1) Tìm n \(\in\)Z :\((2n^3+n-7):(n-2)\)
2) Cho x2 + y2 =7 và xy = 1. Hãy tính x5 + y5
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6: \(-x^2y\left(xy^2-\dfrac{1}{2}xy+\dfrac{3}{4}x^2y^2\right)\)
\(=-x^3y^3+\dfrac{1}{2}x^3y^2-\dfrac{3}{4}x^4y^3\)
7: \(\dfrac{2}{3}x^2y\cdot\left(3xy-x^2+y\right)\)
\(=2x^3y^2-\dfrac{2}{3}x^4y+\dfrac{2}{3}x^2y^2\)
8: \(-\dfrac{1}{2}xy\left(4x^3-5xy+2x\right)\)
\(=-2x^4y+\dfrac{5}{2}x^2y^2-x^2y\)
9: \(2x^2\left(x^2+3x+\dfrac{1}{2}\right)=2x^4+6x^3+x^2\)
10: \(-\dfrac{3}{2}x^4y^2\left(6x^4-\dfrac{10}{9}x^2y^3-y^5\right)\)
\(=-9x^8y^2+\dfrac{5}{3}x^6y^5+\dfrac{3}{2}x^4y^7\)
11: \(\dfrac{2}{3}x^3\left(x+x^2-\dfrac{3}{4}x^5\right)=\dfrac{2}{3}x^3+\dfrac{2}{3}x^5-\dfrac{1}{2}x^8\)
12: \(2xy^2\left(xy+3x^2y-\dfrac{2}{3}xy^3\right)=2x^2y^3+6x^3y^3-\dfrac{4}{3}x^2y^5\)
13: \(3x\left(2x^3-\dfrac{1}{3}x^2-4x\right)=6x^4-x^3-12x^2\)
Bài 1: Ta có 200920 = (20092)10 = (2009.2009)10
2009200910 = (10001.2009)10
Mà 2009 < 10001 ➩ (2009.2009)10 < (10001.2009)10
Vậy 200920 < 2009200910
a) \(a\left(b+1\right)=3\left(a;b\inℤ\right)\)
\(\Rightarrow a;\left(b+1\right)\in U\left(3\right)=\left\{-1;1;-3;3\right\}\)
\(\Rightarrow\left(a;b\right)\in\left\{\left(-1;-4\right);\left(1;2\right);\left(-3;-2\right);\left(3;0\right)\right\}\)
b) \(2n+7⋮n+1\left(n\inℤ\right)\)
\(\Rightarrow2n+7-2\left(n+1\right)⋮n+1\)
\(\Rightarrow2n+7-2n-2⋮n+1\)
\(\Rightarrow5⋮n+1\)
\(\Rightarrow n+1\in U\left(5\right)=\left\{-1;1;-5;5\right\}\)
\(\Rightarrow n\in\left\{-2;0;-6;4\right\}\)
c) \(xy+x-y=6\left(x;y\inℤ\right)\)
\(\Rightarrow x\left(y+1\right)-y-1+1=6\)
\(\Rightarrow x\left(y+1\right)-\left(y+1\right)=5\)
\(\Rightarrow\left(x-1\right)\left(y+1\right)=5\)
\(\Rightarrow\left(x-1\right);\left(y+1\right)\in U\left(5\right)=\left\{-1;1;-5;5\right\}\)
\(\Rightarrow\left(x;y\right)\in\left\{\left(-0;-6\right);\left(2;4\right);\left(-4;-2\right);\left(6;0\right)\right\}\)
10:
Vì n là số lẻ nên n=2k-1
Số số hạng là (2k-1-1):2+1=k(số)
Tổng là (2k-1+1)*k/2=2k*k/2=k^2 là số chính phương
11:
n^3-n^2+2n+7 chia hết cho n^2+1
=>n^3+n-n^2-1+n+8 chia hết cho n^2+1
=>n+8 chia hết cho n^2+1
=>n^2-64 chia hết cho n^2+1
=>n^2+1-65 chia hết cho n^2+1
=>n^2+1 thuộc {1;5;13;65}
=>\(n\in\left\{0;2;-2;2\sqrt{3};-2\sqrt{3};8;-8\right\}\)
Bài 11:
Ta có: \(n^3-n^2+2n+7⋮n^2+1\)
\(\Leftrightarrow n^3+n-n^2-1+n+8⋮n^2+1\)
\(\Leftrightarrow n^2-64⋮n^2+1\)
\(\Leftrightarrow n^2+1\in\left\{1;5;13;65\right\}\)
\(\Leftrightarrow n^2\in\left\{0;4;64\right\}\)
hay \(n\in\left\{0;-2;2;8;-8\right\}\)