq+t

qs-ps+p+q=pq+t

s(p-q)+p+q=pq+t

s(p-q)=pq+t-p-q

s=(pq+t-p-q)/(p-q)

i think is like that~

p+qs
———— –t ＝ pq 【s】
ps-q

(p+qs)/(ps-q)=pq+t
p+qs=(ps-q)(pq+t)
p+qs=p^2qs+pts-pq^2-qt
p+pq^2-qt=(p^2q+pt-q)s
s= (p^2q+pt-q) / (p+pq^2-qt)

2007-01-01 19:35:32 補充：
更正︰(p+qs)/(ps-q)=pq+tp+qs=(ps-q)(pq+t)p+qs=p^2qs+pts-pq^2-qtp+pq^2+qt=(p^2q+pt-q)ss= (p+pq^2+qt) / (p^2q+pt-q)