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13 tháng 9 2017

a) 5ay - 3bx + ax - 15by

= (5ay + ax) - (3bx + 15by)

= a (5y + x) - 3b (x + 5y)

= (5y + x) (a - 3b)

b) x^3 + x^2 - x - 1

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

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

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

c) (2a + b)^2 - (2b + a)^2

= 4a^2 + 4ab + b^2 - 4b^2 - 4ab - a^2

= 3a^2 - 3b^2

= 3 (a^2 - b^2)

d) (8a^3 - 27b^3) - 2a (4a^2 - 9b^2)

= 8a^3 - 27b^3 - 8a^3 + 18ab^2

= 27b^3 + 18ab^2

= 9b^2 (3b + 2a)

24 tháng 9 2020

a) 4a2b3 - 6a3b2 = 2a2b2( 2b - 3a )

b) ( a - b )2 - ( b - a ) = ( a - b )2 + ( a - b ) = ( a - b )( a - b + 1 )

c) ( 8a3 - 27b3 ) - 2a( 4a2 - 9b2 ) = 8a3 - 27b3 - 8a3 + 18ab2 = 18ab2 - 27b3 = 9b2( 2a - 3b )

d) 10x2 + 10xy + 5x + 5y = 10x( x + y ) + 5( x + y ) = ( x + y )( 10x + 5 ) = 5( x + y )( 2x + 1 )

e) 5ay - 3bx + ax - 15by = 5y( a - 3b ) + x( a - 3b ) = ( a - 3b )( 5y + x )

24 tháng 9 2020

a) \(4a^2.b^3-6a^3.b^2=2a^2.b^2\left(2b-3a\right)\)

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

\(=\left(a-b\right).\left(a-b+1\right)\)

c) \(8a^3-27b^3-2a.\left(4a^2-9b^2\right)=8a^3-27b^3-8a^3+18ab^2\)

\(=-27b^3+18ab^2=18ab^2-27b^3=9b^2.\left(2a-3b\right)\)

d) \(10x^2+10xy+5x+5y=5.\left(2x^2+2xy+x+y\right)\)

\(=5.\left[\left(2x^2+2xy\right)+\left(x+y\right)\right]=5.\left[2x\left(x+y\right)+\left(x+y\right)\right]\)

\(=5\left(x+y\right)\left(2y+1\right)\)

e) \(5ay-3bx+ax-15by=\left(5ay-15by\right)-\left(3bx-ax\right)\)

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

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

24 tháng 7 2017

a, \(\left(8a^3-27b^3\right)-2a\left(4a^2-9b^2\right)\)

\(=\left(2a-3b\right)\left[\left(2a\right)^2+2a.3b+\left(3b\right)^2\right]-2a\left(2a-3b\right)\left(2a+3b\right)\)

\(=\left(2a-3b\right)\left[4a^2+6ab+9b^2-2a\left(2a+3b\right)\right]\)

\(=\left(2a-3b\right)\left(4a^2+6ab+9b^2-4a^2-6ab\right)\)

\(=\left(2a-3b\right).9b^2\)

b, \(\left(x^3-y^3\right)+\left(x-y\right)^2\)

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

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

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

c, \(\left(m^3+n^3\right)+\left(m+n\right)^2\)

\(=\left(m+n\right)\left(m^2-mn+n^2\right)+\left(m+n\right)^2\)

\(=\left(m+n\right)\left(m^2-mn+n^2+m+n\right)\)

Chúc bạn học tốt!!!

1. \(4x^2-17xy+13y^2=4x^2-4xy-13xy+13y^2=4x\left(x-y\right)-13y\left(x-y\right)=\left(x-y\right)\left(4x-13y\right)\)

2. \(2x\left(x-5\right)-x\left(3+2x\right)=26\Leftrightarrow2x^2-10x-3x-2x^2=26\Leftrightarrow-13x=26\Leftrightarrow x=-2\)

3. \(A=\left(2a-3b\right)^2+2\left(2a-3b\right)\left(3a-2b\right)+\left(2b-3a\right)^2\)

\(\Leftrightarrow\left(2a-3b\right)^2-2\left(2a-3b\right)\left(2b-3a\right)+\left(2b-3a\right)^2=\left(2a-3b-2b+3a\right)^2=\left(5a-5b\right)^2\)

\(=25\left(a-b\right)^2=25\cdot100=2500\)

28 tháng 9 2017

bạn viết so mu nhu the nao day ?

28 tháng 9 2017

a)x2-2xy+y2+3x-3y-10

=(x2-2xy+y2)+(3x-3y)-10

=(x-y)2+3(x-y)-10

=(x-y).(x-y+3)-10

26 tháng 5 2017

1. (a2+b2+ab)2-a2b2-b2c2-c2a2

=a4+b4+a2b2+2(a2b2+ab3+a3b)-a2b2-b2c2-c2a2

=a4+b4+2a2b2+2ab3+2a3b-b2c2-c2a2

=(a2+b2)2+2ab(a2+b2)-c2(a2+b2)

=(a2+b2)[(a+b)2-c2]

=(a2+b2)(a+b+c)(a+b-c)

2. a4+b4+c4-2a2b2-2b2c2-2a2c2=(a2-b2-c2)2

3. a(b3-c3)+b(c3-a3)+c(a3-b3)

=ab3-ac3+bc3-ba3+ca3-cb3

=a3(c-b)+b3(a-c)+c3(b-a)

=a3(c-b)-b3(c-a)+c3(b-a)

=a3(c-b)-b3(c-b+b-a)+c3(b-a)

=a3(c-b)-b3(c-b)-b3(b-a)+c3(b-a)

=(c-b)(a-b)(a2+ab+b2)-(b-a)(b-c)(b2+bc+c2)

=(a-b)(c-b)(a2+ab+2b2+bc+c2)

4. a6-a4+2a3+2a2=a4(a+1)(a-1)+2a2(a+1)=(a+1)(a5-a4+2a2)=a2(a+1)(a3-a2+2)

5. (a+b)3-(a-b)3=(a+b-a+b)[(a+b)2+(a+b)(a-b)+(a-b)2]

=2b(3a2+b2)

6. x3-3x2+3x-1-y3=(x-1)3-y3=(x-1-y)[(x-1)2+(x-1)y+y2]

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

7. xm+4+xm+3-x-1=xm+3(x+1)-(x+1)=(x+1)(xm+3-1)

(Đúng nhớ like nhá !)

26 tháng 5 2017

Minh Hải,Lê Thiên Anh,Nguyễn Huy Tú,Ace Legona,...giúp mk vs mai mk đi hk rùi

28 tháng 8 2018

a) \(\left(x+1\right)^4+\left(x^2+x+1\right)^2\)

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

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

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

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

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

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

b) \(\left(a+b-2c\right)^3+\left(b+c-2a\right)^3+\left(c+a-2b\right)^3\)

Đặt \(x=a+b-2c\)

\(y=b+c-2a\)

\(z=c+a-2b\)

\(\Rightarrow x+y+z=a+b-2c+b+c-2a+c+a-2b\)

\(\Rightarrow x+y+z=0\)

\(\Rightarrow x+y=-z\left(1\right)\)

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

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

\(\Rightarrow x^3+y^3+z^3+3xy.\left(-z\right)=0\) ( Vì x + y = -z )

\(\Rightarrow x^3+y^3+z^3-3xyz=0\)

\(\Rightarrow x^3+y^3+z^3=3xyz\)

\(\Rightarrow\left(a+b-2c\right)^3+\left(b+c-2a\right)^3+\left(c+a-2b\right)^3=3\left(a+b-2c\right)\left(b+c-2a\right)\left(c+a-2b\right)\)

c) \(\left(x^2-x+2\right)^2-\left(x-2\right)^2\)

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

\(=x^4-2x^3+6x^2-8x+8\)

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

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

d) \(\left(x^2-8\right)^2+36\)

\(=x^4-16x^2+64+36\)

\(=x^4-16x^2+100\)

\(=x^4+20x^2+10^2-36x^2\)

\(=\left(x^2+10\right)^2-\left(6x\right)^2\)

\(=\left(x^2+10-6x\right)\left(x^2+10+6x\right)\)