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设(x+√(x^2+1))(y+√(y^2+4))=9则x√(y^2+4)+y√(x^2+1)=?
题目详情
设(x+√(x^2+1))(y+√(y^2+4))=9 则x√(y^2+4)+y√(x^2+1)=?
▼优质解答
答案和解析
(x+√(x^2+1))(y+(√y^2+4))=9 ………a
方程两边同时乘以:(x-√(x^2+1))(y-(√y^2+4))
得:(x^2-(√(x^2+1))^2) (y^2-(√(y^2+4))^2)
=9(x-√(x^2+1))(y-(√y^2+4))
(-1)(-4)=9(x-√(x^2+1))(y-(√y^2+4))
(x-√(x^2+1))(y-(√y^2+4))=4/9…………b
a展开-b展开
xy+ x√(y^2+4)+ y√(x^2+1)+√(y^2+4)√(x^2+1)=9
xy-x√(y^2+4)- y√(x^2+1)+√(y^2+4)√(x^2+1)=4/9
2(x√(y^2+4)+ y√(x^2+1))=9-4/9
∴x√(y^2+4)+ y√(x^2+1)=77/18
方程两边同时乘以:(x-√(x^2+1))(y-(√y^2+4))
得:(x^2-(√(x^2+1))^2) (y^2-(√(y^2+4))^2)
=9(x-√(x^2+1))(y-(√y^2+4))
(-1)(-4)=9(x-√(x^2+1))(y-(√y^2+4))
(x-√(x^2+1))(y-(√y^2+4))=4/9…………b
a展开-b展开
xy+ x√(y^2+4)+ y√(x^2+1)+√(y^2+4)√(x^2+1)=9
xy-x√(y^2+4)- y√(x^2+1)+√(y^2+4)√(x^2+1)=4/9
2(x√(y^2+4)+ y√(x^2+1))=9-4/9
∴x√(y^2+4)+ y√(x^2+1)=77/18
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