√(x1^2+1)-ax1-√(x^2+1)+ax2=(x1^2-x2^2)/[√(x1^12+1)+√(x2^2+1)]-a(x1-x2)怎么来的,

Question

√(x1^2+1)-ax1-√(x^2+1)+ax2 =(x1^2-x2^2)/[√(x1^12+1)+√(x2^2+1)]-a(x1-x2) 怎么来的,

Answer 1

原式=√(x1^2+1)-√(x2^2+1)-(ax1-ax2)
=[√(x1^2+1)-√(x2^2+1)]*[√(x1^2+1)+√(x2^2+1)]/[√(x1^2+1)+√(x2^2+1)]-a(x1-x2)【分子有理化；提取公因式】
={[√(x1^2+1)]^2-[√(x2^1+1)]^2}/[√(x1^2+1)+√(x2^1+1)]-a(x1-x2)【平方差公式】
=[(x1^2+1)-(x2^2+1)]/[√(x1^2+1)+√(x2^2+1)]-a(x1-x2)
=(x1^-x2^2)/[√(x1^2+1)+√(x2^2+1)]-a(x1-x2)