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\(1-\dfrac{3}{n\left(n+2\right)}=\dfrac{n\left(n+2\right)-3}{n\left(n+2\right)}=\dfrac{\left(n-1\right)\left(n+3\right)}{n\left(n+2\right)}\)
\(\Rightarrow M=\dfrac{1.5}{2.4}.\dfrac{2.6}{3.5}.\dfrac{3.7}{4.6}...\dfrac{\left(n-1\right)\left(n+3\right)}{n\left(n+2\right)}\)
\(=\dfrac{1.2.3...\left(n-1\right)}{2.3.4...n}.\dfrac{5.6.7...\left(n+3\right)}{4.5.6...\left(n+2\right)}\)
\(=\dfrac{1}{n}.\dfrac{n+3}{4}=\dfrac{n+3}{4n}=\dfrac{1}{4}+\dfrac{3}{4n}>\dfrac{1}{4}\) (đpcm)
Thừa số tổng quát:
\(1+\dfrac{1}{n^2+2n}=\dfrac{n^2+2n+1}{n^2+2n}=\dfrac{\left(n+1\right)^2}{\left(n+1\right)^2-1}\)
Đặt: \(\left(n+1\right)^2=t\ge0\) biểu thức được phát biểu dưới dạng: \(\dfrac{t}{t-1}\) Thay vào bài toán tìm được giá trị.
Lời giải:
Ta có:
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\)
\(\Leftrightarrow \frac{a+b}{ab}=\frac{1}{a+b+c}-\frac{1}{c}=\frac{-(a+b)}{c(a+b+c)}\)
\(\Leftrightarrow (a+b)\left(\frac{1}{ab}+\frac{1}{c(a+b+c)}\right)=0\)
\(\Leftrightarrow (a+b).\frac{ab+c(a+b+c)}{abc(a+b+c)}=0\)
\(\Leftrightarrow (a+b).\frac{(c+a)(c+b)}{abc(a+b+c)}=0\)
\(\Leftrightarrow (a+b)(b+c)(c+a)=0\)
Ta sẽ cm \(\frac{1}{a^{2n+1}}+\frac{1}{b^{2n+1}}+\frac{1}{c^{2n+1}}=\frac{1}{a^{2n+1}+b^{2n+1}+c^{2n+1}}(*)\)
Thật vậy: \((*)\Leftrightarrow \frac{a^{2n+1}+b^{2n+1}}{(ab)^{2n+1}}=\frac{1}{a^{2n+1}+b^{2n+1}+c^{2n+1}}-\frac{1}{c^{2n+1}}\)
\(\Leftrightarrow \frac{a^{2n+1}+b^{2n+1}}{(ab)^{2n+1}}=\frac{-(a^{2n+1}+b^{2n+1})}{c^{2n+1}(a^{2n+1}+b^{2n+1}+c^{2n+1})}\)
\(\Leftrightarrow (a^{2n+1}+b^{2n+1})\left(\frac{1}{(ab)^{2n+1)}}+\frac{1}{c^{2n+1}(a^{2n+1}+b^{2n+1}+c^{2n+1})}\right)=0\)
\(\Leftrightarrow (a^{2n+1}+b^{2n+1}).\frac{c^{2n+1}(a^{2n+1}+b^{2n+1}+c^{2n+1})+(ab)^{2n+1}}{(abc)^{2n+1}(a^{2n+1}+b^{2n+1}+c^{2n+1})}=0\)
\(\Leftrightarrow \frac{(a^{2n+1}+b^{2n+1})(c^{2n+1}+b^{2n+1})(c^{2n+1}+a^{2n+1})}{abc^{2n+1}(a^{2n+1}+b^{2n+1}+c^{2n+1})}=0\)
Thấy rằng
\((a^{2n+1}+b^{2n+1})(b^{2n+1}+c^{2n+1})(c^{2n+1}+a^{2n+1})=(a+b).X.(b+c).Y.(c+a).Z\)
\(=0\) (do \((a+b)(b+c)(c+a)=0\) )
Do đó đẳng thức $(*)$ cần chứng minh đúng.
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Ta tiếp tục chứng minh \(\frac{1}{a^{2n+1}+b^{2n+1}+c^{2n+1}}=\frac{1}{(a+b+c)^{2n+1}}(**)\)
\(\Leftrightarrow a^{2n+1}+b^{2n+1}+c^{2n+1}=(a+b+c)^{2n+1}\)
Thật vậy:
\((a+b)(b+c)(c+a)=0\)\(\Rightarrow \left[\begin{matrix} a+b=0\\ b+c=0\\ c+a=0\end{matrix}\right.\)
Không mất tổng quát giả sử \(a+b=0\)
\(\Rightarrow \left\{\begin{matrix} a^{2n+1}+b^{2n+1}+c^{2n+1}=(-b)^{2n+1}+b^{2n+1}+c^{2n+1}=c^{2n+1}\\ (a+b+c)^{2n+1}=(0+c)^{2n+1}=c^{2n+1}\end{matrix}\right.\)
\(\Rightarrow a^{2n+1}+b^{2n+1}+c^{2n+1}=(a+b+c)^{2n+1}\)
Do đó $(**)$ đúng
Từ $(*)$ và $(**)$ ta có đpcm.
Ta có:
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a+b+c}\)
\(\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=-b\\b=-c\\c=-a\end{matrix}\right.\)
Xét \(a=-b\) thì ta có
\(\left\{{}\begin{matrix}\dfrac{1}{a^{2n+1}}+\dfrac{1}{b^{2n+1}}+\dfrac{1}{c^{2n+1}}=\dfrac{1}{c^{2n+1}}\\\dfrac{1}{a^{2n+1}+b^{2n+1}+c^{2n+1}}=\dfrac{1}{c^{2n+1}}\\\dfrac{1}{\left(a+b+c\right)^{2n+1}}=\dfrac{1}{c^{2n+1}}\end{matrix}\right.\)
\(\Rightarrow\dfrac{1}{a^{2n+1}}+\dfrac{1}{b^{2n+1}}+\dfrac{1}{c^{2n+1}}=\dfrac{1}{a^{2n+1}+b^{2n+1}+c^{2n+1}}=\dfrac{1}{\left(a+b+c\right)^{2n+1}}\)
Tương tự cho 2 bộ số còn lại ta được ĐPCM.
a: \(=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{\left(2n-1\right)\left(2n+1\right)}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{2n+1-1}{2n+1}=\dfrac{1}{2}\cdot\dfrac{2n}{2n+1}=\dfrac{n}{2n+1}\)
b: \(=\dfrac{1}{4}\left(\dfrac{4}{1\cdot5}+\dfrac{4}{5\cdot9}+...+\dfrac{4}{\left(4n-3\right)\left(4n+1\right)}\right)\)
\(=\dfrac{1}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{4n-3}-\dfrac{1}{4n+1}\right)\)
\(=\dfrac{1}{4}\cdot\dfrac{4n}{4n+1}=\dfrac{n}{4n+1}\)
Ta có \(\dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+\dfrac{1}{6\cdot8}+...+\dfrac{1}{2n\left(2n+2\right)}=\dfrac{1009}{4038}\)
\(\Leftrightarrow\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+\dfrac{2}{6\cdot8}+...+\dfrac{2}{2n\left(2n+2\right)}=\dfrac{1009}{2019}\)
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{2n}-\dfrac{1}{2n+2}=\dfrac{1009}{2019}\)
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{2n+2}=\dfrac{1009}{2019}\)
\(\Leftrightarrow\dfrac{n}{2n+2}=\dfrac{1009}{2019}\)
\(\Leftrightarrow2019n=1009\left(2n+2\right)\)
\(\Leftrightarrow2019n=2018n+2018\)
\(\Leftrightarrow n=2018\)