Question

这个数学题。第三题

Answer 1

正解如下：
4<lx²-5xl≤6
当x²-5x≥0即x≥5或x≤0时：
等价于 4<x²-5x≤6
x²-5x-4>0, x>(5+√41)/2或x<(5-√41)/2
x²-5x-6≤0， (x-6)(x+1)≤0， -1≤x≤6
得 (5+√41)/2<x≤6或-1≤x<(5-√41)/2
所以 此时解是 (5+√41)/2<x≤6或-1≤x<(5-√41)/2
当x²-5x<0即0<x<5时，等价于-6≤x²-5x<-4
x²-5x+6≥0， x≥3或x≤2
x²-5x+4<0, 1<x<4
得 3≤x<4或1<x≤2
将两种情况求并集：
-1≤x<(5-√41)/2或1<x≤2或3≤x<4或 (5+√41)/2<x≤6

Answer 2

(3)
4<|x^2-5x|≤6
=>
|x^2-5x|≤6 (1) and
|x^2-5x|>4 (2)
from (1)
|x^2-5x|≤6
-6≤x^2-5x≤6
x^2-5x+6≥0 and x^2-5x-6≤0
(x-2)(x-3)≥0 and (x-6)(x+1)≤0
"x≥3 or x≤2" and -1≤x≤6
3≤x≤6 (3)
from (2)
|x^2-5x|>4
x^2-5x>4 or x^2-5x<-4
x^2-5x-4>0 or x^2-5x+4<0
"x> (5+√41)/2 or x<(5+√41)/2" or (x-1)(x-4)<0
"x> (5+√41)/2 or x<(5-√41)/2" or 1<x<4
x<(5-√41)/2 or 1<x<4 or x> (5+√41)/2 (4)
--------------
4<|x^2-5x|≤6
=>
(3) and (4)
3≤x≤6 and "x<(5-√41)/2 or 1<x<4 or x> (5+√41)/2"
3≤x<4 or (5+√41)/2<x≤6

Answer 3

拆成
-6≤x²-5x≤-4
4≤ x²-5x≤6