Chứng minh 1/2!+2/3!+3/4!+...+99/100! < 1

Bài 1:

Ta có:

\(\dfrac{1}{2!}+\dfrac{2}{3!}+\dfrac{3}{4!}+...+\dfrac{99}{100!}\)

\(=\dfrac{2-1}{2!}+\dfrac{3-1}{3!}+\dfrac{4-1}{4!}+...+\dfrac{100-1}{100!}\)

\(=\dfrac{2}{2!}-\dfrac{1}{2!}+\dfrac{3}{3!}-\dfrac{1}{3!}+...+\dfrac{100}{100!}-\dfrac{1}{100!}\)

\(=\dfrac{1}{1!}-\dfrac{1}{2!}+\dfrac{1}{2!}-\dfrac{1}{3!}+...+\dfrac{1}{99!}-\dfrac{1}{100!}\)

\(=1-\dfrac{1}{100!}\)

Mà \(1-\dfrac{1}{100!}< 1\)

Nên \(\dfrac{1}{2!}+\dfrac{2}{3!}+\dfrac{3}{4!}+...+\dfrac{99}{100!}< 1\) (Đpcm)

Bài 2:

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{a+b-c}{c}=\dfrac{b+c-a}{a}=\dfrac{c+a-b}{b}=\dfrac{a+b-c+b+c-a+c+a-b}{a+b+c}=\dfrac{a+b+c}{a+b+c}=1\)

\(\Rightarrow\left\{{}\begin{matrix}a+b-c=c\\b+c-a=a\\c+a-b=b\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}a+b=2c\\b+c=2a\\c+a=2b\end{matrix}\right.\)

Thay vào biểu thức ta có:

\(B=\left(1+\dfrac{b}{a}\right)\left(1+\dfrac{a}{c}\right)\left(1+\dfrac{c}{b}\right)\)

\(=\dfrac{a+b}{a}.\dfrac{c+a}{c}.\dfrac{b+c}{b}\)

\(=\dfrac{2a.2b.2c}{abc}\)

\(=\dfrac{8\left(abc\right)}{abc}=8\)

Vậy \(B=8\)