初二数学因式分解。请写下这几题的过程,我需要对答案。
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-a^5 + a
= -a[ (a^4) - 1 ]
= -a( a” + 1 )( a” - 1 )
= -a( a" + 1 )( a + 1 )( a - 1 )
3( m + n )" - 27n"
= 3[ ( m + n )" - 9n" ]
= 3[ ( m + n )" - (3n)" ]
= 3( m + n + 3n )( m + n - 3n )
= 3( m + 4n )( m - 2n )
16( x + y )” - 25( x - y )"
= [ 4( x + y ) ]" - [ 5( x - y ) ]"
= ( 4x + 4y + 5x - 5y )( 4x + 4y - 5x + 5y )
= ( 9x - y )( -x + 9y )
= -( 9x - y )( x - 9y )
a"(a - b) + b"(b - a)
= a"( a - b ) - b"( a - b )
= ( a - b )( a" - b" )
= ( a - b )"( a + b )
= -a[ (a^4) - 1 ]
= -a( a” + 1 )( a” - 1 )
= -a( a" + 1 )( a + 1 )( a - 1 )
3( m + n )" - 27n"
= 3[ ( m + n )" - 9n" ]
= 3[ ( m + n )" - (3n)" ]
= 3( m + n + 3n )( m + n - 3n )
= 3( m + 4n )( m - 2n )
16( x + y )” - 25( x - y )"
= [ 4( x + y ) ]" - [ 5( x - y ) ]"
= ( 4x + 4y + 5x - 5y )( 4x + 4y - 5x + 5y )
= ( 9x - y )( -x + 9y )
= -( 9x - y )( x - 9y )
a"(a - b) + b"(b - a)
= a"( a - b ) - b"( a - b )
= ( a - b )( a" - b" )
= ( a - b )"( a + b )
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解:(1)原式=a(1-a^4)
=a(1+a^2)(1-a^2)
=a(1+a^2)(1+a)(1-a)
(2)原式=3[(m+n)^2-9n^2]
=3(m+n+3n)(m+n-3n)
=3(m+4n)(m-2n)
(3)原式=[4(x+y)]^2-[5(x-y)]^2
=[4(x+y)+5(x-y)][4(x+y)-5(x-y)]
=(9x-y)(-x+9y)
(4)原式=a^2(a-b)-b^2(a-b)
=(a-b)(a^-b^2)
=(a-b)(a+b)(a-b)
=(a+b)(a-b)^2
=a(1+a^2)(1-a^2)
=a(1+a^2)(1+a)(1-a)
(2)原式=3[(m+n)^2-9n^2]
=3(m+n+3n)(m+n-3n)
=3(m+4n)(m-2n)
(3)原式=[4(x+y)]^2-[5(x-y)]^2
=[4(x+y)+5(x-y)][4(x+y)-5(x-y)]
=(9x-y)(-x+9y)
(4)原式=a^2(a-b)-b^2(a-b)
=(a-b)(a^-b^2)
=(a-b)(a+b)(a-b)
=(a+b)(a-b)^2
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追问
第四小题的第二步是
(a-b)(a^2-b^2)
吧
追答
是的.
第四小题的第二步是(a-b)(a^2-b^2).
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