Question

问题求解

Answer 1

3x^2-5xy+3y^2
=3x^2-6xy+3y^2+xy
=3(x^2-2xy+y^2)+xy
=3(x-y)^2+xy
=3[(√3-√2)/(√3+√2)-(√3+√2)/(√3-√2)]^2+(√3-√2)/(√3+√2)*(√3+√2)/(√3-√2)
=3[(√3-√2)^2/(√3+√2)(√3-√2)-(√3+√2)^2/(√3-√2)(√3+√2)]^2+1
=3[(√3-√2)^2-(√3+√2)^2]^2+1
=3[(√3-√2-√3-√2)(√3-√2+√3+√2)]^2+1
=3(-2√2*2√3)^2+1
=3(-4√6)^2+1
=3*96+1
=289

[√(x+1)-√(x-1)]/[√(x+1)+√(x-1)]-[√(x+1)+√(x-1)]/[√(x+1)-√(x-1)]
=[√(x+1)-√(x-1)]^2/[√(x+1)+√(x-1)][√(x+1)-√(x-1)]-[√(x+1)+√(x-1)]^2/[√(x+1)-√(x-1)][√(x+1)+√(x-1)]
=[√(x+1)-√(x-1)]^2/[√(x+1)^2-√(x-1)^2]-[√(x+1)+√(x-1)]^2/[√(x+1)^2-√(x-1)^2]
=[√(x+1)-√(x-1)]^2/[x+1-x+1)]-[√(x+1)+√(x-1)]^2/[x+1-x+1)]
=[√(x+1)-√(x-1)]^2/2-[√(x+1)+√(x-1)]^2/2
={[√(x+1)-√(x-1)]^2-[√(x+1)+√(x-1)]^2}/2
={[√(x+1)-√(x-1)]-[√(x+1)+√(x-1)]}{[√(x+1)-√(x-1)]+[√(x+1)+√(x-1)]}/2
=[√(x+1)-√(x-1)-√(x+1)-√(x-1)][√(x+1)-√(x-1)+√(x+1)+√(x-1)]}/2
=[-2√(x-1)*2√(x+1)]/2
=-2√[(x-1)(x+1)]
=-2√[x^2-1]
=-2√[(√5/2)^2-1]
=-2√[5/4-1]
=-2√(1/4)
=-2*1/2
=-1

Answer 2