1. Solve the inequality

$$
2^{2 x} \leq 3 \cdot 2^{x+\sqrt{x}}+4 \cdot 2^{2 \sqrt{x}}
$$
1. The equation makes sense under the condition $x \geq 0$.

2 points Then we have:

$$
\begin{gathered}
4 \cdot 2^{2 \sqrt{x}} + 3 \cdot 2^{x + \sqrt{x}} - 2^{2 x} \geq 0 \\
4 \cdot 2^{2 \sqrt{x}} + 4 \cdot 2^{x + \sqrt{x}} - 2^{x + \sqrt{x}} - 2^{2 x} \geq 0 \\
4 \cdot 2^{\sqrt{x}} (2^{\sqrt{x}} + 2^{x}) - 2^{x} (2^{\sqrt{x}} + 2^{x}) \geq 0 \\
(2^{\sqrt{x}} + 2^{x}) (4 \cdot 2^{\sqrt{x}} - 2^{x}) \geq 0
\end{gathered}
$$

Since $2^{\sqrt{x}} + 2^{x} > 0$ for all $x \in \mathbf{R}$, we need to solve

$$
\begin{gathered}
4 \cdot 2^{\sqrt{x}} - 2^{x} \geq 0 \\
2^{\sqrt{x} + 2} \geq 2^{x} \\
\sqrt{x} + 2 \geq x \\
x - \sqrt{x} - 2 \leq 0 \\
(\sqrt{x} + 1)(\sqrt{x} - 2) \leq 0
\end{gathered}
$$

From this, it follows that $\sqrt{x} \in [-1, 2]$, or $\sqrt{x} \in [0, 2]$, so $x \in [0, 4]$ due to the condition $x \geq 0$.

5 points