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Ta có N = 2 x n ( 3 x n + 2 – 1 ) – 3 x n + 2 ( 2 x n – 1 )
N = 2 x n ( 3 x n + 2 – 1 ) – 3 x n + 2 ( 2 x n – 1 )
= 2 x n .3 x n + 2 − 2 x n .1 − 3 x n + 2 .2 x n − 3 x n + 2 . − 1
= 6 x n + n + 2 – 2 x n – 6 . x n + 2 + n + 3 x n + 2 = 6 x 2 n + 2 – 6 x 2 n + 2 – 2 x n + 3 x n + 2 = – 2 x n + 3 x n + 2
Vậy N = – 2 x n + 3 x n + 2
Đáp án cần chọn là: C
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\(\dfrac{3x^{n+1}y^2-2x^5y^n+x^4y^2}{2x^4y^{n-2}}=\dfrac{3}{4}x^{n+1-4}\cdot y^{2-n+2}-x^{5-4}\cdot y^{n-n+2}+\dfrac{1}{2}x^{4-4}\cdot y^{2-n+2}\)
\(=\dfrac{3}{4}x^{n-3}y^{4-n}-xy^2+\dfrac{1}{2}y^{4-n}\)
Để đây là phép chia hết thì n-3>=0 và 4-n>=0
=>3<=n<=4
=>n=3;n=4
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\(=3^{2n}-3x^{n-2}y^{n+2}+3x^{n-2}y^{n-2}-y^{2n}\)
\(=3x^{2n}-y^{2n}\)
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x(x – y) + y(x – y)
= x.x – x.y + y.x – y.y
= x2 – xy + xy – y2
= x2 – y2 + (xy – xy)
= x2 – y2
\(3x^{n-2+n-2}-3x^{n-2}y^{n+2}+3x^{n-2}y^{n+2}-y^{n+2+n-2}\)
\(3x^{2n}-y^{2n}\)
Ta có:
\(3x^{n-2}.\left(x^{n-2}-y^{n+2}\right)+y^{n+2}\left(3x^{n-2}-y^{n-2}\right)\)
\(\Rightarrow3x^{n-2+n-2}-3x^{n-2}y^{n+2}+3x^{n-2}y^{n+2}-y^{n+2+n-2}\)
\(\Rightarrow3x^{2n}-y^{2n}\)