Question

问一道数学题 已知xy=-2,x+y=3,求式子（3xy+10y+[+5x-(2xy+2y-3x)] 怎么写啊

要写详细一点，然后说为什么要镇么写啊，一定详细。很急！！！！！！！！！今晚10点内回答。

Answer 1

（3xy+10y)+[+5x-(2xy+2y-3x)]
=3xy+10y+[+5x-2xy-2y+3x],[原式去括号]
=3xy+10y+5x-2xy-2y+3x,[去括号]
=3xy-2xy+5x+3x+10y-2y,[合并同类项]
=xy+8x+8y
=xy+8(x+y),[提取公因式8]
=-2+8(3) ,[将xy=-2,x+y=3,代入]
=-2+24
=22

Answer 2

3xy+10y+[+5x-(2xy+2y-3x)]
=3xy+10y+[+5x-2xy-2y+3x]……去小括号
=3xy+10y+5x-2xy-2y+3x……去中括号
=xy+8x+8y……合并同类项
=xy+8(x+y)……提取公因式
=-2+8×3……代入数值
=22

Answer 3

原式=3xy-2xy+10y-2y+5x+3x=xy+8x+8y=xy+8(x+y)=-2+8*3=22

Answer 4

（3xy+10y+[+5x-(2xy+2y-3x)] =3xy+10y+5x-2xy-2y+3x
=xy+8x+8y
=-2+8*(x+y)
=-2+8*3
=22

Answer 5

（3xy+10y+[+5x-(2xy+2y-3x)] 去括号
=3xy+10y+5x-2xy-2y+3x
=xy+8(x+y)
=-2+3*8
=22