[tex]\sqrt[3]{4-x} -\sqrt[3]{x} \geq \sqrt[3]{4+2x}[/tex]

Предыдущий
вопрос Следующий
вопрос

First, let's rewrite the inequality using the fact that the cube root is the same as raising to the power of 1/3:

tex^{1/3} - x^{1/3} \geq (4+2x)^{1/3}[/tex]

Next, we can raise both sides of the inequality to the power of 3 to get rid of the cube roots:

[tex]4-x - 3\sqrt[3]{(4-x)x} - 3\sqrt[3]{x(4-x)} - x \geq 4+2x[/tex]

Now, move all terms involving cube roots to one side and simplify:

[tex]-2x - 2\sqrt[3]{x(4-x)} \geq 0[/tex]

At this point, it is not straightforward to directly solve this inequality algebraically. One way to approach this problem is by trial and error or by using graphical methods to determine the interval where the inequality holds true.