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20 tháng 8 2017

T a   c ó : x 2 + y 2 = a 2 + b 2 ⇔ x 2 - a 2 = b 2 - y 2 ⇔ x - a x + a = b - y b + y M à   x + y   = a + b ⇔ x - a = b - y   n ê n   t a   c ó x - a x + a = x - a b + y ⇔ x - a x + a - x - a b + y = 0 ⇔ x - a x + a - b - y = 0 ⇔ x - a = 0 x + a - b - y = 0 ⇔ x = a x - y = b - a

+) Với x = a thay vào x + y = a + b ta có: a + y = a + b

Suy ra y = b

Do đó:   x n + y n = a n + b n

+) Với x - y = b - a suy ra x = b - a + y thay vào x + y = a + b ta có:

 b - a + y  + y = a + b

2y = 2a

y = a

Suy ra x - a = b - a hay x = b

Do đó:  x n + y n = b n + a n = a n + b n

Vậy  x n + y n = a n + b n

Đáp án cần chọn là C

 

b: \(=3x^{n-2+n+2}-3x^{n-2}y^{n+2}+3x^{n-2}y^{n+2}-y^{n+2+n-2}\)

\(=3x^{2n}-y^{2n}\)

c: \(=a^3+ab^2+ac^2-a^2b-abc-a^2c+a^2b+b^3+bc^2-ab^2-b^2c-abc+a^2c+b^2c+c^3-abc-bc^2-ca^2\)

\(=a^3+b^3+c^3-3acb\)

a: \(4x^2\left(3x^{n+1}-2x^n\right)\)

\(=4x^2\cdot3x^{n+1}-4x^2\cdot2x^n\)

\(=12x^{n+3}-8x^{n+2}\)

b: \(2\left(x^{2n}+2x^ny^n+y^{2n}\right)-y^n\left(4x^n+2y^n\right)\)

\(=2x^{2n}+4x^ny^n+2y^{2n}-4x^ny^n-2y^{2n}\)

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

c: \(=\left(x^{3n}-y^{3n}\right)\left(x^{3n}+y^{3n}\right)\)

\(=x^{6n}-y^{6n}\)

d: \(=4^n\cdot4-3\cdot4^n=4^n\)

29 tháng 5 2022

a: 4x2(3xn+1−2xn)4x2(3xn+1−2xn)

=4x2⋅3xn+1−4x2⋅2xn=4x2⋅3xn+1−4x2⋅2xn

=12xn+3−8xn+2=12xn+3−8xn+2

b: 2(x2n+2xnyn+y2n)−yn(4xn+2yn)2(x2n+2xnyn+y2n)−yn(4xn+2yn)

=2x2n+4xnyn+2y2n−4xnyn−2y2n=2x2n+4xnyn+2y2n−4xnyn−2y2n

=2x2n=2x2n

c: =(x3n−y3n)(x3n+y3n)=(x3n−y3n)(x3n+y3n)

=x6n−y6n=x6n−y6n

d: =4n⋅4−3⋅4n=4n

30 tháng 9 2019

a) x^2yz - x^3y^3z + xyz^2

=xyz(x-x2y2+z)

b) 4x^3 + 24x^2 - 12xy^2

= 4(x3+6x2-3xy2)

c) x^2( m + n) - 3y^2 (m + n)

=(m+5)(x2-3y2)

d) 4x^2(x - y) + 9y^2( y - x)

= 4x2(x-y)-9y2(x-y)

= (x-y)(4x2-9y2)

e) x^2(a - b) + 2( b - a)

= x2(a-b)-2(a-b)

= (a-b)(x2-2)

g) 50x^2(x - y)^2 - 8y^2(y - x)^2

= 50x2(x-y)2+8y2(x-y)2

= 2(x-y)2(25x2+4y2)

f)10x^2( a - 2b)^2 - (x^2 + 2)(2b - a)^2

= 10x2(a-2b)2+(x2-2)(a-2b)2

= (a-2b)2(10+x2-2)

= (a-2b)2(8+x2)

h) 15am+nb - 45amb( m thuộc N*)

= 15am.15anb - 45amb

= 15amb(15an-3)

13 tháng 10 2019

Thanks^^

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 !

7 tháng 10 2017

Bài 1:

Ta có:

\(a+b+c=0\\ \Leftrightarrow a^3+b^3+c^3+3\left(a^2b+a^2c+b^2a+b^2c+c^2a+c^2b+2abc\right)=0\\ \Leftrightarrow a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\\ \Leftrightarrow a^3+b^3+c^3-3abc=0\\ \Leftrightarrow a^3+b^3+c^3=3abc\left(dpcm\right)\)

Bài 1:

a) Ta có: \(a^2-b^2-2a+2b\)

\(=\left(a-b\right)\left(a+b\right)-2\left(a-b\right)\)

\(=\left(a-b\right)\left(a+b-2\right)\)

b) Ta có: \(3x-3y-5x\left(y-x\right)\)

\(=3\left(x-y\right)+5x\left(x-y\right)\)

\(=\left(x-y\right)\left(3+5x\right)\)

c) Ta có: \(\left(x-y+4\right)^2-\left(2x+3y-1\right)^2\)

\(=\left(x-y+4-2x-3y+1\right)\left(x-y+4+2x+3y-1\right)\)

\(=\left(-x-2y+5\right)\left(3x+2y+3\right)\)

d) Ta có: \(16-x^2+4xy-4y^2\)

\(=16-\left(x^2-4xy+4y^2\right)\)

\(=16-\left(x-2y\right)^2\)

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

e) Ta có: \(\left(x+3\right)^3+\left(x-3\right)^3\)

\(=\left(x+3+x-3\right)\left[\left(x+3\right)^2-\left(x+3\right)\left(x-3\right)+\left(x-3\right)^2\right]\)

\(=2x\cdot\left(x^2+6x+9-x^2+9+x^2-6x+9\right)\)

\(=2x\cdot\left(x^2+27\right)\)

f) Ta có: \(x^4+x^3+2x^2+x+1\)

\(=\left(x^4+2x^2+1\right)+\left(x^3+x\right)\)

\(=\left(x^2+1\right)^2+x\left(x^2+1\right)\)

\(=\left(x^2+1\right)\left(x^2+1+x\right)\)

g) Ta có: \(9x^2-3xy+y-6x+1\)

\(=\left(9x^2-6x+1\right)-\left(3xy-y\right)\)

\(=\left(3x-1\right)^2-y\left(3x-1\right)\)

\(=\left(3x-1\right)\left(3x-1-y\right)\)

h) Ta có: \(x^3-4x^2+12x-27\)

\(=\left(x^3-27\right)-4x\left(x-3\right)\)

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

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

\(=\left(x-3\right)\left(x^2-x+9\right)\)

Bài 2:

Ta có: \(x^3+x^2z+y^2z-xyz+y^3\)

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

\(=\left(x+y\right)\left(x^2-xy+y^2\right)+z\left(x^2-xy+y^2\right)\)

\(=\left(x^2-xy+y^2\right)\left(x+y+z\right)\)

\(=0\cdot\left(x^2-xy+y^2\right)=0\)(đpcm)