Task 1: abs(7-3x) + 4x > 5
Task 2: sqrt(7-3x) <= x -1
Task 3: sqrt(5-2x) > 1 - x
Task 4: x^2*(1 + 4x^2) < abs(x - 2x^2) + 6 + 4x^3
If 7- 3x< 0 (that is, if x> 7/3) this is 3x- 7+ 4x> 5. Can you solve that? If $\displaystyle x\ge 7/3$ so that $\displaystyle 7x- 3\ge 0$, this is $\displaystyle 7- 3x+ 4x> 5$. Can you solve that?
$\displaystyle \sqrt{7- 3x}$ is positive and if a< b, then $\displaystyle a^2< b^2$ so this is the same as [itex]7- 3x\le (x- 1)^2[/itex]. Can you solve that?Task 2: sqrt(7-3x) <= x -1
Same thing as task 2.]Task 3: sqrt(5-2x) > 1 - x
As in task 1, consider the cases $\displaystyle x- 2x^2\ge 0$ and $\displaystyle x- 2x^2< 0$ separately. The equation $\displaystyle x- 2x^2= x(1- 2x)= 0$ has solutions x= 0 and x= 1/2. If x= -1, $\displaystyle x- 2x^2= -1- 2= -3< 0$ so $\displaystyle x- 2x^2< 0$ for all x < 0. If x= 1/4, then $\displaystyle x- 2x^2= 1/2- 2/16= 4/8- 1/8= 3/8> 0$ so $\displaystyle x- 2x^2> 0$ for all x between 0 and 1/2. Finally, if x= 2, $\displaystyle x- 2x^2= 2- 8= -6< 0$ so that $\displaystyle x- 2x^2< 0$ for all x> 1.Task 4: x^2*(1 + 4x^2) < abs(x - 2x^2) + 6 + 4x^3
That is, there are two cases:
1) x< 0 or x> 1/2. In this case, $\displaystyle x- 2x^2< 0$ so $\displaystyle |x- 2x^2|= 2x^2- x$ and the inequality becomes $\displaystyle x^2*(1 + 4x^2) < 2x^2- x + 6 + 4x^3$ which is the same as $\displaystyle 4x^4+ x^2< 4x^3+ 2x^2- x+ 6$ which is itself the same as $\displaystyle 4x^4- 4x^3+ x^2+ x- 6< 0$. The simplest way to solve that is to find the solutions to $\displaystyle 4x^4- 4x^3+ x^2+ x- 6= 0$ and check points between those solutions to see where they are "> 0" and where "< 0".
2) 0< x< 1/2. In this case, $\displaystyle x- 2x^2> 0$ so $\displaystyle |x- 2x^2|= x- 2x^2$ and the inequality becomes $\displaystyle x^2*(1 + 4x^2) < x- 2x^2 + 6 + 4x^3$ which is the same as $\displaystyle 4x^4+ x^2< 4x^3- 2x^2+ x+ 6$ which is itself the same as $\displaystyle 4x^4- 4x^3+ 3x^2- x- 6< 0$.