Question

怎么解方程x2≥y2?

Answer 1

原题x^2≥y^2
当y=0时,
x^2≥0,x=R（任敬肆意实数庆核）
当y≠0时,x^2-y^2≥0,即(x+y)(x-y)≥0
当誉稿掘x+y≥0且x-y≥0
X1≥-y,X2≥y,
当x+y≤0且x-y≤0
X3≤-y,X4≤y,