\(a,\)\(2\left(x-y\right)\left(x+y\right)+\left(x-y\right)^2+\left(x+y\right)^2.\)

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

\(b,\)\(\left(2x-3\right)\left(4x^2+6x+9\right)-\left(54+8x\right)\)

\(=8x^2-27-54-8x=8x^2-8x-81\)

\(c,\)\(\left(3x+y\right)\left(9x^2-3xy+y^2\right)-\left(3x-y\right)\left(9x^2+3xy+y^2\right)\)

\(=27x^3+y^3-\left(27x^3-y^3\right)=2y^3\)

\(d,\)\(\left(a+b+c\right)^2-\left(a-c\right)^2-2ab+2bc\)

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

\(=b^2+4bc+4ac\)

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

\(A=x^3+8-x^3+2\)

\(A=10\)

b) \(B=\left(x-1\right)\left(x^2+x+1\right)-\left(x+1\right)\left(x^2-x+1\right)\)

\(B=x^3-1-\left(x^3+1\right)\)

\(B=x^3-1-x^3-1\)

\(B=-2\)

c) \(C=\left(2x-y\right)\left(4x^2+2xy+y^2\right)+\left(y-3x\right)\left(y^2+3xy+9x^2\right)\)

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

\(C=8x^3-y^3+y^3-27x^3\)

\(C=-19x^3\)

a)

\(A=\left(x+2\right)\left(x-2\right)\left(x-2\right)-x^3+2\\ =\left(x^2-4\right)\left(x-2\right)-x^3+2\\ =x^3-2x^2-4x+8-x^3+2\\ =-2x^2-4x+10\)

b)

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

c)

\(C=\left(2x\right)^3-y^3+\left(y\right)^3-\left(3x\right)^3\\ =8x^3-y^3+y^3-27x^3\\ =-19x^3\)

A) 2x²(1-3x)+6x³

=2x²*(1-3x)+2x²*3x

=2x²*(1-3x+3x)

=2x²

B) (x-y)²+(x+y)²+2(x-y)(x+y)

=2(x²-y²)+x²+2xy+y²+x²-2xy+y²

=2x²-2y²+x²+2xy+y²+x²-2xy+y²

=4x²

Tham khảo nha \(\)

1. Rút gọn:

a/ \(\left(x-3\right)\left(x^2+3x+9\right)+\left(54+x^3\right)\)

= \(x^3+3x^2+9x-3x^2-9x-27+54+x^3\)

= \(2x^3+27\)

b/ \(\left(3x+y\right)\left(9x^2-3xy+y^2\right)-\left(3x-y\right)\left(9x^2+3xy+y^2\right)\)

\(=27x^3-9x^2y+3xy^2+9x^2y-3xy^2+y^3-27x^3+9x^2y+3xy^2-9x^2y-3xy^2-y^3\)

\(=\left(27x^3-y^3\right)-\left(27x^3+y^3\right)\)

\(=27x^3-y^3-27x^3-y^3=-2y^3\)

2.Chứng minh rằng:

a/ \(a^3+b^3=\left(a+b\right)^3-3ab\left(a+b\right)\)

Xét VP có:

\(=a^3+3a^2b+3ab^2+b^3-3a^2b-3ab^2\)

\(=a^3+b^3\)

=> VT=VP

=> \(a^3+b^3=\left(a+b\right)^3-3ab\left(a+b\right)\)

b/ \(a^3-b^3=\left(a-b\right)^3+3ab\left(a-b\right)\)

Xét VP có:

\(=a^3-3a^2b+3ab^2-b^3+3a^2b-3ab^2\)

\(=a^3-b^3\)

=> VT=VP

=> \(a^3-b^3=\left(a-b\right)^3+3ab\left(a-b\right)\)

2

a, ta có (a+b)³-3ab(a+b)

= a³+3a²b+3ab²+b³-3a²b-2ab²

= a³+b³(đpcm)

b, ta có (a-b)³ +3ab(a+b)

= a³ -3a²b+3ab²-b³+3a²b -2ab²

= a³-b³ (đpcm)

áp dụng

a³+b³=(a+b)³-3ab(a+b)

=(-5)³-3.6(-5)

=-35

Lời giải:

a. $=(x-y)(x+y)=[(-1)-(-3)][(-1)+(-3)]=2(-4)=-8$
b. $=3x^4-2xy^3+x^3y^2+3x^2y+12xy+15y-12xy-12$

$=3x^4-2xy^3+x^3y^2+3x^2y+15y-12$
=3-2.1(-2)^3+1^3.(-2)^2+3.1^2(-2)+15(-2)-12$
$=-25$
c.

$=2x^4+3x^3y-4x^3y-12xy+12xy=2x^4-x^3y$

$=x^3(2x-y)=(-1)^3[2(-1)-2]=-1.(-4)=4$

d. 

$=2x^2y+4x^2-5xy^2-10x+3xy^2-3x^2y$

$=(2x^2y-3x^2y)+4x^2+(-5xy^2+3xy^2)-10x$

$=-x^2y+4x^2-2xy^2-10x$

$=-3^2.(-2)+4.3^2-2.3(-2)^2-10.3=0$

Ta có 2x(2x + 1)² - 3x(x + 3)(x - 3) - 4x(x + 1)²

= 2x(4x² + 4x + 1) - 3x(x² - 9) - 4x(x² + 2x + 1) 

= 8x³ + 8x² + 2x - 3x³ + 27x - 4x³ - 8x² - 4x

= 8x³ - 3x³ - 4x³ + 8x² - 8x² + 2x + 27x - 4x

= x³ + 25x 

a oi hinh nhu sai r con +16x² nua co a , anh tinh lai ho e duoc kh

Kiều Trinh Vũ ko có + 16x² nhá vì 8x² - 8x² = 0 nhá 

1)a)=>x²+y²+2xy-4(x²-y²-2xy)

=>x²+y²+2xy-4.x²+4y²+8xy

=>-3.x²+5y²+10xy

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

\(=\left(2x+y\right).3y\)

b) \(\left(x+1\right)^3+\left(x-1\right)^3\)

\(=\left(x+1+x-1\right)\left[\left(x+1\right)^2-\left(x+1\right)\left(x-1\right)+\left(x-1\right)^2\right]\)

\(=2x\left[\left(x+1\right)^2-\left(x^2-1\right)+\left(x-1\right)^2\right]\)

c) \(9x^2-3x+2y-4y^2\)

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

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

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

d) \(4x^2-4xy+2x-y+y^2\)

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

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

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

e) \(x^3+3x^2+3x+1-y^3\)

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

\(=\left(x+1-y\right)\left[\left(x+1\right)^2+y\left(x+1\right)+y^2\right]\)

g) \(x^3-2x^2y+xy^2-4x\)

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

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

\(=x\left[\left(x-y\right)^2-4\right]\)

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

a) (x + 2y)² - (x - y)²

= (x + 2y - x + y)(x + 2y + x - y)

= 3y(2x + y)

b) (x + 1)³ + (x - 1)³

= (x + 1 + x - 1)[(x + 1)² - (x + 1)(x - 1) + (x - 1)²]

= 2x(x² + 2x + 1 - x² + 1 + x² - 2x + 1)

= 2x(x² + 3)

c) 9x² - 3x + 2y - 4y²

= (9x² - 4y²) - (3x - 2y)

= (3x - 2y)(3x + 2y) - (3x - 2y)

= (3x - 2y)(3x + 2y - 1)

d) 4x² - 4xy + 2x - y + y²

= (4x² - 4xy + y²) + (2x - y)

= (2x - y)² + (2x - y)

= (2x - y)(2x - y + 1)

e) x³ + 3x² + 3x + 1 - y³

= (x³ + 3x² + 3x + 1) - y³

= (x + 1)³ - y³

= (x + 1 - y)[(x + 1)² + (x + 1)y + y²]

= (x - y + 1)(x² + 2x + 1 + xy + y + y²)

g) x³ - 2x²y + xy² - 4x

= x(x² - 2xy + y² - 4)

= x[(x² - 2xy + y²) - 4]

= x[(x - y)² - 2²]

= x(x - y - 2)(x - y + 2)

bai2 :cmr

a, a^3+b^3=(a+b)^3-3ab.(a+b)

VP= \(\left(a+b\right)^3-3ab\left(a+b\right)\)

=\(a^3+b^3+3a^2b+3ab^2-3a^2b-3ab^2=a^3+b^3\)

=VT

b.a^3-b^3=(a-b)^3+3ab,(a-b)

\(VP=\left(a-b\right)^3+3ab\left(a-b\right)\)

=\(a^3-3a^2b+ab^2.3-b^3+3a^2b-3ab^2=a^3-b^3\)

=VT

=> ĐPCM

bài 1.

a) = 8x^3+4x^2y+2xy^2-4x^2y-2xy^2-y^3-(8x^3-4x^2y+2xy^2+4x^2y-2xy^2+y^3)

= 8x³+4x_²y+2xy_²-4x²y-2xy²-y³ - 8x³+4x²y-2xy²-4x²y+2xy²-y³

=-8x²y-6y³

b) = 27x³-18x²y+12xy²+18x²y-12xy²+8y³-27x³

=8y