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Ta có:
\(f\left(a\right)+f\left(b\right)=f\left(a\right)+f\left(1-a\right)\\ =\dfrac{100^a}{100^a+10}+\dfrac{100^{1-a}}{100^{1-a}+10}\\ =\dfrac{100^a}{100^a+10}+\dfrac{\dfrac{100}{100^a}}{\dfrac{100}{100^a}+10}\\ =\dfrac{100^a}{100^a+10}+\dfrac{100}{100^a}.\dfrac{100^a}{100+10.100^a}\\ =\dfrac{100^a}{100^a+10}+\dfrac{10}{10+100^a}\\ =\dfrac{100^a+10}{10+100^a}=1\left(đpcm\right)\)
Bài 4:
\(f\left(5\right)-f\left(4\right)=2019\)
=>\(125a+25b+25c+d-64a-16b-4c-d=2019\)
=>\(61a+9b+21c=2019\)
\(f\left(7\right)-f\left(2\right)\)
\(=343a+49b+7c+d-8a-4b-2c-d\)
\(=335a+45b+5c\)
\(=5\left(61a+9b+21c\right)=5\cdot2019\) là hợp số
tham khảo thôi nhé ko giống y sì đâu
https://olm.vn/hoi-dap/detail/213882782299.html
Ta có \(f\left(x\right)=\frac{100^x}{100^x+10}\)
\(\Rightarrow\left\{\begin{matrix}f\left(a\right)=\frac{100^a}{100^a+10}\\f\left(b\right)=\frac{100^b}{100^b+10}\end{matrix}\right.\)
\(\Rightarrow f\left(a\right)+f\left(b\right)=\frac{100^a}{100^a+10}+\frac{100^b}{100^b+10}\)
\(=\frac{100^a\left(100^b+10\right)+100^b\left(100^a+10\right)}{100^b\left(100^a+10\right)+10\left(100^a+10\right)}\)
\(=\frac{100^a.100^b+100^a.10+100^b,100^a+100^b.10}{100^b.100^a+100^b.10+100^a.10+100}\)
\(=\frac{100^{a+b}+100^a.10+100^{b+a}+100^b.10}{100^{b+a}+100^b.10+100^a.10+100}\)
Thế \(a+b=1\)
\(\Rightarrow\frac{100+100^a.10+100+100^b.10}{100+100^b.10+100^a.10+100}=1\)
\(\Leftrightarrow f\left(a\right)+f\left(b\right)=1\)