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a, (n+3)2-(n-1)2
= n2+6n+9-n2+2n-1
= 8n + 8
= 8(n+1) chia hết cho 8
Tiếp câu b nha
\(A=\frac{n^5}{120}+\frac{n^4}{10}+\frac{7n^3}{24}+\frac{5n^2}{12}+\frac{n}{5}\)
\(=\frac{n^5+10n^4+35n^3+50n^2+24n}{120}\)
Ta có:\(n^5+10n^4+35n^3+50n^2+24n\)
\(=n\left(n^4+10x^3+35x^2+50x+24\right)\)
\(=n\left(n^4+2n^3+8n^3+16n^2+19n^2+38n+12n+4\right)\)
\(=n\left(n+3\right)\left(n^3+3n^2+5n^2+15n+4n+12\right)\)
\(=n\left(n+2\right)\left(n+3\right)\left(n+4n+n+4\right)\)
\(=n\left(n+1\right)\left(n+2\right)\left(n+3\right)\left(n+4\right)⋮3;5;8\)
Mà \(ƯC\left(3;5;8\right)=1\)
\(\Rightarrow n\left(n+1\right)\left(n+2\right)\left(n+3\right)\left(n+4\right)⋮120\)
Vậy A chia hết cho 120
\(\frac{n}{12}+\frac{n^2}{8}+\frac{n^3}{24}=\frac{2n+3n^2+n^3}{24}=\frac{n^3+2n^2+n^2+2n}{24}=\frac{n^2\left(n+2\right)+n\left(n+2\right)}{24}\)
\(=\frac{\left(n^2+n\right)\left(n+2\right)}{24}=\frac{n\left(n+1\right)\left(n+2\right)}{24}\)
Do n chẵn nên n=2k (k nguyên) => n+2=2k+2=2(k+1) => n(n+2)=2k.2(k+1)=4k(k+1)
k(k+1) là 2 số nguyên liên tiếp, trong đó có ít nhất 1 số chẵn nên k(k+1) chia hết cho 2 => 4k(k+1) chia hết cho 8
=>n(n+2) chia hết cho 8=>n(n+1)(n+2) chia hết cho 8 (1)
Mặt khác n;n+1;n+2 là 3 số nguyên liên tiếp nên trong đó có ít nhất 1 số chia hết cho 3 (tự chứng minh hoặc xem cách chứng minh trên mạng nhé)
=>n(n+1)(n+2) chia hết cho 3 (2)
Từ (1) và (2) và (3;8)=1 => n(n+1)(n+2) chia hết cho 3.8=24
=>\(\frac{n\left(n+1\right)\left(n+2\right)}{24}\) nguyên => đpcm
vì bài dài quá nên mình làm từng bài 1 nhé
1. Ta thấy : \(\frac{1}{n^3}< \frac{1}{n^3-n}=\frac{1}{\left(n-1\right)n\left(n+1\right)}=\frac{1}{2}.\frac{\left(n+1\right)-\left(n-1\right)}{\left(n-1\right)n\left(n+1\right)}=\frac{1}{2}.\left[\frac{1}{\left(n-1\right)n}-\frac{1}{n\left(n+1\right)}\right]\)
Do đó :
\(B< \frac{1}{2}.\left[\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{\left(n-1\right)n}-\frac{1}{n\left(n+1\right)}\right]< \frac{1}{2}.\frac{1}{6}=\frac{1}{12}\)
2.
Nhận xét : \(1+\frac{1}{n\left(n+2\right)}=\frac{\left(n+1\right)^2}{n\left(n+2\right)}\)
Do đó :
\(A=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}...\frac{\left(n+1\right)^2}{n\left(n+2\right)}=\frac{2.3...\left(n+1\right)}{1.2...n}.\frac{2.3...\left(n+1\right)}{3.4...\left(n+2\right)}=\frac{n+1}{1}.\frac{2}{n+2}< 2\)
\(\frac{x+7}{3}+\frac{x+5}{4}=\frac{x+3}{5}+\frac{x+1}{6}\)
\(\Rightarrow\frac{x+7}{3}+2+\frac{x+5}{4}+2=\frac{x+3}{5}+2+\frac{x+1}{6}+2\)
\(\Rightarrow\frac{x+13}{3}+\frac{x+13}{4}=\frac{x+13}{5}+\frac{x+13}{6}\)
\(\Rightarrow\frac{x+13}{3}+\frac{x+13}{4}-\frac{x+13}{5}-\frac{x+13}{6}=0\)
\(\Rightarrow\left(x+13\right)\left(\frac{1}{3}+\frac{1}{4}-\frac{1}{5}-\frac{1}{6}\right)=0\)
Vì \(\left(\frac{1}{3}>\frac{1}{4}>\frac{1}{5}>\frac{1}{6}\right)\Rightarrow\)\(\left(\frac{1}{3}+\frac{1}{4}-\frac{1}{5}-\frac{1}{6}\right)>0\)
\(\Rightarrow x+13=0\Leftrightarrow x=-13\)
\(\frac{x+m}{n+p}+\frac{x+n}{p+m}+\frac{x+p}{n+m}+3=0\)
\(\Rightarrow\frac{x+m}{n+p}+1+\frac{x+n}{p+m}+1+\frac{x+p}{n+m}+1=0\)
\(\Rightarrow\frac{x+m+n+p}{n+p}+\frac{x+m+n+p}{p+m}+\frac{x+m+n+p}{n+m}=0\)
\(\Rightarrow\left(x+m+n+p\right)\left(\frac{1}{n+p}+\frac{1}{p+m}+\frac{1}{n+m}\right)=0\)
Vì m,n,p là số dương nên \(\left(\frac{1}{n+p}+\frac{1}{p+m}+\frac{1}{n+m}\right)>0\)
\(\Rightarrow x+m+n+p=0\Rightarrow x=-\left(m+n+p\right)\)
\(\frac{5x+\frac{3x-4}{5}}{15}=\frac{\frac{3-x}{15}+7x}{5}+1-x\)
\(\Rightarrow\frac{\frac{25x+3x-4}{5}}{15}=\frac{\frac{3-x+105x}{15}}{5}+1-x\)
\(\Rightarrow\frac{\frac{28x-4}{5}}{15}=\frac{\frac{3+104x}{15}}{5}+1-x\)
\(\Rightarrow\frac{28x-4}{75}=\frac{3+104x}{75}+1-x\)
\(\Rightarrow\frac{28x-4}{75}=\frac{3+104x+75-75x}{75}\)
\(\Rightarrow\frac{28x-4}{75}=\frac{78+29x}{75}\)
\(\Rightarrow28x-4=78+29x\)
\(\Rightarrow x=-82\)
a; A = \(\dfrac{1}{2^2}\) + \(\dfrac{1}{4^2}\) + \(\dfrac{1}{6^2}\) + ... + \(\dfrac{1}{\left(2n\right)^2}\)
A = \(\dfrac{1}{2^2}\).(\(\dfrac{1}{1^2}\) + \(\dfrac{1}{2^2}\) + \(\dfrac{1}{3^2}\) + ... + \(\dfrac{1}{n^2}\))
A = \(\dfrac{1}{4}\).(\(\dfrac{1}{1}\) + \(\dfrac{1}{2.2}\) + \(\dfrac{1}{3.3}\) + ... + \(\dfrac{1}{n.n}\))
Vì \(\dfrac{1}{2.2}\) < \(\dfrac{1}{1.2}\); \(\dfrac{1}{3.3}\) < \(\dfrac{1}{2.3}\); ...; \(\dfrac{1}{n.n}\) < \(\dfrac{1}{\left(n-1\right)n}\)
nên A < \(\dfrac{1}{4}\).(\(\dfrac{1}{1}\) + \(\dfrac{1}{1.2}\) + \(\dfrac{1}{2.3}\) + ... + \(\dfrac{1}{\left(n-1\right)n}\))
A < \(\dfrac{1}{4.}\)(1 + \(\dfrac{1}{1}\) - \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{n-1}\) - \(\dfrac{1}{n}\))
A < \(\dfrac{1}{4}\).(1 + 1 - \(\dfrac{1}{n}\))
A < \(\dfrac{1}{4}\).(2 - \(\dfrac{1}{n}\))
A < \(\dfrac{1}{2}\) - \(\dfrac{1}{4n}\) < \(\dfrac{1}{2}\) (đpcm)
+ Ta có : \(n^5-n=n\left(n^2-1\right)\left(n^2+1\right)\)
\(=n\left(n-1\right)\left(n+1\right)\left(n^2-4+5\right)\)
\(=\left(n-2\right)\left(n-1\right)n\left(n+1\right)\left(n+2\right)+5\left(n-1\right)n\left(n+1\right)\)
+ \(\left(n-2\right)\left(n-1\right)n\left(n+1\right)\left(n+2\right)\)là tích 5 số nguyên liên tiếp
\(\Rightarrow\left(n-2\right)\left(n-1\right)n\left(n+1\right)\left(n+2\right)⋮5\)
\(\Rightarrow\left(n-2\right)\left(n-1\right)n\left(n+1\right)\left(n+2\right)+5\left(n-1\right)n\left(n+1\right)⋮5\)
\(\Rightarrow n^5-n⋮5\)
+ \(n^3-n=\left(n-1\right)n\left(n+1\right)⋮3\)
\(B=\frac{n^5-n}{5}+\frac{n^3-n}{3}+\frac{7n}{15}+\frac{n}{5}+\frac{n}{3}\)
\(=\frac{n^5-n}{5}+\frac{n^3-n}{3}+\frac{15n}{15}\)
=> B là số nguyên
\(A=\frac{n^5+10n^4+35n^3+50n^2+24n}{120}\) \(=\frac{n\left[n^3\left(n+1\right)+9n^2\left(n+1\right)+26n\left(n+1\right)+24\left(n+1\right)\right]}{120}\)
\(=\frac{n\left(n+1\right)\left[n^3+9n^2+26n+24\right]}{120}\) \(=\frac{n\left(n+1\right)\left[n^2\left(n+2\right)+7n\left(n+2\right)+12\left(n+2\right)\right]}{120}\)
\(=\frac{n\left(n+1\right)\left(n+2\right)\left(n^2+7n+12\right)}{120}\) \(=\frac{n\left(n+1\right)\left(n+2\right)\left(n+3\right)\left(n+4\right)}{120}\)
+ \(n\left(n+1\right)\left(n+2\right)\left(n+3\right)\left(n+4\right)\)là tích 5 số nguyên liên tiếp\
\(\Rightarrow\left\{{}\begin{matrix}n\left(n+1\right)\left(n+2\right)\left(n+3\right)\left(n+4\right)⋮3\\n\left(n+1\right)\left(n+2\right)\left(n+3\right)\left(n+4\right)⋮5\end{matrix}\right.\) (1)
+ trong 5 số nguyên liên tiếp tồn tại ít nhất 2 số chẵn liên tiếp
\(\Rightarrow n\left(n+1\right)\left(n+2\right)\left(n+3\right)\left(n+4\right)⋮8\) ( do tích 2 số chẵn liên tiếp chia hết cho 8 ) (2)
+ Từ (1) và (2) => \(n\left(n+1\right)\left(n+2\right)\left(n+3\right)\left(n+4\right)⋮120\)
=> đpcm
+ \(C=\frac{n^3+3n^2+2n}{24}=\frac{n\left(n+1\right)\left(n+2\right)}{24}\)
+ \(n\left(n+1\right)\left(n+2\right)\) là tích 3 số nguyên liên tiếp
\(\Rightarrow n\left(n+1\right)\left(n+2\right)⋮3\) (3)
+ n và n + 2 là 2 số chẵn liên tiếp
\(\Rightarrow n\left(n+2\right)⋮8\Rightarrow n\left(n+1\right)\left(n+2\right)⋮8\) (4)
+ Từ (3) và (4) \(\Rightarrow n\left(n+1\right)\left(n+2\right)⋮24\)
=> C là số nguyên