因式分解r^2+10r+25-^2

r^2+10r+25-s^2
=(r+5)^2-(s)^2
=[(r+5)+s][(r+5)-s]
=(r+s+5)(r-s+5)

r^2+10r+25-s^2
=r^2+2(5)(r)+5^2-s^2
=(r+5)^2-s^2..........................呢到用左a^2+2ab+b^2=(a+b)^2
=(r+5+s)(r+5-s)......................呢到用左a^2-b^2=(a+b)(a-b)

因為，
r2 + 10r + 25
= (r+5)2

所以，
r2 + 10r + 25﹣s2
= (r+5)2﹣s2
= (r+5+s)(r+5﹣s)

根據恆等式a2﹣b2= (a﹣b)(a+b)

2008-01-02 15:12:51 補充：
r^2 + 10r + 25= r^2 + 10r + 5^2= (r+5)^2這是根據恆等式：a^2+2ab+b^2 = (a+b)^2