Question

So sánh A và B biết:

\(A=\dfrac{10^{49}+1}{10^{51}+1};B=\dfrac{10^{48}+1}{10^{50}+1}\)

Còn mỗi bài này thui mai e phải đi hok mong anh/cj giúp đỡ ạ!!!!!!!!!!

Thanks

Answer 1

\(\dfrac{a}{b}< 1\Leftrightarrow\dfrac{a+m}{b+m}\left(m\in N\right)\)

\(A=\dfrac{10^{49}+1}{10^{51}+1}< 1\)

\(A< \dfrac{10^{49}+1+9}{10^{51}+1+9}< \dfrac{10^{49}+10}{10^{51}+10}< \dfrac{10\left(10^{48}+1\right)}{10\left(10^{50}+1\right)}< \dfrac{10^{48}+1}{10^{50}+1}=B\)

\(\Leftrightarrow A< B\)

Answer 2

Ta có:

\(10^2A=\dfrac{10^{51}+1+99}{10^{51}+1}=1+\dfrac{99}{10^{51}+1}\)

\(10^2B=\dfrac{10^{50}+1+99}{10^{50}+1}=1+\dfrac{99}{10^{50}+1}\)

Vì \(1=1\) mà \(\dfrac{99}{10^{51}+1}< \dfrac{99}{10^{50}+1}\) (do \(99=99\); \(10^{51}+1>10^{50}+1\))

nên \(10^2A< 10^2B\)

\(\Rightarrow A< B\)

Vậy \(A< B\)

Chúc bạn học tốt!!!

Answer 3

Ta có :

* 100A = 10²A = \(10^2\dfrac{\left(10^{49}+1\right)}{10^{51}+1}\) = \(\dfrac{10^{51}+100}{10^{51}+1}=\dfrac{10^{51}+1}{10^{51}+1}+\dfrac{99}{10^{51}+1}\) = \(1+\dfrac{99}{10^{51}+1}\)

* 100B = 10²B = \(10^2\dfrac{\left(10^{48}+1\right)}{10^{50}+1}=\dfrac{10^{50}+100}{10^{50}+1}=\dfrac{10^{50}+1}{10^{50}+1}+\dfrac{99}{10^{50}+1}\)

= \(1+\dfrac{99}{10^{50}+1}\)

Vì \(\dfrac{99}{10^{50}+1}>\dfrac{99}{10^{51}+1}\Rightarrow1+\dfrac{99}{10^{50}+1}>1+\dfrac{99}{10^{51}+1}\)

Hay 100B > 100A => B>A