CMR \(A=\dfrac{n^3}{3}+\dfrac{n^5}{5}+\dfrac{7n}{15}\)ϵ Z với nϵZ
Những câu hỏi liên quan
Bài 1: CMR với n ϵ Z các phân số sau tối giản
a) \(\dfrac{n}{2n+1}\)
b) \(\dfrac{n+5}{n+6}\)
c) \(\dfrac{n+1}{2n+3}\)
d) \(\dfrac{3n+2}{5n+3}\)
e)\(\dfrac{1}{7n+1}\)
Các bạn giải chi tiết cho mình nhé. Thanks all !
CMR nếu n ∈ Z thì \(\dfrac{n^5}{5}+\dfrac{n^3}{3}+\dfrac{7n}{15}\) là số nguyên
a)cmr:
\(\dfrac{n^5}{5}=\dfrac{n^3}{3}=\dfrac{7n}{15}\) là số nguyên với mọi n \(\in Z\)
b)cmr:với n chẵn thì \(\dfrac{n}{12}+\dfrac{n^2}{8}+\dfrac{n^3}{24}\) là số nguyên
\(\frac{a^5}{5}+\frac{a^3}{3}+\frac{7a}{15}\left(n\Rightarrow a\text{ }nha\right)=\frac{a^5}{5}+\frac{a^3}{3}+\frac{7a}{15}=\frac{a^5}{5}+\frac{a^3}{3}+\frac{15a-5a-3a}{15}=\frac{a^5-a}{5}+\frac{a^3-a}{3}+\frac{15a}{15}=\frac{a^5-a}{5}+\frac{a^3-a}{3}+a;a^k-a⋮k\left(a\in Z;1< k\in N\right)\left(fecmat\right)\Rightarrow\left\{{}\begin{matrix}a^5-a⋮5\\a^3-a⋮3\end{matrix}\right.\Rightarrow dpcm\)
Đúng 0
Bình luận (0)
\(\frac{a}{12}+\frac{a^2}{8}+\frac{a^3}{24}\left(n\Rightarrow a\text{ nha}\right)=\frac{a^3+3a^2+2a}{24}=\frac{\left(a+2\right)\left(a+1\right)a}{24}.a=2k\left(k\in N\right)\Rightarrow;\frac{a\left(a+1\right)\left(a+2\right)}{24}=\frac{2k.\left(2k+1\right)\left(2k+2\right)}{24}=\frac{k\left(k+1\right)\left(2k+1\right)}{6}\Leftrightarrow k\left(k+1\right)\left(2k+1\right)⋮6\)
Đúng 1
Bình luận (0)
chứng minh: \(\dfrac{n^5}{5}+\dfrac{n^3}{3}+\dfrac{7n}{15}\) \(\in Z\)với \(\forall n\in Z\)
chứng minh: \(\dfrac{n^5}{5}+\dfrac{n^3}{3}+\dfrac{7n}{15}\) \(\in Z\)với \(\forall n\in Z\)
Cho A = \(\left(\dfrac{x-5\sqrt{x}}{x-25}-1\right):\left(\dfrac{25-x}{x+2\sqrt{x}-15}-\dfrac{\sqrt{x}+3}{\sqrt{x}+5}+\dfrac{\sqrt{x}-5}{\sqrt{x}-3}\right)\)
a ) Rút gọn A
b) Tìm x ϵ Z để A ϵ Z
a) Ta có: \(A=\left(\dfrac{x-5\sqrt{x}}{x-25}-1\right):\left(\dfrac{25-x}{x+2\sqrt{x}-15}-\dfrac{\sqrt{x}+3}{\sqrt{x}+5}+\dfrac{\sqrt{x}-5}{\sqrt{x}-3}\right)\)
\(=\left(\dfrac{\sqrt{x}\left(\sqrt{x}-5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}-1\right):\left(\dfrac{25-x}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}-\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}+\dfrac{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\right)\)
\(=\left(\dfrac{\sqrt{x}}{\sqrt{x}+5}-1\right):\left(\dfrac{25-x-\left(x-9\right)+x-25}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\right)\)
\(=\left(\dfrac{\sqrt{x}}{\sqrt{x}+5}-\dfrac{\sqrt{x}+5}{\sqrt{x}+5}\right):\left(\dfrac{25-x-x+9+x-25}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\right)\)
\(=\dfrac{\sqrt{x}-\sqrt{x}-5}{\sqrt{x}+5}:\dfrac{x+9}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{-5}{\sqrt{x}+5}\cdot\dfrac{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}{x+9}\)
\(=\dfrac{-5\left(\sqrt{x}-3\right)}{x+9}\)
Đúng 1
Bình luận (0)
Cho A=[\(\dfrac{n}{2}\)]+[\(\dfrac{n+1}{2}\)]
B=[\(\dfrac{n}{3}\)]+[\(\dfrac{n+1}{3}\)]+[\(\dfrac{n+2}{3}\)]
Với giá trị nào của nϵZ thì
A⋮2
B⋮3
25 tháng 7 2021 lúc 10:20
câu 1 : tìm a,b ϵ Z biết : \(\dfrac{a}{3}=\dfrac{b}{2}=\dfrac{c}{5}\) và a - b + 2c = 77
câu 2 : (x\(^n\))\(^m\) = ?
Câu 1
Ta có: \(\dfrac{a}{3}=\dfrac{b}{2}=\dfrac{2c}{10}\) và a-b+2c=77
\(\dfrac{a-b+2c}{3-2+10}=\dfrac{77}{11}=7\)
\(\dfrac{a}{3}=7\) ⇒ a=21
\(\dfrac{b}{2}=7\) ⇒ b=14
\(\dfrac{c}{5}=7\) ⇒ c=35
Đúng 0
Bình luận (0)
Xem thêm câu trả lời
Tìm n ϵ Z sao cho n là số nguyên
\(\dfrac{2n-1}{n-1};\dfrac{3n+5}{n+1};\dfrac{4n-2}{n+3};\dfrac{6n-4}{3n+4};\dfrac{n+3}{2n-1};\dfrac{6n-4}{3n-2};\dfrac{2n+3}{3n-1};\dfrac{4n+3}{3n+2}\)