
已知a=根号3-根号2,b=根号3+根号2,求(1)a^2b+ab^2;(2)a^2/1+b^2/1
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(1)a²b+ab²
=ab(a+b)
=(√3-√2)(√3+√2)(√3-√2+√3+√2)
=1×2√3
=2√3
(2)1/a²+1/b²
=1/(√3-√2)²+1/(√3+√2)²
=1/(5-2√6)+1/(5+2√6)
=(5+2√6+5-2√6)/(5-2√6)(5+2√6)
=10/(25-24)
=10
=ab(a+b)
=(√3-√2)(√3+√2)(√3-√2+√3+√2)
=1×2√3
=2√3
(2)1/a²+1/b²
=1/(√3-√2)²+1/(√3+√2)²
=1/(5-2√6)+1/(5+2√6)
=(5+2√6+5-2√6)/(5-2√6)(5+2√6)
=10/(25-24)
=10
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