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16 tháng 9 2019

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

\(=xy^2-xz^2+yz^2-x^2y+zx^2-zy^2\)

\(=xy^2-xz^2+yz^2-x^2y+zx^2-zy^2-xyz+xyz\)

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

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

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

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

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

10 tháng 7 2017

a) xy(x + y) + yz(z + y) + zx(z + x) + 3xyz

= [xy(x + y) + xyz] + [yz(z + y) + xyz] + [zx(z + x) + xyz]

= xy(x + y + z) + yz(x + y + z) + zx(x + y + z)

= (xy + yz + zx)(x + y + z)

b) Vô câu hỏi tương tự 

26 tháng 7 2017

a) xy(x + y) + yz(z + y) + zx(z + x) + 3xyz

= [xy(x + y) + xyz] + [yz(z + y) + xyz] + [zx(z + x) + xyz]

= xy(x + y + z) + yz(x + y + z) + zx(x + y + z)

= (xy + yz + zx)(x + y + z)

b) tương tự 

2 tháng 7 2021

a) xy(x + y) + yz(y + z) + xz(z + x) + 3xyz

= xy(X + y + z)  + yz(x + y + z) + xz(X + y + z)

= (x + y +z)(xy + yz+ xz)

b) xy(x + y) - yz(y + z) - xz(z - x)

= x2y + xy2 - y2z - yz2 - xz2 + x2z

= x2(y + z) - yz(y + z) + x(y2 - z2)

= x2(y + z) - yz(y + z) + x(y + z)(y - z)

= (y + z)(x2 - yz + xy - xz)

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

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

c) x(y2 - z2) + y(z2 - x2) + z(x2 - y2)

 = x(y - z)(y + z) + yz2 - yx2 + x2z - y2z

= x(y - z)(y + z) - yz(y - z) - x2(y - z)

= (y - z)((xy + xz - yz - x2)

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

= (y - z)(x - z)(y -x) 

6 tháng 8 2017

a,Từ giả thiết ta có

(x2+y2+z2)(x+y+z)2+(xy+yz+zx)2

=(x2+y2+z2)(x2+y2+z2+2xy+2yz+2zx)+(xy+yz+zx)2

Đặt x2+y2+z2=a

xy+yz+zx=b

=>(x2+y2+z2)(x2+y2+z2+2xy+2yz+2zx)+(xy+yz+zx)2

=a(a+2b)+b2

=a2+2ab+b2

=(a+b)2

=(x2+y2+z2+xy+yz+zx)2

câu b hơi dài mình gửi sau nhé

6 tháng 8 2017

Ta có: 2(x^4+y^4+z^4)-(x^2+y^2+z^2)^2-2(x^2+y^2+z^2)(x+y+z)^2+(x+y+z)^4

Gọi x^4+y^4+z^4=a

x^2+y^2+z^2=b

x+y+z=c

=>2(x^4+y^4+z^4)-(x^2+y^2+z^2)^2-2(x^2+y^2+z^2)(x+y+z)^2+(x+y+z)^4=2a-b^2-2bc^2+c^4

=2a-2b^2+b^2-2bc^2+c^4

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

Ta có

2(a-b2)=2[x^4+y^4+z^4-(x^2+y^2+z^2)2]

=2[x^4+y^4+z^4-x^4-y^4-z^4-2x2y2-2y2z2-2z2x2]

=2.(-2)(x2y2+y2z2+z2x2)

=-4(x2y2+y2z2+z2x2)

Lại có

(b+c^2)^2

=[(x^2+y^2+z^2)+(x+y+z)2]2

=[(x^2+y^2+z^2)-(x^2+y^2+z^2)-2(xy+yz+zx)]2

=4(xy+yz+zx)2

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

=-4(x2y2+y2z2+z2x2)+4(xy+yz+zx)2

=8xyz(x+y+z)

25 tháng 7 2017

a/ \(\left(x-y\right)\left(z-x\right)\left(z-y\right)\)

b/ \(\left(1-y\right)\left(y-x\right)\)

25 tháng 7 2017

a. \(\left(x-y\right)\left(z-x\right)\left(z-y\right)\)

b. \(\left(1-y\right)\left(y-x\right)\)

22 tháng 7 2015

A ) xy(z+y)+yz(y+z)+zx(z+x)

=y.[x(z+y)+z(y+z)]+zx(z+x)

=y.(xz+xy+zy+z2)+zx(z+x)

=y.(xz+z2+xy+zy)+zx(z+x)

=y.[z.(z+x)+y.(z+x)]+zx(z+x)

=y.(z+x)(z+y)+zx(z+x)

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

=(z+x)(yz+y2+zx)

B )xy(x+y)-yz(y+z)-zx(z-x)

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

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

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

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

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

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

=(x-z)(yx+yz+y2+zx)

=(x-z)(yx+zx+yz+y2)

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

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

 

30 tháng 6 2021

b. \(\text{ xy(x+y)-yz(y+z)-xz(z-x) =xy(x+y+z-z)+yz(y+z)+xz(x-z) =xy(x-z)+xy(y+z)+yz(y+z)+xz(x-z) =(x+y)(y+z)(x-z) }\)

25 tháng 10 2020

a) \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-8\)

\(=\left[\left(x+1\right)\left(x+4\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]-8\)

\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-8\)

\(=\left(x^2+5x+5\right)^2-1-8\)

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

\(=\left(x^2+5x+2\right)\left(x^2+5x+8\right)\)

b) \(xy\left(x-y\right)+yz\left(y-z\right)+zx\left(z-x\right)\)

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

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

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

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

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

25 tháng 10 2020

a) ( x + 1 )( x + 2 )( x + 3 )( x + 4 ) - 8

= [ ( x + 1 )( x + 4 ) ][ ( x + 2 )( x + 3 ) ] - 8

= ( x2 + 5x + 4 )( x2 + 5x + 6 ) - 8

Đặt t = x2 + 5x + 5

bthuc ⇔ ( t - 1 )( t + 1 ) - 8

           = t2 - 1 - 8

           = t2 - 9

           = ( t - 3 )( t + 3 )

           = ( x2 + 5x + 5 - 3 )( x2 + 5x + 5 + 3 )

           = ( x2 + 5x + 2 )( x2 + 5x + 8 )

b) xy( x - y ) + yz( y - z ) + zx( z - x )

= x2y - xy2 + y2z - yz2 + zx( z - x )

= ( y2z - xy2 ) - ( yz2 - x2y ) + zx( z - x )

= y2( z - x ) - y( z2 - x2 ) + zx( z - x )

= ( z - x )( y2 + zx ) - y( z - x )( z + x )

= ( z - x )( y2 + zx - yz - yx )

= ( z - x )[ ( y2 - yx ) - ( yz - zx ) ]

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

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