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24 tháng 7 2017

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

28 tháng 12 2017

a) \(\dfrac{2x-6}{x^2-x-6}\)

\(=\dfrac{2\left(x-3\right)}{x^2-3x+2x-6}\)

\(=\dfrac{2\left(x-3\right)}{x\left(x-3\right)+2\left(x-3\right)}\)

\(=\dfrac{2\left(x-3\right)}{\left(x-3\right)\left(x+2\right)}\)

\(=\dfrac{2}{x+2}\)

b) \(\dfrac{6x^2-x-2}{4x^2-1}\)

\(=\dfrac{6x^2+3x-4x-2}{\left(2x\right)^2-1^2}\)

\(=\dfrac{3x\left(2x+1\right)-2\left(2x+1\right)}{\left(2x-1\right)\left(2x+1\right)}\)

\(=\dfrac{\left(2x+1\right)\left(3x-2\right)}{\left(2x-1\right)\left(2x+1\right)}\)

\(=\dfrac{3x-2}{2x-1}\)

28 tháng 12 2017

\(c,\dfrac{x^3-x^2+3x-3}{x^3+2x^2+3x+6}\)

\(=\dfrac{x^2\left(x-1\right)+3\left(x-1\right)}{x^2\left(x+2\right)+3\left(x+2\right)}\)

\(=\dfrac{\left(x-1\right)\left(x^2+3\right)}{\left(x+2\right)\left(x^2+3\right)}=\dfrac{x-1}{x+2}\)

d,Sửa đề :

\(\dfrac{a^2-b^2+c^2+2ac}{a^2+b^2-c^2+2ab}\)

\(=\dfrac{\left(a^2+2ac+c^2\right)-b^2}{\left(a^2+2ab+b^2\right)-c^2}\)

\(=\dfrac{\left(a+c\right)^2-b^2}{\left(a+b\right)^2-c^2}\)

\(=\dfrac{\left(a-b+c\right)\left(a+b+c\right)}{\left(a+b-c\right)\left(a+b+c\right)}\)

\(=\dfrac{a-b+c}{a+b-c}\)

e,g Đề ko rõ

25 tháng 5 2019

\(x^{n-2}\left(x^2-1\right)-x\left(x^{n-1}-x^{n-3}\right)\)

\(=x-x^{n-2}-x+x^{n-2}\)

\(=0\)

15 tháng 11 2021

\(a,N=\dfrac{x^2+xy+y^2}{\left(x-y\right)\left(x+y\right)}\cdot\dfrac{\left(x-y\right)\left(x^4-y^4\right)}{\left(x-y\right)\left(x^2+xy+y^2\right)}\\ N=\dfrac{\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)}{\left(x-y\right)\left(x+y\right)}=x^2+y^2\\ b,N=\left(x+y\right)^2-2xy=0-2\cdot1=-2\)

15 tháng 11 2021

ĐKXĐ: \(x\ne y\)

a) \(N=\dfrac{x^2+y\left(x+y\right)}{\left(x-y\right)\left(x+y\right)}:\dfrac{\left(x-y\right)\left(x^2+xy+y^2\right)}{x^4\left(x-y\right)-y^4\left(x-y\right)}=\dfrac{x^2+xy+y^2}{\left(x-y\right)\left(x+y\right)}.\dfrac{\left(x-y\right)^2\left(x+y\right)\left(x^2+y^2\right)}{\left(x-y\right)\left(x^2+xy+y^2\right)}=x^2+y^2\)

b) \(x+y=0\Leftrightarrow\left(x+y\right)^2=0\Leftrightarrow x^2+y^2-2xy=0\)

\(\Leftrightarrow N=x^2+y^2=0+2xy=2.1=2\)

 

15 tháng 11 2021

Sửa lại ĐKXĐ là \(x\ne\pm y\) nha

15 tháng 7 2019

1)\(n^2\left(n-1\right)\left(n+1\right)-\left(n^2+2\right)\left(n^2-2\right)=n^2\left(n^2-1\right)-\left(n^4-4\right)=n^4-n^2-n^4+4\)

\(=-n^2+4\)

2)\(\left(y+3\right)\left(y-3\right)\left(y^2+9\right)-\left(y^2-4\right)\left(y^2+4\right)=\left(y^2-9\right)\left(y^2+9\right)-\left(y^4-16\right)\)

\(=y^4-81-y^4+16=-65\)

3)\(\left(x-2y+3\right)\left(x+2y-3\right)-\left(x-2y\right)\left(x+2y\right)=\left(x+3\right)^2-4y^2-\left(x^2-4y^2\right)\)

\(=x^2+6x+9-4y^2-x^2+4y^2=6x+9\)

4)\(\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2bc+2ac\)

5)\(\left(a+b-c\right)^2=a^2+b^2+c^2+2ab-2bc-2ac\)

6)\(\left(a-b-c\right)^2=a^2+b^2+c^2-2ab+2bc-2ac\)

Học tốt nha bạn !