已知x=√2+√5y=√5-√2求x²+xy=y²和(x+y)²-(x-y)²的
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已知x=√2+√5y=√5-√2求x²+xy=y²和(x+y)²-(x-y)²的值
是x²+xy+y²
是x²+xy+y²
x²+xy+y²
=(√2+√5)²+(√2+√5)(√5-√2)+(√5-√2)²
=(√2+√5)²+2(√2+√5)(√5-√2)+(√5-√2)²-(√2+√5)(√5-√2)
=[(√2+√5)+(√5-√2)]²-(√5+√2)(√5-√2)
=(2√5)²-2√5
=20-2√5
(x+y)²-(x-y)²
=[(√2+√5)+(√5-√2)]²-[(√2+√5)-(√5-√2)]²
=(2√5)²-(2√2)²
=20-8
=12
=(√2+√5)²+(√2+√5)(√5-√2)+(√5-√2)²
=(√2+√5)²+2(√2+√5)(√5-√2)+(√5-√2)²-(√2+√5)(√5-√2)
=[(√2+√5)+(√5-√2)]²-(√5+√2)(√5-√2)
=(2√5)²-2√5
=20-2√5
(x+y)²-(x-y)²
=[(√2+√5)+(√5-√2)]²-[(√2+√5)-(√5-√2)]²
=(2√5)²-(2√2)²
=20-8
=12
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