2 x 3 + x 2 − x − 2 x 2 − 2 \dfrac{2x^3+x^2-x-2}{x^2-2} x 2 − 2 2 x 3 + x 2 − x − 2
= 2 x ( x 2 − 2 ) + ( x 2 − 2 ) + 3 x x 2 − 2 =\dfrac{2x(x^2-2)+(x^2-2)+3x}{x^2-2} = x 2 − 2 2 x ( x 2 − 2 ) + ( x 2 − 2 ) + 3 x
= 2 x + 1 + 3 x x 2 − 2 =2x+1+\dfrac{3x}{x^2-2} = 2 x + 1 + x 2 − 2 3 x Given n ∈ Z + n\in \Z^+ n ∈ Z +
n = 1 , 2 ( 1 ) + 1 + 3 ( 1 ) ( 1 ) 2 − 2 = 0 , 0 ∈ Z n=1, 2(1)+1+\dfrac{3(1)}{(1)^2-2}=0, 0\in \Z n = 1 , 2 ( 1 ) + 1 + ( 1 ) 2 − 2 3 ( 1 ) = 0 , 0 ∈ Z
n ≥ 2 , 3 x x 2 − 2 ≥ 1 n\ge 2, \dfrac{3x}{x^2-2}\ge1 n ≥ 2 , x 2 − 2 3 x ≥ 1
x 2 − 3 x − 2 x 2 − 2 ≤ 0 \dfrac{x^2-3x-2}{x^2-2}\le0 x 2 − 2 x 2 − 3 x − 2 ≤ 0
x 2 − 3 x − 2 x 2 − 2 ≤ 0 \dfrac{x^2-3x-2}{x^2-2}\le0 x 2 − 2 x 2 − 3 x − 2 ≤ 0
x 2 − 3 x − 2 = 0 x^2-3x-2=0 x 2 − 3 x − 2 = 0
x 1 = 3 − 17 2 , x 2 = 3 + 17 2 x_1=\dfrac{3-\sqrt{17}}{2}, x_2=\dfrac{3+\sqrt{17}}{2} x 1 = 2 3 − 17 , x 2 = 2 3 + 17
x ∈ ( − 2 , 3 − 17 2 ] ∪ ( 2 , 3 + 17 2 ] x\in(-\sqrt{2}, \dfrac{3-\sqrt{17}}{2}]\cup (\sqrt{2}, \dfrac{3+\sqrt{17}}{2}] x ∈ ( − 2 , 2 3 − 17 ] ∪ ( 2 , 2 3 + 17 ] Since n n n is positive integer, we consider n = 2 , 3. n=2,3. n = 2 , 3.
n = 2 , 3 n n 2 − 2 = 3 ( 2 ) ( 2 ) 2 − 2 = 3 , i n t e g e r n=2,\dfrac{3n}{n^2-2}=\dfrac{3(2)}{(2)^2-2}=3, integer n = 2 , n 2 − 2 3 n = ( 2 ) 2 − 2 3 ( 2 ) = 3 , in t e g er
n = 3 , 3 n n 2 − 2 = 3 ( 3 ) ( 3 ) 2 − 2 = 9 7 , i s n o t i n t e g e r n=3,\dfrac{3n}{n^2-2}=\dfrac{3(3)}{(3)^2-2}=\dfrac{9}{7}, is\ not \ integer n = 3 , n 2 − 2 3 n = ( 3 ) 2 − 2 3 ( 3 ) = 7 9 , i s n o t in t e g er
n = 1 , n = 3. n=1,n=3. n = 1 , n = 3.
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