Cho \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a+b+c}\)
Chứng minh rằng: \(\dfrac{1}{a^{2n+1}}+\dfrac{1}{b^{2n+1}}+\dfrac{1}{c^{2n+1}}=\dfrac{1}{a^{2n+1}+b^{2n+1}+c^{2n+1}}=\dfrac{1}{\left(a+b+c\right)^{2n+1}}\)
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Lời giải:
a/
Gọi ƯCLN(n+1, 2n+3)=d$
Khi đó:
$n+1\vdots d\Rightarrow 2n+2\vdots d(1)$
$2n+3\vdots d(2)$
Từ $(1); (2)\Rightarrow (2n+3)-(2n+1)\vdots d$ hay $1\vdots d$
$\Rightarrow d=1$
Vậy $n+1, 2n+3$ nguyên tố cùng nhau nên phân số đã cho tối giản.
Câu b,c làm tương tự.
\(P=\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+...+\dfrac{1}{\left(2n+1\right)\left(2n+3\right)}\\ 2P=\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{\left(2n+1\right)\left(2n+3\right)}\\ =\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2n+1}-\dfrac{1}{2n+3}\\ =1-\dfrac{1}{2n+3}\\ =\dfrac{2\left(n+1\right)}{2n+3}\\ P=\dfrac{2\left(n+1\right)}{2n+3}:2\\ =\dfrac{n+1}{2n+3}\)
a) Vế trái \(=\dfrac{1.3.5...39}{21.22.23...40}=\dfrac{1.3.5.7...21.23...39}{21.22.23....40}=\dfrac{1.3.5.7...19}{22.24.26...40}\)
\(=\dfrac{1.3.5.7....19}{2.11.2.12.2.13.2.14.2.15.2.16.2.17.2.18.2.19.2.20}\\ =\dfrac{1.3.5.7.9.....19}{\left(1.3.5.7.9...19\right).2^{20}}=\dfrac{1}{2^{20}}\left(đpcm\right)\)
b) Vế trái
\(=\dfrac{1.3.5...\left(2n-1\right)}{\left(n+1\right).\left(n+2\right).\left(n+3\right)...2n}\\ =\dfrac{1.2.3.4.5.6...\left(2n-1\right).2n}{2.4.6...2n.\left(n+1\right)\left(n+2\right)...2n}\\ =\dfrac{1.2.3.4...\left(2n-1\right).2n}{2^n.1.2.3.4...n.\left(n+1\right)\left(n+2\right)...2n}\\ =\dfrac{1}{2^n}.\\ \left(đpcm\right)\)
\(\dfrac{1}{\left(3n-1\right)\left(3n+2\right)}=\dfrac{1}{3}\left(\dfrac{1}{3n-1}-\dfrac{1}{3n+2}\right)\)
\(\Rightarrow A=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{3n-1}-\dfrac{1}{3n+2}\right)\)
\(\Rightarrow A=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{3n+2}\right)\)
\(\Rightarrow A=\dfrac{3n}{6\left(3n+2\right)}=\dfrac{n}{6n+4}\)
\(\dfrac{1}{\left(2n-1\right)\left(2n+1\right)\left(2n+3\right)}=\dfrac{1}{4}\left(\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}-\dfrac{1}{\left(2n+1\right)\left(2n+3\right)}\right)\)
\(\Rightarrow B=\dfrac{1}{4}\left(\dfrac{1}{1.3}-\dfrac{1}{3.5}+\dfrac{1}{3.5}-\dfrac{1}{3.7}+...+\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}-\dfrac{1}{\left(2n+1\right)\left(2n+3\right)}\right)\)
\(\Rightarrow B=\dfrac{1}{4}\left(\dfrac{1}{1.3}-\dfrac{1}{\left(2n+1\right)\left(2n+3\right)}\right)\)
\(\Rightarrow B=\dfrac{n\left(n+2\right)}{3\left(2n+1\right)\left(2n+3\right)}\)
\(\sqrt{1+\dfrac{1}{n^2}+\dfrac{1}{\left(n+1\right)^2}}=\sqrt{\dfrac{n^2\left(n+1\right)^2+\left(n+1\right)^2+n^2}{n^2\left(n+1\right)^2}}\)
\(=\sqrt{\dfrac{n^2\left(n+1\right)^2+2n^2+2n+1}{n^2\left(n+1\right)^2}}=\sqrt{\dfrac{n^2\left(n+1\right)^2+2n\left(n+1\right)+1}{n^2\left(n+1\right)^2}}\)
\(=\sqrt{\dfrac{\left[n\left(n+1\right)+1\right]^2}{n^2\left(n+1\right)^2}}=\dfrac{n\left(n+1\right)+1}{n\left(n+1\right)}=1+\dfrac{1}{n\left(n+1\right)}=1+\dfrac{1}{n}-\dfrac{1}{n+1}\)
\(\Rightarrow C=1+\dfrac{1}{1}-\dfrac{1}{2}+1+\dfrac{1}{2}-\dfrac{1}{3}+1+\dfrac{1}{3}-\dfrac{1}{4}+...+1+\dfrac{1}{2018}-\dfrac{1}{2019}\)
\(\Rightarrow C=2019-\dfrac{1}{2019}\)
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.