Question 2 (x)
Add:
x2−y2−1, y2−1−x2, 1−x2−y2
x2−y2−1, y2−1−x2, 1−x2−y2=x2−y2−1+y2−1−x2+1−x2−y2 =x2−x2−x2−y2+y2−y2−1−1+1 =(1−1−1)x2+(−1+1−1)y2−1−1+1 =−x2−y2−1
Question 70
From the sum of x2−y2−1,y2−x2−1 and 1−x2−y2, subtract −(1+y2).
Let f(x,y)=√x2+y2+√x2+y2−2x+1+√x2+y2−2y+1+√x2+y2−6x−8y+25∀x,yϵR, then