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 \(B=8\)

Ta có: (a+b+c)^3 = a^3 + b^3 + c^3 + 3(ab+bc+ca)  [ Cái này tự cm nhé, nếu k biết pm mình ]

      <=> 9^3 = 53 + 3(ab+bc+ca)

      <=> 3(ab+bc+ca) = 9^3 - 53 

Chúc làm bài tốt nhé !

Ta có:

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

\(\Rightarrow100-1-\dfrac{1}{2}-...-\dfrac{1}{100}=\dfrac{1}{2}+\dfrac{2}{3}+...+\dfrac{99}{100}\)

\(\Rightarrow100=1+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{2}{3}+...+\dfrac{1}{100}+\dfrac{99}{100}\)

\(\Rightarrow100=1+1+1+...+1\) (\(100\) số \(1\))

\(\Rightarrow100=100\)

Vậy \(100-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}\right)=\dfrac{1}{2}+\dfrac{2}{3}+...+\dfrac{99}{100}\) (Đpcm)

Ta có:

\(S=3+\dfrac{3}{2}+\dfrac{3}{2^2}+...+\dfrac{3}{2^9}\)

\(\Rightarrow S=3\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\right)\)

\(\Rightarrow2S=3\left(2+1+\dfrac{1}{2}+...+\dfrac{1}{2^8}\right)\)

\(\Rightarrow2S-S=3\left[\left(2+1+\dfrac{1}{2}+...+\dfrac{1}{2^8}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\right)\right]\)

\(\Rightarrow S=3\left(2-\dfrac{1}{2^9}\right)\)

\(\Rightarrow S=3.\dfrac{1023}{512}=\dfrac{3069}{512}\)

Vậy \(S=\dfrac{3069}{512}\)

Giải:

Ta có:

\(3^{n+2}-2^{n+4}+3^n+2^n\)

\(=3^n.9-2^n.16+3^n+2^n\)

\(=3^n\left(9+1\right)-2^n\left(16-1\right)\)

\(=3^n.10+2^n.15\)

\(=3^{n-1}.3.10-2^{n-1}.2.15\)

\(=3^{n-1}.30-2^{n-1}.30\)

\(=30\left(3^{n-1}-2^{n-1}\right)\)

Mặt khác \(n\) là số nguyên dương nên \(n-1\) là số tự nhiên

\(\Rightarrow30\left(3^{n-1}-2^{n-1}\right)⋮30\)

Hay \(3^{n+2}-2^{n+4}+3^n+2^n⋮30\forall n\) nguyên dương (Đpcm)

Ta có:

\(C=\dfrac{1}{1.2.3.4}+\dfrac{1}{2.3.4.5}+...+\dfrac{1}{27.28.29.30}\)

\(\Rightarrow3C=\dfrac{3}{1.2.3.4}+\dfrac{3}{2.3.4.5}+...+\dfrac{3}{27.28.29.30}\)

\(\Rightarrow3C=\dfrac{1}{1.2.3}-\dfrac{1}{2.3.4}+\dfrac{1}{2.3.4}-\dfrac{1}{3.4.5}+...+\dfrac{1}{27.28.29}-\dfrac{1}{28.29.30}\)

\(\Rightarrow3C=\dfrac{1}{1.2.3}-\dfrac{1}{28.29.30}\)

\(\Rightarrow3C=\dfrac{1353}{8120}\)

\(\Rightarrow C=\dfrac{1353}{\dfrac{8120}{3}}=\dfrac{451}{8120}\)

Vậy \(C=\dfrac{451}{8120}\)

Ta có:

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

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

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

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

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

Mà \(1-\dfrac{1}{100!}< 1\)

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

Giải:

Áp dụng BĐT \(\left|a\right|-\left|b\right|\le\left|a-b\right|\) ta có:

\(\left|x-1004\right|-\left|x+1003\right|\le\left|x-1004-x-1003\right|=2007\)

Dấu "=" xảy ra khi \(\Leftrightarrow x=-1013\)

Vậy \(MAX_A=2007\) tại \(x=-1013\)

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