Hello, Showcase_22!
Here's the first one.
. . Your punchlines are a bit off . . .
$\displaystyle x - 4 \;< \;x(x-4) \;<\;5$
The left side: .$\displaystyle x - 4 \:<\:x(x-4)$
We have: .$\displaystyle x - 4 \:<\:x^2-4x \quad\Rightarrow\quad 0 \:< \:x^2-5x + 4 \quad\Rightarrow\quad x^2-5x + 4 \:>\:0$
Factor: .$\displaystyle (x - 1)(x - 4) \:>\:0$
There are two cases: [1] both factors are positive, [2] both factors are negative.
. . $\displaystyle [1]\;\;\begin{Bmatrix}x - 1 &> &0 & \Rightarrow & x & > & 1 \\ x - 4 & > & 0 & \Rightarrow & x & > & 4 \end{Bmatrix} \quad\Rightarrow\quad x \:>\:4$
. . $\displaystyle [2]\;\;\begin{Bmatrix}x - 1 &<&0 & \Rightarrow & x &<& 1 \\ x - 4 &<& 0 & \Rightarrow & x &<& 4 \end{Bmatrix} \quad\Rightarrow\quad x \:<\: 1 $
Therefore: .$\displaystyle \boxed{x < 1\;\text{ or }\;x > 4}$
The right side: .$\displaystyle x(x-4) \:<\:5 \quad\Rightarrow\quad x^2 - 4x - 5 \:<\:0$
Factor: .$\displaystyle (x+1)(x-5) \:<\:0$
There are two cases: [1] the factors are (-)(+), [2] the factors are (+)(-).
$\displaystyle [1]\;\;\begin{Bmatrix}x + 1 &<& 0 & \Rightarrow & x &<& \text{-}1 \\
x - 5 &>& 0 & \Rightarrow & x &>& 5 \end{Bmatrix}\quad\Rightarrow\quad \text{impossible} $
$\displaystyle [2]\;\;\begin{Bmatrix}x + 1 &>& 0 & \Rightarrow & x &>& \text{-}1 \\
x-5 &<&0 & \Rightarrow & x &<& 5 \end{Bmatrix} \quad\Rightarrow\quad \boxed{\text{-}1 \:< \:x \:<\:5}$
To determine the final solution, I had to make pretty pictures . . . Code:
x < 1 or x > 4
= = = = o - - - - - o = = = =
1 4
-1 < x < 5
- - o = = = = = = = = = = = o - -
-1 5
Combining them, we have:
Code:
- - o = = o - - - - - o = = o - -
-1 1 4 5
Therefore: .$\displaystyle \boxed{\;(\text{-}1 \:<\: x\:<\:1)\;\text{ or }\;(4 \:<\:x\:<\:5)\;} $