Question

设[x+√(x2+1）][（y+√(y2+4)]=9,则x√（y2+4)+y√（x2+1)=

尽快！！

Answer 1

设
x+√(x2+1）= u (1)
y+√(y2+4) = v (2)
那么uv = 9
由(1)(2)可以得到
2x = u - 1/u
2y = v - 4/v
2√(x2+1）= u + 1/u
2√(y2+4) = v + 4/v
所以4*[x√（y2+4)+y√（x2+1)] = (v - 4/v)(u + 1/u) + (v + 4/v)(u - 1/u)
=2uv-8/uv = 18-8/9
所以x√（y2+4)+y√（x2+1)= 9/2-2/9 = 77/18

Answer 2

设
x+√(x2+1）= u (1)
y+√(y2+4) = v (2)
那么uv = 9
由(1)(2)可以得到
2x = u - 1/u
2y = v - 4/v
2√(x2+1）= u + 1/u
2√(y2+4) = v + 4/v
所以4*[x√（y2+4)+y√（x2+1)] = (v - 4/v)(u + 1/u) + (v + 4/v)(u - 1/u)
=2uv-8/uv = 18-8/9
所以x√（y2+4)+y√（x2+1)= 9/2-2/9 = 77/18