(10y+x)^2-(10x+y)=138 用一元2次方程解y=x+2
问题描述:
(10y+x)^2-(10x+y)=138 用一元2次方程解y=x+2
答
y=x+2代入(10y+x)^2-(10x+y)=138 并化简:121x^2+429x+260=0解得x1=-2.7696,x2=-0.7758,y1=-0.7696,y2=1.2304