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(√2+√3-√6)²-(√2-√3+√6)²
=(√2+√3-√6+√2-√3+√6)(√2+√3-√6-√2+√3-√6)
=2√2*(2√3-2√6)
=4√6-8√3
求x²+3x-1
=(√2-1)²+3(√2-1)-1
=2-2√2+1+3√2-3-1
=√2-1
=(√2+√3-√6+√2-√3+√6)(√2+√3-√6-√2+√3-√6)
=2√2*(2√3-2√6)
=4√6-8√3
求x²+3x-1
=(√2-1)²+3(√2-1)-1
=2-2√2+1+3√2-3-1
=√2-1
2012-02-18 · 知道合伙人教育行家
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(√2+√3-√6)²-(√2-√3+√6)²
= (√2+√3-√6+√2-√3+√6)(√2+√3-√6-√2+√3-√6)
= (2√2)(2√3-2√6)
= 4(√6-√12)
= 4(√6-2√3)
x=√2-1
x²+3x-1 = (x+1)²+x-2 = (√2-1+1)²+√2-1-2 = (√2)²+√2-1-2 = 2+√2-3 = √2-1
= (√2+√3-√6+√2-√3+√6)(√2+√3-√6-√2+√3-√6)
= (2√2)(2√3-2√6)
= 4(√6-√12)
= 4(√6-2√3)
x=√2-1
x²+3x-1 = (x+1)²+x-2 = (√2-1+1)²+√2-1-2 = (√2)²+√2-1-2 = 2+√2-3 = √2-1
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