Question

Phân tích đa thức thành nhân tử 

a,x(x+2)(x+3)(x+5)+5

b,6x^2-5xy+y^2

Answer 1

a) x( x + 2 )( x + 3 )( x + 5 ) + 5

= [ x( x + 5 ) ][ ( x + 2 )( x + 3 ) ] + 5

= ( x² + 5x )( x² + 5x + 6 ) + 5 (1)

Đặt t = x² + 5x

(1) <=> t( t + 6 ) + 5

       = t² + 6t + 5

       = t² + t + 5t + 5 

       = t( t + 1 ) + 5( t + 1 )

       = ( t + 1 )( t + 5 )

       = ( x² + 5x + 1 )( x² + 5x + 5 )

b) 6x² - 5xy + y² = 6x² - 3xy - 2xy + y² = 3x( 2x - y ) - y( 2x - y ) = ( 2x - y )( 3x - y )

Answer 2

a,\(x\left(x+2\right)\left(x+3\right)\left(x+5\right)+5\)

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

\(=\left(x^2+5x\right)\left(x^2+5x+6\right)+5\)(*)

Đặt \(a=x^2+5x\)ta đc:

(*)=\(a\left(a+6\right)+5\)

\(=a^2+6a+5\)

\(=a^2+a+5a+5\)

\(=a\left(a+1\right)+5\left(a+1\right)\)

\(=\left(a+5\right)\left(a+1\right)\)

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

b,\(6x^2-3xy-2xy+y^2\)

\(=3x\left(2x-y\right)-y\left(2x-y\right)\)

\(=\left(3x-y\right)\left(2x-y\right)\)

Answer 3

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Answer 4

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Answer 5

a) \(x\left(x+2\right)\left(x+3\right)\left(x+5\right)+5\)

\(=\left[x\left(x+5\right)\right]\cdot\left[\left(x+2\right)\left(x+3\right)\right]+5\)

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

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

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

Answer 6

b) \(6x^2-5xy+y^2\)

\(=\left(6x^2-6xy\right)-\left(xy-y^2\right)\)

\(=6x\left(x-y\right)-y\left(x-y\right)\)

\(\left(x-y\right)\left(6x-y\right)\)