Question

phân tích đa thức thành nhân tử:

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

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

 

Answer 1

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

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

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

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

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

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

Answer 2

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 )