Hello, calmo11!
$\displaystyle (c)\;\;e^{x^2} \:\ge\:e^{5x}$
We have: .$\displaystyle x^2 \:\ge\:5x \quad\Rightarrow\quad x^2 - 5x \:\ge\:0$
. . . . . . . . $\displaystyle x(x-5)\:\ge\:0$
It says: the product of two number is positive.
This is true if both factors are positive or both factors are negative.
We must consider these two cases.
Both positive:
. . $\displaystyle x \,\ge\, 0 \:\text{ and }\:x-5\,\ge\,0 \quad\Rightarrow\quad x \,\ge\,5$
. . Both are true if $\displaystyle x\,\ge\,5$
Both negative:
. . $\displaystyle x\,\le\,0\:\text{ and }\:x-5 \,\le\,0 \quad\Rightarrow\quad x\,\le\,5$
. . Both are true if $\displaystyle x\,\le\,0$
Answer: .$\displaystyle (x\,\ge\,5)\,\text{ or }\,(x\,\le\,0)$
$\displaystyle (d)\;\;|4x-5| \:>\:1$
The absolute inequality gives us two statements:
. . $\displaystyle [1]\;4x-5 \,>\,1\;\text{ or }\;[2]\;4x-5\,<\,-1$
Solve them separately:
. . $\displaystyle [1]\;4x - 5 \,>\,1 \quad\Rightarrow\quad 4x\,>\.6 \quad\Rightarrow\quad x \,>\,\tfrac{3}{2}$
. . $\displaystyle [2]\;4x-5 \,<\,-1 \quad\Rightarrow\quad 4x \,<\,4 \quad\Rightarrow\quad x \,<\,1$
Answer: .$\displaystyle (x\,<\,1)\:\text{ or }\:(x\,>\,\tfrac{3}{2})$