已知x=√3-√2⼀√3+√2,且xy=1,求3x^2+5xy+3y^2的值

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2024-12-19 10:07:45
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回答1:

x=√3-√2/√3+√2,xy=1
所以y=√3+√2/√3-√2
x分母有理化后得:5-2√6
同理得:y=5+2√6
x+y=10,xy=1
所以,原式得:3x^2+3y^2+5xy
=3(x^2+y^2)+5xy
=3(x^2+2xy+y^2-2xy)+5xy
=3[(x+y)^2-2xy]+5xy
=3[10^2-2*1]+5*1
=3[100-2]+5
=300-6+5
=299