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Lời giải:
Ta có: \(\frac{1}{k(k+1)(k+2)}=\frac{1}{2}.\frac{2}{k(k+1)(k+2)}=\frac{1}{2}.\frac{(k+2)-k}{k(k+1)(k+2)}\)
\(=\frac{1}{2}\left(\frac{k+2}{k(k+1)(k+2)}-\frac{k}{k(k+1)(k+2)}\right)=\frac{1}{2}\left(\frac{1}{k(k+1)}-\frac{1}{(k+1)(k+2)}\right)\)
Áp dụng vào bài toán:
\(\frac{1}{1.2.3}=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}\right)\)
\(\frac{1}{2.3.4}=\frac{1}{2}\left(\frac{1}{2.3}-\frac{1}{3.4}\right)\)
\(\frac{1}{3.4.5}=\frac{1}{2}\left(\frac{1}{3.4}-\frac{1}{4.5}\right)\)
.......
\(\frac{1}{n(n+1)(n+2)}=\frac{1}{2}\left(\frac{1}{n(n+1)}-\frac{1}{(n+1)(n+2)}\right)\)
\(\Rightarrow B=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{n(n+1)}-\frac{1}{(n+1)(n+2)}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{(n+1)(n+2)}\right)=\frac{1}{4}-\frac{1}{2(n+1)(n+2)}\)
Bài 1 :
Để \(\dfrac{x^3+x^2-x-1}{x^3+2x-3}=0\) thì \(x^3+x^2-x-1=0\)
\(\Leftrightarrow x^2\left(x+1\right)-\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2-1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+1=0\\x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=1\end{matrix}\right.\)
Vậy,.........
\(B=\dfrac{1}{2}\left(\dfrac{2}{1\cdot2\cdot3}+\dfrac{2}{2\cdot3\cdot4}+...+\dfrac{2}{n\left(n+1\right)\left(n+2\right)}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{1\cdot2}-\dfrac{1}{2\cdot3}+\dfrac{1}{2\cdot3}-\dfrac{1}{3\cdot4}+...+\dfrac{1}{n\cdot\left(n+1\right)}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{n^2+3n+2-2}{2\left(n+1\right)\left(n+2\right)}=\dfrac{n\left(n+3\right)}{4\left(n+1\right)\left(n+2\right)}\)
\(M=\dfrac{1}{\left(x-1\right)\left(x-2\right)}+\dfrac{1}{\left(x-2\right)\left(x-3\right)}+\dfrac{1}{\left(x-3\right)\left(x-4\right)}+\dfrac{1}{\left(x-4\right)\left(x-5\right)}\)
\(M=\dfrac{1}{x-1}-\dfrac{1}{x-2}+\dfrac{1}{x-2}-\dfrac{1}{x-3}+\dfrac{1}{x-3}-\dfrac{1}{x-4}+\dfrac{1}{x-4}-\dfrac{1}{x-5}\)
\(M=\dfrac{1}{x-1}-\dfrac{1}{x-5}\)
\(M=\dfrac{x-5-x+1}{\left(x-5\right)\left(x-1\right)}=-\dfrac{4}{x^2-6x+5}\)
Bài 4:
a: \(K=\dfrac{2x-9}{\left(x-2\right)\left(x-3\right)}-\dfrac{x+3}{x-2}+\dfrac{2x+1}{x-3}\)
\(=\dfrac{2x-9-x^2+9+\left(2x+1\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}\)
\(=\dfrac{-x^2+2x+2x^2-4x+x-2}{\left(x-2\right)\left(x-3\right)}\)
\(=\dfrac{x^2-x-2}{\left(x-2\right)\left(x-3\right)}=\dfrac{x+1}{x-3}\)
b: Để K<1 thì K-1<0
\(\Leftrightarrow\dfrac{x+1-x+3}{x-3}< 0\)
=>x-3<0
hay x<3
c: Để K là số nguyên thì \(x+1⋮x-3\)
\(\Leftrightarrow x-3+4⋮x-3\)
\(\Leftrightarrow x-3\in\left\{1;-1;2;-2;4;-4\right\}\)
hay \(x\in\left\{4;5;1;7;-1\right\}\)
1/
A= \(\dfrac{2x+6}{\left(x+3\right)\left(x-2\right)}\) = 0 ;(ĐKXĐ : x ≠ -3; x ≠ 2)
⇔ A = \(\dfrac{2\left(x+3\right)}{\left(x+3\right)\left(x-2\right)}\) = 0
⇔ A = \(\dfrac{2}{x-2}\) = 0
⇒ x = 2 (loại) ⇒ pt vô nghiệm
Ta có:
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=k\)
\(\Rightarrow\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2=k^2\)
\(\Rightarrow\dfrac{1}{a^2}+\dfrac{2}{ab}+\dfrac{1}{b^2}+\dfrac{2}{bc}+\dfrac{1}{c^2}+\dfrac{2}{ac}=k^2\)
\(\Rightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{1\left(a+b+c\right)}{abc}=k^2\)
\(\Rightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=k^2-k\)
b.
\(B=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{\left(n-1\right)n\left(n+1\right)}\\ =\dfrac{1}{2}\left(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+....+\dfrac{2}{\left(n-1\right).n.\left(n+1\right)}\right)\\ =\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{\left(n-1\right).n}-\dfrac{1}{n\left(n+1\right)}\right)\\ =\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{n\left(n+1\right)}\right)=\dfrac{1}{4}-\dfrac{1}{2n\left(n+1\right)}\)
Ta có: \(m+n+k=0\)
\(\Leftrightarrow m+n=-k\)
\(\Leftrightarrow\left(m+n\right)^2=\left(-k\right)^2\)
\(\Leftrightarrow m^2+2mn+n^2=k^2\)
\(\Leftrightarrow m^2+n^2-k^2=-2mn\)
Tương tự, ta có: \(n^2+k^2-m^2=-2nk\)
\(k^2+m^2-n^2=-2km\)
Thay \(m^2+n^2-k^2=-2mn;n^2+k^2-m^2=-2nk;\)\(k^2+m^2-n^2=-2km\) vào biểu thức M ta có:
M = \(\dfrac{1}{-2mn}+\dfrac{1}{-2nk}+\dfrac{1}{-2km}=\dfrac{-1}{2}\left(\dfrac{1}{mn}+\dfrac{1}{nk}+\dfrac{1}{km}\right)\)
M = \(\dfrac{-1}{2}\left(\dfrac{nk^2m+m^2nk+mn^2k}{m^2n^2k^2}\right)\)
\(M=\dfrac{-1}{2}\left(\dfrac{mnk\left(k+m+n\right)}{m^2n^2k^2}\right)\)
M = \(\dfrac{-1}{2}.\dfrac{0}{mnk}\)\(=0\)