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28 tháng 7 2017
Đặt \(B=\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{n}\)
Đặt \(A=\dfrac{n-1}{1}+\dfrac{n-2}{2}+...+\dfrac{n-\left(n-2\right)}{n-2}+\dfrac{n-\left(n-1\right)}{n-1}\)
\(=\dfrac{n}{1}+\dfrac{n}{2}+...+\dfrac{n}{n-2}+\dfrac{n}{n-1}-1-1-...-1\)
\(=n+\dfrac{n}{2}+\dfrac{n}{3}+...+\dfrac{n}{n-1}-\left(n-1\right)\)
\(=\dfrac{n}{2}+\dfrac{n}{3}+...+\dfrac{n}{n-1}+\dfrac{n}{n}=n\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2}\right)=n.B\)
\(A:B=n\)
NT
25 tháng 12 2017
b.
\(B=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{\left(n-1\right)n\left(n+1\right)}\\ =\dfrac{1}{2}\left(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+....+\dfrac{2}{\left(n-1\right).n.\left(n+1\right)}\right)\\ =\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{\left(n-1\right).n}-\dfrac{1}{n\left(n+1\right)}\right)\\ =\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{n\left(n+1\right)}\right)=\dfrac{1}{4}-\dfrac{1}{2n\left(n+1\right)}\)
HN
26 tháng 3 2017
Ta có: \(1-\dfrac{1}{n^2}=\dfrac{\left(n-1\right)\left(n+1\right)}{n^2}\)
Thế vô bài toán ta được
\(\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)...\left(1-\dfrac{1}{n^2}\right)=\dfrac{1.3}{2.2}.\dfrac{2.4}{3.3}...\dfrac{\left(n-1\right)\left(n+1\right)}{n.n}=\dfrac{1}{2}.\dfrac{n+1}{n}\)
Ta thấy
\(\dfrac{1}{2}.\dfrac{n}{n}< \dfrac{1}{2}.\dfrac{n+1}{n}< \dfrac{1}{2}.\dfrac{n+n}{n}\)
\(\Rightarrow\dfrac{1}{2}< \dfrac{1}{2}.\dfrac{n+1}{n}< 1\)
\(\Rightarrow\)ĐPCM
28 tháng 5 2017
Nguyễn Trần Thành ĐạtXuân Tuấn TrịnhHung nguyenHoang HungQuan Ace Legona giúp với
14 tháng 6 2022
a: \(=\dfrac{x^2+2x+1-x^2+2x-1}{\left(x-1\right)\left(x+1\right)}:\left(\dfrac{1}{x+1}+\dfrac{x}{x-1}+\dfrac{2}{\left(x-1\right)\left(x+1\right)}\right)\)
\(=\dfrac{4x}{\left(x-1\right)\left(x+1\right)}:\dfrac{x-1+x^2+x+2}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{4x}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{\left(x-1\right)\left(x+1\right)}{x^2+2x+1}=\dfrac{4x}{x^2+2x+1}\)
b: \(=\dfrac{x+2}{-\left(x-2\right)}\cdot\dfrac{\left(x-2\right)^2}{4x^2}\cdot\left(\dfrac{2}{2-x}-\dfrac{4}{\left(x+2\right)\left(x^2-2x+4\right)}\cdot\dfrac{x^2-2x+4}{2-x}\right)\)
\(=\dfrac{-\left(x+2\right)\left(x-2\right)}{4x^2}\cdot\left(\dfrac{2}{2-x}-\dfrac{4}{\left(x+2\right)\left(2-x\right)}\right)\)
\(=\dfrac{-\left(x+2\right)\left(x-2\right)}{4x^2}\cdot\dfrac{2x+4-4}{\left(2-x\right)\left(x+2\right)}\)
\(=\dfrac{2x}{4x^2}=\dfrac{1}{2x}\)
14 tháng 12 2018
a) \(\dfrac{x}{x-3}+\dfrac{9-6x}{x^2-3x}=\dfrac{x^2}{x\left(x-3\right)}+\dfrac{9-6x}{x\left(x-3\right)}=\dfrac{x^2-6x+9}{x\left(x-3\right)}=\dfrac{\left(x-3\right)^2}{x\left(x-3\right)}=\dfrac{x-3}{x}\)
AT
17 tháng 7 2017
\(=>P=\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\sqrt{3}\)
CHÚC BẠN HỌC TỐT..........
6 tháng 12 2022
\(=\dfrac{n^3+3n^2+2n}{24}=\dfrac{n\left(n+1\right)\left(n+2\right)}{24}\)
\(=\dfrac{2k\left(2k+1\right)\left(2k+2\right)}{24}=\dfrac{4k\left(2k+1\right)\left(k+1\right)}{24}\)
\(=\dfrac{4k\left(k+1\right)\left(k+2+k-1\right)}{24}\)
\(=\dfrac{4k\left(k+1\right)\left(k+2\right)+4k\left(k+1\right)\left(k-1\right)}{24}=\dfrac{k\left(k+1\right)\left(k+2\right)+k\left(k+1\right)\left(k-1\right)}{6}\)
Vì k;k+1;k+2 là ba số liên tiếp
nen k(k+1)(k+2) chia hết cho 3!=6
k;k+1;k-1 là ba số liên tiếp
nên k(k+1)(k-1) chia hết cho 3!=6
=>A chia hêt cho 6
6 tháng 7 2022
a: \(A=\dfrac{4x\left(2-x\right)+8x^2}{\left(2+x\right)\left(2-x\right)}:\dfrac{x-1-2x+4}{x\left(x-2\right)}\)
\(=\dfrac{8x-4x^2+8x^2}{\left(x+2\right)\cdot\left(-1\right)\cdot\left(x-2\right)}\cdot\dfrac{x\left(x-2\right)}{-x+3}\)
\(=\dfrac{8x+4x^2}{\left(x+2\right)\cdot\left(-1\right)}\cdot\dfrac{x}{-x+3}\)
\(=\dfrac{4x\left(x+2\right)}{\left(x+2\right)\left(x+3\right)}\cdot x=\dfrac{4x^2}{x+3}\)
b: \(=\left(n^2+3n+1+1\right)\left(n^2+3n+1-1\right)\)
\(=\left(n^2+3n+2\right)\left(n^2+3n\right)\)
\(=n\left(n+1\right)\left(n+2\right)\left(n+3\right)⋮4!=24\)
\(N=\dfrac{20125}{20126^2-20125.20127}\)
\(=\dfrac{20125}{20126-\left(20126-1\right)\left(20126+1\right)}\)
\(=\dfrac{20125}{20126^2-\left(20126^2-1\right)}\)
\(=\dfrac{20125}{20126^2-20126^2+1}\)
\(=\dfrac{20125}{1}\)
\(=20125\)