数学问题,快啊,快!111111!1
问题描述：
数学问题,快啊,快!111111!1
用方程解应用题
已知X^2+XY=12,XY+Y^2=8,求X^2-Y^2与X^2+2XY+Y^2的值
已知X-Y=5,XY=3,求(6X-4Y+5XY)-(3X-Y+2XY)的值
谢谢了111!1!
答
1、X^2-Y^2
=(X^2+XY)-(XY+Y^2)
=12-8
=4
X^2+2XY+Y^
=(X^2+XY)+(XY+Y^2)
=12+8
=20
2、(6X-4Y+5XY)-(3X-Y+2XY)
=3x-3y+3xy
=3(x-y+xy)
=3*(5+3)
=24