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11 tháng 6 2017

sửa đề:\(\left(x+y+z\right)^3-x^3-y^3-z^3\)

giải:

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

11 tháng 6 2017

b,W = \(x^4+x^2+1+2009x^2+2009x+2009\)

\(=\left(x^4+2x^2+1\right)-x^2+2009\left(x^2+x+1\right)\)

\(=\left(x^2+1\right)^2-x^2+2009\left(x^2+x+1\right)\)

\(=\left(x^2+1-x\right)\left(x^2+1+x\right)+2009\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left(x^2-x+2010\right)\)

1 tháng 9 2023

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

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

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

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

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

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

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

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

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

#\(Urushi\text{☕}\)

1 tháng 9 2023

Áp dụng (a+b)3 = a3+b3+3ab(a+b), ta có:

(x+y+z)3-x3-y3-z3

=[(x+y)+z]3-x3-y3-z3

=(x+y)3+z3+3z(x+y)(x+y+z)-x3-y3-z3

=x3+y3+3xy(x+y)+z3+3z(x+y)(x+y+z)-x3-y3-z3

=3(x+y)(xy+xz+yz+z2)

=3(x+y)[x(y+z)+z(y+z)]

=3(x+y)(y+z)(x+z)

13 tháng 8 2020

Ta có: ( x - y) z3 + ( y - z ) x3 + ( z - x ) y 

= ( x - y ) z3 + ( y - z )x3 + ( z - y)y3 + ( y - x ) y3

= ( x - y ) ( z3 - y) + ( y - z ) ( x3 - y3

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

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

= ( x - y ) ( y - z ) [ ( x2 - z2) + ( xy - zy) ]

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

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

13 tháng 8 2020

(x - y).z3 + (y - z).x3 + (z - x).y3

= z3(x - y) + x3y - x3z + y3z - xy3

= z3(x - y) + xy(x2 - y2) - z(x3 - y3)

= z3(x - y) + xy(x - y)(x + y) - z(x - y)(x2 + xy + y2)

= (x - y)(z3 + x2y + xy2 - x2z - xyz - y2z)

= (x - y)[z(z2 - x2) + xy(x - z) + y2(x - z)]

= (x - y)[z(z - x)(z  + x) - xy(z- x) - y2(z - x)]

= (x - y)(z - x)(z2 + xz - xy - y2)

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

= (x  - y)(z - x)(y - z)(y + z - x)

21 tháng 3 2017

23 tháng 7 2021

( x + y + z)3 - x3 - y3 - z3=x3+y3+z3+3(a+b)(a+c)(b+c)- x3 - y3 - z3

                                              = 3(a+b)(b+c)(a+c)

11 tháng 12 2019

a) 12x.                           b) 4xy

c) 2y(3 x 2   +   y 2 ).

d) (x + y + z)( x 2   +   y 2   + z 2  – xy – xz - yz).

8 tháng 8 2019

a) 16(12 t 2  +1).

b) Gợi ý x 3   +   y 3 = ( x   +   y ) 3  - 3xy(x + y)

(x + y - z)( x 2   +   y 2   +   z 2  - xy + xz + yz).

a: =(x+y)^3+z^3-3xy(x+y)-3xyz

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

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

b: a+b+c<>0

A=(a+b+c)^3-a^3-b^3-c^3/a+b+c

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

=a^2+b^2+c^2-ab-ac-bc

=1/2[a^2-2ab+b^2+b^2-2bc+c^2+a^2-2ac+c^2]

=1/2[(a-b)^2+(b-c)^2+(a-c)^2]>=0

a: =(x+y)^3+z^3-3xy(x+y)-3xyz

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

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

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

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

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

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

c: \(=\left(x^2+x\right)^2+3\left(x^2+x\right)+2-12\)

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

=(x^2+x+5)(x^2+x-2)

=(x^2+x+5)(x+2)(x-1)

d: =b^2c+bc^2+ac^2-a^2c-a^2b-ab^2

=b^2c-b^2a+bc^2-a^2b+ac^2-a^2c

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

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

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

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

=(c-a)[b(b+c)+a(b+c)]

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