如何解下列方程组(1) x+y=12 (2)x²-y²-5x-5y=-12xy=35 2x+y=0
问题描述:
如何解下列方程组
(1) x+y=12 (2)x²-y²-5x-5y=-12
xy=35 2x+y=0
答
(1) x+y=12 xy=35 =5x7x=5 y=7 或x=7 y=5y=35/xx+35x=12x^2-12x+35=0(x-7)(x-5)=0x1=7 y1=5x2=5 y1=7(2)y=-2xx^2-4x^2-5x+10x=-123x^2-5x-12=0(3x+4)(x-3)=0x1=-4/3 y1=8/3x2=3 y2=-6