数学求值域问题！！！！！

y=x+√(10x-x^2-23)
y-x=√(10x-x^2-23)
(y-x)^2=10x-x^2-23
y^2+x^2-2xy-10x+x^2+23=0
2x^2-2x(y+5)+23+y^2=0
△=4(y+5)^2-4*2*(23+y^2)>=0
△=4y^2+10y+25-184-8y^2>=0
△=-4y^2+10y-159>=0
解不等式即可
就可得到值域
即[5-√2,7]

y=x+√(10x-x^2-23)
=x+√[2-(x-5)^2]
令：x-5=√2cosa a属于[0,π]
则：y=5+√2cosa+√2sina
=5+2sin(a+π/4)
a+pi/4属于[π/4, 5π/4]
故：最大值是7,最小值是5-√2