Question

若正数a,b满足ab=a+b+3,求ab的取值范围

Answer 1

a+b=ab-3

a>0,b>0
a+b>=2√ab
所以ab-3>=2√ab
ab-2√ab-3>=0
(√ab-3)(√ab+1)>=0
√ab>0,√ab+1>0
所以√ab-3>=0
√ab>=3
ab>=9

Answer 2

ab=a+b+3
∴ab-b＝b(a-1)＝a+3>3，故a>1
b＝(a+3)/(a-1)＝1+4/(a-1)
ab＝a*[1+4/(a-1)]
＝a+4a/(a-1)
＝a+4+4/(a-1)
＝(a-1)+4/(a-1)+5
≥2√[(a-1)*4/(a-1)]+5＝9
∴ab≥9，当且仅当a-1=4/(a-1)jf ，即a=b=3时取得

Answer 3

a+b≧2√ab
a+b+3≧3+2√ab
因为ab=a+b+3
所以：ab≧3+2√ab
令√ab=t
则t²≧3+2t
t²-2t-3≧0
(t-3)(t+1)≧0
t≧3或t≦-1
因为t=√ab
所以显然t=√ab≧3
所以：ab≧9

Answer 4

ab=a+b+3>3
ab>=2√(ab)+3
(√ab-3)(√ab+1)>=0
√ab>0,√ab+1>0
所以√ab-3>=0
√ab>=3
ab>=9