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\(\frac{1}{21}+\frac{1}{77}+\frac{1}{165}+...+\frac{1}{n^2+4n}=\frac{56}{673}\)
<=> \(\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+...+\frac{1}{n.\left(n+4\right)}=\frac{56}{673}\)
<=> \(4.\left(\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+...+\frac{1}{n.\left(n+4\right)}\right)=4.\frac{56}{673}\)
<=> \(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{n\left(n+4\right)}=\frac{224}{673}\)
<=> \(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{n}-\frac{1}{n+4}=\frac{224}{673}\)
<=> \(\frac{1}{3}-\frac{1}{n+4}=\frac{224}{673}\)
<=> \(\frac{n+4-3}{3.\left(n+4\right)}=\frac{224}{673}\Leftrightarrow\frac{n}{3.\left(n+4\right)}=\frac{224}{673}\)
<=> 673n = 224.3(n+4)
<=> 673n = 224.3.n + 224.3.4
<=> 673n = 672n + 2688
<=> 673n - 672n = 2688
<=> n = 2688
\(\frac{n}{6}=\frac{1}{2}-\frac{1}{m}=\frac{m-2}{2m}\)
\(\Rightarrow6.\frac{n}{6}=6.\left(\frac{m-2}{2m}\right)\)
\(\Rightarrow n=6.\frac{m-2}{2m}\)\(=\frac{3m-6}{m}\)
\(\Rightarrow n=3-\frac{6}{m}\)
Để m ; n \(\in\) Z thì m là Ư ( 6 ) = { -1 ; 1 ; -2 ; 2 ; -3 ; 3 ; -6 ; 6 } => n = ( 9 ; -3 ; 6 ; 0 ; 5 ; 1 ; 4 ; 2 )
Để A nguyên thì n-2\(⋮\)n+1.
Ta có:n-2=n+1-1-2=(n+1)-3
Vì (n+1)\(⋮\)(n+1)\(\Rightarrow\)3\(⋮\)n+1\(\Rightarrow\)n+1\(\in\) Ư(3)={\(\pm\)1,\(\pm\)3}
\(\Leftrightarrow\left[{}\begin{matrix}n+1=1\\n+1=-1\\n+1=3\\n+1=-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}n=1-1\\n=-1-1\\n=3-1\\n=-3-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}n=0\\n=-2\\n=2\\n=-4\end{matrix}\right.\)
a)\(\frac{1}{n}-\frac{1}{n+1}=\frac{n+1}{n\left(n+1\right)}-\frac{n}{n\left(n-1\right)}=\frac{n+1-n}{n\left(n+1\right)}=\frac{1}{n\left(n+1\right)}\)
\(\Rightarrow\frac{1}{n\left(n+1\right)}=\frac{1}{n}.\frac{1}{n+1}\)
b) \(C=\frac{1}{2}.\frac{1}{3}+\frac{1}{3}.\frac{1}{4}+\frac{1}{4}.\frac{1}{5}+\frac{1}{5}.\frac{1}{6}+\frac{1}{6}.\frac{1}{7}+\frac{1}{7}.\frac{1}{8}\)
\(=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\)
\(=\frac{1}{2}+0+0+0+0+0-\frac{1}{8}\)
\(=\frac{1}{2}-\frac{1}{8}=\frac{4}{8}-\frac{1}{8}=\frac{4-1}{8}=\frac{3}{8}\)
Để M thuộc Z thì x + 1 chia hết cho 3
=> \(x=3.k+2\left(k\in Z\right)\)
Vậy với \(x=3.k+2\left(k\in Z\right)\)thì \(M=\frac{x+1}{3}\in Z\)
(x+1) / 3 thuộc Z
=> x+1 chia hết cho 3
=> x+1=3k ( k E Z )
x=3k-1
Với x=3k-1 thì (x+1) / 3 thuộc Z
\(\frac{x+1}{x-1}=\frac{x-1+2}{x-1}=1+\frac{2}{x-1}\)
\(\Rightarrow x-1\inƯ\left(2\right)\)
\(\Rightarrow x-1=\left\{-1;1-2;2\right\}\)
\(\Rightarrow x-1=-1\Rightarrow x=0\)
...........
Tự thay nha
Để \(M\in Z\)thì x + 1 chia hết cho x - 1
=> x - 1 + 2 chia hết cho x - 1
Do x - 1 chia hết cho x - 1 => 2 chia hết cho x - 1
=> \(x-1\in\left\{1;-1;2;-2\right\}\)
=> \(x\in\left\{2;0;3;-1\right\}\)
\(\frac{1}{m}+\frac{n}{6}=\frac{1}{2}\)
\(\Leftrightarrow\frac{6+mn}{6m}=\frac{1}{2}\)
\(\Rightarrow12+2mn-6m=0\)
\(\Leftrightarrow m\left(n-3\right)=-6\)
Do \(m,n\inℤ\)nên \(m,n-3\)là các ước của \(-6\).
Ta có bảng giá trị: