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

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

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

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

Bài làm:

a) \(x^2-2xy+y^2-zx+yz\)

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

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

a/ \(x^2-2xy+y^2-zx+yz.\)

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

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

c/ \(x^2-y^2-2x-2y.\)

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

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

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

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

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

sử dụng hằng thành thạo = ez 

\(a,x^2-2xy+y^2-zx+yz\)

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

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

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

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

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

phân tích đa thức thành nhân tử đặt biến phụ

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

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

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

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

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

= (x2 + y2 + z2 + xy + yz + zx)2

Đảm bảo ko phân tích tiếp đc nữa đâu ^^, đây tuy ko phải cách đặt biến phụ nhưng cách này chắc ngắn hơn cách đặt biến phụ.

  bởi Bùi Xuân Chiến 

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)

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

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

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

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

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

a. 

b. 

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

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

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

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

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

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

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 )