y=3x+5/5x-3（x>1)的值域; y=x+根号(1+2x)的值域; y=x/x^2-2x+4的值域;

题目

y=3x+5/5x-3（x>1)的值域; y=x+根号(1+2x)的值域; y=x/x^2-2x+4的值域;

答案

1.
y=(3x+5)/(5x-3)
5yx-3y=3x+5
(5y-3)x=5+3y
x=(5+3y)/(5y-3)＞1
当5y-3＞0时,即y＞3/5时,
5+3y＞5y-3
y＜4
所以3/5＜y＜4;
当5y-3＜0时,即y＜3/5时,
5+3y＜5y-3
y＞4(舍去)
所以3/5＜y＜4.
2.
y=x+√(1+2x)
1+2x≥0
x≥-1/2
y=x+√(1+2x)
2y=(2x+1)+2√(1+2x)-1
=[√(1+2x)+1]^2-2
≥{√[1+2(-1/2)]+1}^2-2
=-1
y≥-1/2 .
3.
y=x/(x^2-2x+4)
(x^2-2x+4)y-x=0
yx^2-(2y+1)x+4y=0
当y=0时,x=0
当y≠0时,
△=(2y+1)^2-16y^2
=-12y^2+4y+1≥0
(-2y+1)(6y+1)≥0
(2y-1)(6y+1)≤0
-1/6≤y≤1/2.