因式分解,求求求过程

发布网友 发布时间：2023-08-02 22:14

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热心网友 时间：2023-09-14 12:26

x” + (k+2)x + (k+1)
= x” + x + (k+1)x + (k+1)
= x( x + 1 ) + (k+1)( x + 1 )
= ( x + 1 )( x + k + 1 )

(x”+3x)” - 8(x”+3x) - 20
= (x”+3x)” + 2(x”+3x) - 10(x”+3x) - 20
= (x”+3x)(x”+3x+2) -10(x”+3x+2)
= (x”+3x+2)(x”+3x-10)
= (x”+x+2x+2)(x”+5x-2x-10)
= [ x(x+1) +2(x+1) ][ x(x+5) -2(x+5) ]
= ( x + 1 )( x + 2 )( x + 5 )( x - 2 )

(t”+4t)” - 2(t”+4t) - 24
= (t”+4t)” + 4(t”+4t) - 6(t”+4t) - 24
= (t”+4t)(t”+4t+4) - 6(t”+4t+4)
= (t”+4t+4)(t”+4t-6)
= ( t + 2 )”( t” + 4t - 6 )
= ( t + 2 )”( t” + 4t + 4 - 10 )
= ( t + 2 )”[ (t+2)” - (√10)” ]
= ( t + 2 )”( t + 2 - √10 )( t + 2 + √10 )

4a”b” - ( a” + b” )”
= -[ ( a” + b” )” - ( 2ab )" ]
= -( a” - 2ab + b” )( a” + 2ab + b” )
= -( a - b )”( a + b )”

热心网友 时间：2023-09-14 12:27

解：x^2+ (k+2)x + (k+1)
= x”^2+ x + (k+1)x + (k+1)
= x( x + 1 ) + (k+1)( x + 1 )
= ( x + 1 )( x + k + 1 )

(x^2+3x)^2 - 8(x^2+3x) - 20
= (x^2+3x)^2 + 2(x^2+3x) - 10(x^2+3x) - 20
= (x^2+3x)(x^2+3x+2) -10(x^2+3x+2)
= (x^2+3x+2)(x^2+3x-10)
= (x^2+x+2x+2)(x^2+5x-2x-10)
= [ x(x+1) +2(x+1) ][ x(x+5) -2(x+5) ]
= ( x + 1 )( x + 2 )( x + 5 )( x - 2 )

(t^2+4t)^2- 2(t^2+4t) - 24
= (t^2+4t)^2 + 4(t^2+4t) - 6(t^2+4t) - 24
= (t^2+4t)(t^2+4t+4) - 6(t^2+4t+4)
= (t^2+4t+4)(t^2+4t-6)
= ( t + 2 )^2( t^2 + 4t - 6 )
= ( t + 2 )^2[ (t+2)^2 - (√10)”^2]
= ( t + 2 )^2( t + 2 - √10 )( t + 2 + √10 )

4a^2b^2 - ( a^2 + b”^2)^2
= -[ ( a^2 + b”^2)”^2- ( 2ab )^2 ]
= -( a^2 - 2ab + b^2 )( a^2 + 2ab + b”^2)
= -( a - b )^2( a + b )^2