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In a math problem
( 1)(a+3)/(c-b)=k a+3=kc-bk

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.