đại số

T

thaoteen21

tl

f(x)=$\dfrac{4^x}{4^x+2}$
a) a+b=1 thì f(a)+f(b)=1; ta có:
f(a)=$\dfrac{4^a}{4^a+2}$
f(b)=$\dfrac{4^b}{4^b+2}$
\Rightarrow f(a)+ f(b)=$\dfrac{4^a}{4^a+2}$+$\dfrac{4^b}{4^b+2}$
= $\dfrac{4^(a+b)+2.4^a+4^(a+b)+2.4^b}{ 4+2.4^a+2.4^b+4}$=1 (vì a+b=1 nên $4^(a+b)$=4) '' 4 mũ (a+b)''
\Rightarrow đ f cm
:):):):):):):)
 
T

thaoteen21

tl

câu b : ta có a+b=1 thì f(a)+f(b)=1
\Rightarrow f($\dfrac{1}{2015}$)+f($\dfrac{2014}{2015}$ )=1
\Rightarrow S= $\dfrac{2014}{2}$=1007
:):):):):):):):)
 
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