软键盘怎么调出来:Csoole之数学题目求助 40分(悬赏+追)

来源:百度文库 编辑:高考问答 时间:2024/05/09 03:45:51
1、X,y为实数,且x^2+y^2/2+4≤xy+2y,则x,y的值分别是x=_____,y=__________

2、设实数x>2,且{x+[5/(x-2)]}^2=4(2x+1)则x=______

40分详求解题步骤,正确率第一详细第二
谢谢各位帮助

在线等谢谢
2题都要详细步骤和正确答案
辛苦大家乐

1.
x^2+y^2/2+4≤xy+2y
x^2-xy+y^2/2-2y+4≤0
x^2-xy+y^2/4+y^2/4-2y+4≤0
(x-y/2)^2+(y/2-2)^2≤0
(x-y/2)^2和(y/2-2)^2非负
x-y/2=0,y/2-2=0
x=2,y=4

2.
{x+[5/(x-2)]}^2=4(2x+1)
{x-2+2+[5/(x-2)]}^2=4(2x+1)
[(x-2)+(2x+1)/(x-2)]^2=4(2x+1)
(x-2)^2+2(2x+1)+[(2x+1)/(x-2)]^2=4(2x+1)
(x-2)^2-2(2x+1)+[(2x+1)/(x-2)]^2=0
[(x-2)-(2x+1)/(x-2)]^2=0
(x-2)=(2x+1)/(x-2)
x^2-4x+4=2x+1
x^2-6x+3=0
x1=3+√6
x2=3-√6<2(舍去)

1.解:
x^2-xy+y^2/2≤xy+2y
(x-(1/2)*y)^2+((1/2)*y-2)^2≤0
即(x-(1/2)*y)^2+((1/2)*y-2)^2=0
所以y=4 x=2

2.解:
((x^2-2x+5)/(x-2))^2=4(2x+1)
((x-2)^2+2x+1)/(x-2)=4(2x+1)
((x-2)+(2x+1)/(x-2))^2=4(2x+1)
(x-2)^2-2(2x+1)+((2x+1)/(x-2))^2=0
((x-2)-((2x+1)/(x-2)))^2=0
即x-2=(2x+1)/(x-2)
x1=3+√6 x2=3-√6(舍去)

解:
x^2-xy+y^2/2≤xy+2y
(x-(1/2)*y)^2+((1/2)*y-2)^2≤0
即(x-(1/2)*y)^2+((1/2)*y-2)^2=0
所以y=4 x=2