一元一次解方程
(1)5-6(2a+a+1/3)(2)2a-(5b-a)+b(3)-3(2x-y)-2(4x+1/2y)+2011(4)-[2m-3(m-n+1)-2]-1...
(1)5-6(2a+a+1/3) (2)2a-(5b-a)+b (3)-3(2x-y)-2(4x+1/2y)+2011 (4)-[2m-3(m-n+1)-2]-1
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(1)5-6(2a+a+1/3)
=5-6(3a+1/3)
=5-18a-2
=5-2-18a
=3-18a
(2)2a-(5b-a)+b
=2a-5b+a+b
=2a+a-5b+b
=3a-4b
(3)-3(2x-y)-2(4x+1/2y)+2011
=-6x+3y-8x-y+2011
=-6x-8x+3y-y+2011
=-14x+2y+2011
(4)-[2m-3(m-n+1)-2]-1
=-[2m-3m+3n-3-2]-1
=-[-m+3n-5]-1
=m-3n+5-1
=m-3n+4
=5-6(3a+1/3)
=5-18a-2
=5-2-18a
=3-18a
(2)2a-(5b-a)+b
=2a-5b+a+b
=2a+a-5b+b
=3a-4b
(3)-3(2x-y)-2(4x+1/2y)+2011
=-6x+3y-8x-y+2011
=-6x-8x+3y-y+2011
=-14x+2y+2011
(4)-[2m-3(m-n+1)-2]-1
=-[2m-3m+3n-3-2]-1
=-[-m+3n-5]-1
=m-3n+5-1
=m-3n+4
来自:求助得到的回答
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(1)5-6(2a+a+1/3) = 3-18a
(2)2a-(5b-a)+b= 3a-4b
(3)-3(2x-y)-2(4x+1/2y)+2011= 14x+2y+2011
(4)-[2m-3(m-n+1)-2]-1= m-3n+4
(2)2a-(5b-a)+b= 3a-4b
(3)-3(2x-y)-2(4x+1/2y)+2011= 14x+2y+2011
(4)-[2m-3(m-n+1)-2]-1= m-3n+4
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