Question

已知x=1/2(√11+√7),y=1/2(√11-√7)求x²-xy+y²的值.!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

(√5-√3+√2)(√5+√3-√2)
(√3a/2b)(√b/a÷2√1/b)

Answer 1

解：
x=1/2(√11+√7)
y=1/2(√11-√7)
x^2-xy+y^2=(x-y)^2+xy
=[1/2(√11+√7)-1/2(√11-√7)]^2+1/2(√11+√7)*1/2(√11-√7)
=(√7)^2+(11-7)/4
=7+1
=8

(√5-√3+√2)(√5+√3-√2)
=[(√5-(√3-√2)][√5+(√3-√2) ]
=5-(√3-√2)^2
=5-(5-2√6)
=2√6

(√3a/2b)(√b/a÷2√1/b) 题目不太看得懂
=(√3a/2b)(√(b/a)*√b/2
=√3/4

Answer 2

x²-xy+y²化成（x-y）^2+xy来做。。。。。把x=1/2(√11+√7),y=1/2(√11-√7) 代入 就可以解出来了。。实践出真知。结果自己算吧。。。。