Tìm \(n\in N\) biết: \(x^3y^4+2x^3y^4+3x^3y^4+...+nx^3y^4=820x^3y^4\)
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MX
2
12 tháng 3 2017
\(x^3y^4+2x^3y^4+3x^3y^4+....+nx^3y^4=820x^3y^4\)
\(\Leftrightarrow x^3y^4\left(1+2+3+....+n\right)=820x^3y^4\)
\(\Leftrightarrow1+2+3+....+n=820\)
\(\Leftrightarrow\frac{n\left(n+1\right)}{2}=820\)
\(\Leftrightarrow n\left(n+1\right)=1640=40.41\)
\(\Rightarrow n=40\)
DT
1
NP
1
4 tháng 1 2020
Ta có: x3y + 2x3y + 3x3y + ... + nx3y = 20100x3y
=> x3y(1 + 2 + 3 + ... + n) = 20100x3y
=> (n + 1)[(n - 1) : 1 + 1] : 2 = 20100
=> (n + 1)n = 40200
=> n2 + n - 40200 = 0
=> n2 + 201n - 200n - 40200 = 0
=> (n + 201)(n - 200) = 0
=> \(\orbr{\begin{cases}n+201=0\\n-200=0\end{cases}}\)
=> \(\orbr{\begin{cases}n=-201\left(ktm\right)\\n=200\left(tm\right)\end{cases}}\)
25 tháng 1 2018
\(a\text{) }\left|2x-5\right|+\left|3y+1\right|=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left|2x-5\right|=0\\\left|3y+1\right|=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}2x-5=0\\3y+1=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}2x=5\\3y=-1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{5}{2}\\y=-\dfrac{1}{3}\end{matrix}\right.\)
Vậy \(\left\{{}\begin{matrix}x=\dfrac{5}{2}\\y=-\dfrac{1}{3}\end{matrix}\right.\)
b) \(\left|3x-4\right|+\left|3y-5\right|=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left|3x-4\right|=0\\\left|3y-5\right|=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}3x-4=0\\3y-5=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}3x=4\\3y=5\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{4}{3}\\y=\dfrac{5}{3}\end{matrix}\right.\)
Vậy \(\left\{{}\begin{matrix}x=\dfrac{4}{3}\\y=\dfrac{5}{3}\end{matrix}\right.\)
c) \(\left|2x-5\right|+\left|xy-3y+2\right|=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left|2x-5\right|=0\\\left|xy-3y+2\right|=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}2x-5=0\\xy-3y+2=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}2x=5\\xy-3y=-2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{5}{2}\\xy-3y=-2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{5}{2}\\\dfrac{5}{2}y-3y=-2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{5}{2}\\\left(\dfrac{5}{2}-3\right)y=-2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{5}{2}\\\left(-\dfrac{1}{2}\right)y=-2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{5}{2}\\y=4\end{matrix}\right.\)
Vậy \(\left\{{}\begin{matrix}x=\dfrac{5}{2}\\\left(-\dfrac{1}{2}\right)y=-2\end{matrix}\right.\)
Đặt \(A=x^3y^4+2x^3y^4+3x^3y^4+...+nx^3y^4\)
\(A=x^3y^4\left(1+2+3+...+n\right)\)
Lại có:\(A=820x^3y^4\)
\(\Rightarrow x^3y^4\left(1+2+3+...+n\right)=820x^3y^4\)
\(\Rightarrow1+2+3+...+n=820\)
\(\Rightarrow\dfrac{\left(n+1\right)n}{2}=820\)
\(\Rightarrow\left(n+1\right)n=1640\)
\(\Rightarrow\left(n+1\right)n=41\cdot40\)(vì \(n\in N\) nên ta không xét trường hợp âm)
\(\Rightarrow n=40\)
Vậy n=40