bk=kc-a-3
b=(kc-a-3)/k
a(a- 1)/(b+c)=k
a(a- 1)=bk+ck
bk=(a^2-a-ck)
b=(a^2-a-ck)/k
So (a 2-a-CK)/k = (KC-a-3)/k
a^2-a-ck=kc-a-3
2kc=a^2+3
k=(a^2+3)/2c
(2)k=(a^2+3)/2c & gt; 0
So (a+3)/(c-b) = k >; 0(a+3 & gt; 0)
c-b & gt; 0
c & gtb
(3)k=(a^2+3)/2c=2
a^2+3=4c
A 2-4a+3 = 4c-4a = 4 (c-a) ... (1)
(a-3)(a- 1)=4(c-a)
a(a- 1)= 2 b+ 2c & lt; 4c =a^2+3
a^2-a<; a^2-3
A>3 So (a-3) (a- 1) = 4 (c-a) is represented by formula (a)>0.
So c> is a c>b.
So c is the maximum.
Teacher's final teaching summary 600 words 1
A semester passes in a hurry, and the county bureau will issue a unified test