K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

19 tháng 7 2018

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

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

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

\(=6a^2b+2b^3\)

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

19 tháng 7 2018

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

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

\(=6ab^2+2b^3\)(rút gọn hết)

b/\(x^3+y^3+z^3-3xyz\)

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

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

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

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

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

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

Hok tốt

5 tháng 10 2020

a) 16x2 - ( x2 + 4 )2

= ( 4x )2 - ( x2 + 4 )2

= [ 4x - ( x2 + 4 ) ][ 4x + ( x2 + 4 ) ]

= ( -x2 + 4x - 4 )( x2 + 4x + 4 )

= [ -( x2 - 4x + 4 ) ]( x + 2 )2

= [ -( x - 2 )2 ]( x + 2 )2

b) ( x + y )3 + ( x - y )3

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

= ( x + y + x - y )[ x2 + 2xy + y2 - ( x2 - y2 ) + x2 - 2xy + y2 ]

= 2x( 2x2 + 2y2 - x2 + y2

= 2x( x2 + 3y2 )

4 tháng 8 2016

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

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

=(b+c)(a2+ab+ac+bc)

b) x3+y3+z3-3xyz

= (x+y)3-3xy(x+y) +z3-3xyz

= (x+y+z)(x2+y2+2xy-xz-yz+z2) - 3xy(x+y+z)

=(x+y+z)( x2+y2+z2-xy-yz-xz)

4 tháng 8 2016

câu a chưa pt hết kìa :V
a, 3(a+b)(b+c)(c+a)
có thẻ dùng hđt : (a+b+c)^3=a^3+b^3+c^3+3(a+b)(b+c)(c+a)

2 tháng 8 2016

a)(x+y)2-(x-y)2

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

=2y.2x=4xy

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

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

=2x.(4x+2)

=4x(2x+1)

c) x3+y3+z3-3xyz

= (x+y)3- 3xy(x+y) +z3-3xyz

=(x+y+z)( x2+2xy+y2-xz-yz+z2)-3xy(x+y+z)

=(x+y+z)(x2+y2+z2-xy-xz-yz)

4 tháng 8 2016

Phân tích đa thức sau thành nhân tử :

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

b) \(x^3+y^3+z^3-3xyz\)

3 tháng 1 2016

\(64-27x^3=4^3-\left(3x\right)^3=\left(4-3x\right)\left(16+12x+9x^2\right)\)

4 tháng 8 2019

Đặt \(x+y-z=a;x-y+z=b;y+z-x=c\)

Ta có:\(A=\left(a+b+c\right)^3-a^3-b^3-c^3\)

\(A=\left[\left(a+b\right)+c\right]^3-a^3-b^3-c^3\)

\(A=\left(a+b\right)^3+3\left(a+b\right)\cdot c\cdot\left(a+b+c\right)+c^3-a^3-b^3-c^3\)

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

\(A=3\left(a+b\right)\left(ab+ac+bc+c^2\right)\)

\(A=3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)

Hay \(A=3\cdot2x\cdot2y\cdot2z\)

\(A=24xyz\)

23 tháng 9 2016

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

\(=x^3+y^3+z^3+3x^2y+3x^2z+3y^2z+3xy^2+3xz^2+3yz^2+6xyz-x^3-y^3-z^2\) 

\(=3x^2y+3xy^2+3x^2z+3xz^2+3y^2z+3yz^2+6xyz\)

\(=3xy\left(x+y\right)+3xz\left(x+z\right)+3yz\left(y+z\right)+6xyz\)

\(=3\left[xy\left(x+y\right)+xz\left(x+z\right)+yz\left(y+z\right)+2xyz\right]\)

\(=3\left[xy\left(x+y\right)+x^2z+xz^2+y^2z+yz^2+2xyz\right]\)

\(=3\left[xy\left(x+y\right)+xz\left(x+y\right)+z^2\left(x+y\right)+yz\left(x+y\right)\right]\)

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

\(=3\left(x+y\right)\left[x\left(y+z\right)+z\left(y+z\right)\right]\)

\(=3\left(x+y\right)\left(y+z\right)\left(x+z\right)\)

23 tháng 9 2016

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

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

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

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

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

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

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

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

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