Tìm \(n\in N\)biết: \(x^3y^4+2x^3y^4+3x^3y^4+...+nx^3y^4=820x^3y^4\)
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Tìm \(n\in N\) biết: \(x^3y^4+2x^3y^4+3x^3y^4+...+nx^3y^4=820x^3y^4\)
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
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Tìm n thuộc N biết :
a) \(\left(7x^2y^3\right).\left(x^ny^5\right)=7x^3y^8\)
b) \(x^3y^4+2x^3y^4+3x^3y^4+...+nx^3y^4=820x^3y^4\)
c)
tìm n thuộc N biết:x^3y^4+2x^3y^4+3x^3y^4+.........+ n.x^3y^4=820x^3y^4
tìm n thuộc N biết
x^3y+ 2x^3y+ 3x^3y+....+nx^3y= 20100x^3y
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
=> \(\orbr{\begin{cases}n+201=0\\n-200=0\end{cases}}\)
=> \(\orbr{\begin{cases}n=-201\left(ktm\right)\\n=200\left(tm\right)\end{cases}}\)
a)(-6x^3y^4+4x^4y^3):2x^3y^3. b)(5x^4y^2-x^3y^2):x^3y^2. c)(27x^3y^5+9x^2y^4-6x^3y^3):(-3x^2y^3)
a: \(\dfrac{-6x^3y^4+4x^4y^3}{2x^3y^3}\)
\(=\dfrac{-6x^3y^4}{2x^3y^3}+\dfrac{4x^4y^3}{2x^3y^3}\)
\(=-3y+2x\)
b: \(\dfrac{5x^4y^2-x^3y^2}{x^3y^2}=\dfrac{5x^4y^2}{x^3y^2}-\dfrac{x^3y^2}{x^3y^2}\)
\(=5x-1\)
c: \(\dfrac{27x^3y^5+9x^2y^4-6x^3y^3}{-3x^2y^3}\)
\(=-\dfrac{27x^3y^5}{3x^2y^3}-\dfrac{9x^2y^4}{3x^2y^3}+\dfrac{6x^3y^3}{3x^2y^3}\)
\(=-9xy^2-3y+2x\)
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a) \(\dfrac{-6x^3y^4+4x^4y^3}{2x^3y^3}\)
\(=\dfrac{2x^3y^3\cdot\left(-3y+2x\right)}{2x^3y^3}\)
\(=-3y+2x\)
\(=2x-3y\)
b) \(\dfrac{5x^4y^2-x^3y^2}{x^3y^2}\)
\(=\dfrac{5x\cdot x^3y^2-x^3y^2\cdot1}{x^3y^2}\)
\(=\dfrac{x^3y^2\cdot\left(5x-1\right)}{x^3y^2}\)
\(=5x-1\)
c) \(\dfrac{27x^3y^5+9x^2y^4-6x^3y^3}{-3x^2y^3}\)
\(=\dfrac{-3x^2y^3\cdot-9xy^2+-3x^2y^3\cdot-3y+-3x^2y^3\cdot2x}{-3x^2y^3}\)
\(=\dfrac{-3x^2y^3\cdot\left(-9xy^2-3y+2x\right)}{-3x^2y^3}\)
\(=-9xy^2-3x+2x\)
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giải các hệ phương trình
9x-6y=4 và 3(4x-3y)=-3x+y+7
3(x+1)+2y=-x và 5(x+y)=-3x+y-5
2(2x+3y)=3(2x-3y)+10 và 4x-3y=4(6y-2x)+3
Bài 1: Thực hiện phép tính
1, (3y +1/3y^4)^2
2, (-3x^2 -1/2x)^2
3, (x^2 +2x -3)^2
4, 3 (x+3) (x-3) - (x-9)^2
5, (x^n +x^n:1)^2
6, (5x-3y)^2 - (5x +3y)^2
Tìm x biết :
a) ( 2x - 5 | + | 3y + 1 | = 0
b) | 3x - 4 | + | 3y - 5 | = 0
c) | 2x - 5 | + | xy - 3y + 2 | = 0
\(a\text{) }\left|2x-5\right|+\left|3y+1\right|=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left|2x-5\right|=0\\\left|3y+1\right|=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}2x-5=0\\3y+1=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}2x=5\\3y=-1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{5}{2}\\y=-\dfrac{1}{3}\end{matrix}\right.\)
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 \(\left\{{}\begin{matrix}x=\dfrac{5}{2}\\\left(-\dfrac{1}{2}\right)y=-2\end{matrix}\right.\)
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C=3x^2y-2xy^2+x^3y^3+3xy^2-2^2y-2x^3y^3
D=15x^2y^3+7y^2-8x^3y^2-12x^2+11x^3y^2-12x^2y^3
E=3x^5+1/3xy^4+3/4x^2y^3-1/2x^5y+2xy^4-x^2y^3
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