CMR \(A=\dfrac{n^3}{3}+\dfrac{n^5}{5}+\dfrac{7n}{15}\)ϵ Z với nϵZ
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
GH
1
S
28 tháng 7 2019
\(\frac{a^5}{5}+\frac{a^3}{3}+\frac{7a}{15}\left(n\Rightarrow a\text{ }nha\right)=\frac{a^5}{5}+\frac{a^3}{3}+\frac{7a}{15}=\frac{a^5}{5}+\frac{a^3}{3}+\frac{15a-5a-3a}{15}=\frac{a^5-a}{5}+\frac{a^3-a}{3}+\frac{15a}{15}=\frac{a^5-a}{5}+\frac{a^3-a}{3}+a;a^k-a⋮k\left(a\in Z;1< k\in N\right)\left(fecmat\right)\Rightarrow\left\{{}\begin{matrix}a^5-a⋮5\\a^3-a⋮3\end{matrix}\right.\Rightarrow dpcm\)
S
28 tháng 7 2019
\(\frac{a}{12}+\frac{a^2}{8}+\frac{a^3}{24}\left(n\Rightarrow a\text{ nha}\right)=\frac{a^3+3a^2+2a}{24}=\frac{\left(a+2\right)\left(a+1\right)a}{24}.a=2k\left(k\in N\right)\Rightarrow;\frac{a\left(a+1\right)\left(a+2\right)}{24}=\frac{2k.\left(2k+1\right)\left(2k+2\right)}{24}=\frac{k\left(k+1\right)\left(2k+1\right)}{6}\Leftrightarrow k\left(k+1\right)\left(2k+1\right)⋮6\)
17 tháng 2 2021
a) Ta có: \(A=\left(\dfrac{x-5\sqrt{x}}{x-25}-1\right):\left(\dfrac{25-x}{x+2\sqrt{x}-15}-\dfrac{\sqrt{x}+3}{\sqrt{x}+5}+\dfrac{\sqrt{x}-5}{\sqrt{x}-3}\right)\)
\(=\left(\dfrac{\sqrt{x}\left(\sqrt{x}-5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}-1\right):\left(\dfrac{25-x}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}-\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}+\dfrac{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\right)\)
\(=\left(\dfrac{\sqrt{x}}{\sqrt{x}+5}-1\right):\left(\dfrac{25-x-\left(x-9\right)+x-25}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\right)\)
\(=\left(\dfrac{\sqrt{x}}{\sqrt{x}+5}-\dfrac{\sqrt{x}+5}{\sqrt{x}+5}\right):\left(\dfrac{25-x-x+9+x-25}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\right)\)
\(=\dfrac{\sqrt{x}-\sqrt{x}-5}{\sqrt{x}+5}:\dfrac{x+9}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{-5}{\sqrt{x}+5}\cdot\dfrac{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}{x+9}\)
\(=\dfrac{-5\left(\sqrt{x}-3\right)}{x+9}\)
Sam Tiểu Thư bn đăng câu hỏi sai nơi rồi nha