问题补充:
已知函数f(x)= | lg(x-1)|,且对实数a,b满足1
答案:
f(a)= f(b/(b-1)),
∴lg(a-1)=土lg[b/(b-1)-1],
∴a=b/(b-1),或(a-1)/(b-1)=1,
1∴后者不成立,
由b/(b-1)b-1>1,∴b>2,∴a-2=b/(b-1)-2=(2-b)/(b-1)∴a
时间:2020-02-11 03:59:09
已知函数f(x)= | lg(x-1)|,且对实数a,b满足1
f(a)= f(b/(b-1)),
∴lg(a-1)=土lg[b/(b-1)-1],
∴a=b/(b-1),或(a-1)/(b-1)=1,
1∴后者不成立,
由b/(b-1)b-1>1,∴b>2,∴a-2=b/(b-1)-2=(2-b)/(b-1)∴a
解答题已知函数f(x)满足:对任意实数a b都有f(a?b)=af(b)+bf(a).
2020-04-14