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ĐỂ x4 - x3 + 6x2 -x \(⋮x^2-x+5\)
\(\Rightarrow x-5=0\Rightarrow x=5\)
b , ta có : \(3x^3+10x^2-5⋮3x+1\)
\(\Rightarrow3x^3+x^2+9x^2+3x-3x-1-4⋮3x+1\)
\(\Rightarrow x\left(3x+1\right)+3x\left(3x+1\right)-\left(3x+1\right)-4⋮3x+1\)
mà : \(\left(3x+1\right)\left(4x-1\right)⋮3x+1\)
\(\Rightarrow4⋮3x+1\Rightarrow3x+1\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)
Nếu : 3x + 1 = 1 => x = 0 ( TM )
3x + 1 = -1 => x = -2/3 ( loại )
3x + 1 = 2 => x = 1/3 ( loại )
3x + 1 = -2 => x = -1 ( TM )
3x + 1 = 4 => x = 1 ( TM )
3x + 1 = -1 => x = -5/3 ( loại )
\(\Rightarrow x\in\left\{0;\pm1\right\}\)
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1: \(\Leftrightarrow3n^3+n^2+9n^2+3n-3n-1-4⋮3n+1\)
\(\Leftrightarrow3n+1\in\left\{1;4;2;-2;-1;-4\right\}\)
\(\Leftrightarrow3n\in\left\{0;3;-3\right\}\)
hay \(n\in\left\{0;1;-1\right\}\)
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\(n^5+1 ⋮n^3+1\)
\(\Rightarrow n^2\left(n^3+1\right)-\left(n^2-1\right)⋮n^3+1\)
\(\Rightarrow\left(n^2-1\right)⋮\left(n+1\right)\left(n^2-n+1\right)\)
\(\Rightarrow\left(n+1\right)\left(n-1\right)⋮\left(n+1\right)\left(n^2-n+1\right)\)
\(\Rightarrow\left(n-1\right)⋮\left(n^2-n+1\right)\)
\(\Rightarrow n\left(n-1\right)-\left(n^2-n+1\right)⋮n^2-n+1\)
\(\Rightarrow-1⋮n^2-n+1\)
Trường hợp 1:
\(n^2-n+1=1\Rightarrow n\left(n-1\right)=0\Rightarrow n=0;n=1\)
Trường hợp 2:
\(n^2-n+1=-1\left(a\right)\)
Vì \(n^2-n+1=n^2-n+\dfrac{1}{4}-\dfrac{1}{4}+1=\left(n-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
\(\Rightarrow\left(a\right)\) vô lý
Vậy \(n=0;n=1\)
n5 + 1 = n5 + n2 - n2 + 1 = n2(n3 + 1) - ( n2 - 1)
=> n2 - 1 = 0
=> n = 1 hoặc -1