Question

Cho hàm số $f(x)=\dfrac{100^x}{100^x+10}$. Chứng minh rằng : nếu a, b là hai số thỏa mãn : $a+b=1$ thì $f(a)+f(b)=1$.

Answer 1

Ta có:

\(f\left(a\right)+f\left(b\right)=f\left(a\right)+f\left(1-a\right)\\ =\dfrac{100^a}{100^a+10}+\dfrac{100^{1-a}}{100^{1-a}+10}\\ =\dfrac{100^a}{100^a+10}+\dfrac{\dfrac{100}{100^a}}{\dfrac{100}{100^a}+10}\\ =\dfrac{100^a}{100^a+10}+\dfrac{100}{100^a}.\dfrac{100^a}{100+10.100^a}\\ =\dfrac{100^a}{100^a+10}+\dfrac{10}{10+100^a}\\ =\dfrac{100^a+10}{10+100^a}=1\left(đpcm\right)\)

Answer 2

Answer 3

Answer 4

Answer 5

 

 

Answer 6

f(a)+f(b)=f(a)+f(1−a)=100a+10100a​+1001−a+101001−a​=100a+10100a​+100a100​+10100a100​​=100a+10100a​+100a100​.100+10.100a100a​=100a+10100a​+10+100a10​=10+100a100a+10​=1(đpcm)

Answer 7

Ta có:

f(a)+f(b)=\(\dfrac{100a}{100a+10}\)+\(\dfrac{100}{100b+10}\)=\(\dfrac{\text{100 a (100 b +10)+100 b (100 a +10) ​ }}{\left(100a+10\right)\left(100b+10\right)}\)

=\(\dfrac{\text{200+10(100 a +100 b )}}{\text{200+10(100 a +100 b )}}\)=1

=>f(a)+f(b)=1

Answer 8

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Answer 9

b=1-a

f(a) + f(1)= 100^a/100^a+10 +100^1-a/100^1-a+10

=100^a/100^a+10 + (100/100^a)/(100/100^a +10)

= 100^a/100^a+10 + 100/100^a. 1/(100+100^a.10)/100^a

⁼100^a/100^a+10 + 100/100^a.100^a/100+100^a.10

⁼100^a/100^a+10 + 10/10+100^a

⁼100^a+10/100^a+10

=1(đpcm)

 

Answer 10

 

Answer 11

Ta có:fa+sb=100a trên 100a+10

Answer 12

1

Answer 13

Con chịu:_)

Answer 14

cô ơi con chx bt lm ạ=))

Answer 15

Em chịu

Answer 16

kk

Answer 17

x=a

\(\Rightarrow\) f(a)=\(\dfrac{100^a}{100^a+10}\)

x=b

\(\Rightarrow\) f(b)=\(\dfrac{100^b}{100^b+10}\)

 f(a)+f(b)=\(\dfrac{100^a}{100^a+10}\)+\(\dfrac{100^b}{100^b+10}\)

=\(\dfrac{100^a.\left(100^b+10\right)+100^b.\left(100^a+10\right)}{\left(100^a+10\right).\left(100^b+10\right)}\)

=\(\dfrac{100^a.100^b+10.100^a+100^b.100^a+10.100^b}{100^a+10.100^b+10}\)

=\(\dfrac{100^{a+b}+10.100^a+100^{b+a}+10.100^b}{100^{a+b}+10.100^a+10.100^b+100}\)

= \(\dfrac{2.100^{a+b}+10.\left(100^a+100^b\right)}{100^{a+b}+10.\left(100^a+100^b\right)+100}\)

nếu a+b=1 thì  f(a)+f(b)=\(\dfrac{200+10.\left(100^a+100^b\right)}{200+10.\left(100^a+100^b\right)}\)=1

Answer 18