Вопрос

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Мысль

Solving inequality

Looks like we're working with two inequalities. The first has square roots and powers of 2 and 3. I’m simplifying terms and ensuring everything is valid. For example, the expression inside the square roots simplifies to absolute values. I’ll need to carefully check whether squaring leads us in the right direction further. Let me move forward with this simplification.

Simplifying the second inequality

For the second part of the inequality, I’m examining the term $2 \cdot 3^{x+2}$ . It simplifies to $18 \cdot 3^x$ , which is close, but not fully a perfect square. I notice I can adjust it with a constant to get a more manageable form with square roots. Now, I'm focusing on how to simplify the combination of square roots while keeping the inequality valid.

Final steps for inequality

I’ve got the simplified inequality now: $(2^x - 5) + \sqrt{(9^x - 2 \cdot 3^{x+2} + 17)} \leq 2^x - 5$ . The square root must be zero, leading to the equation $9^x - 2 \cdot 3^{x+2} + 17 = 0$ . Solving this reveals that $x = 0$ or $x \approx 2.578$ .

Solving for #5 and #6

For inequality #5, I verified $x = \log_3 17$ is the only valid solution, since $x = 0$ doesn't satisfy the inequality.

Now, for #6, I’m trying to work with an expression of powers of 5 and the square root term. I’m considering substitution but it’s tricky with the powers and roots not lining up neatly. I'll keep tweaking the structure.

Solving inequality #6

The inequality simplifies to

Thought for 2m 17s

Ответ

Решим оба неравенства.

5.

\sqrt{4^x-5\cdot2^{x+1}+25}+\sqrt{9^x-2\cdot3^{x+2}+17}\le 2^x-5.

Первая подкоренная:

4^x-5\cdot2^{x+1}+25=(2^x-5)^2 \;\Rightarrow\; \sqrt{\cdot}=|2^x-5|.

Правая часть неотрицательна ⇒ $2^x-5\ge0 \Rightarrow x\ge \log_2 5$ . Тогда $|2^x-5|=2^x-5$ , и неравенство превращается в

(2^x-5)+\sqrt{9^x-2\cdot3^{x+2}+17}\le 2^x-5 \;\Rightarrow\; \sqrt{9^x-2\cdot3^{x+2}+17}\le0.

Следовательно

9^x-2\cdot3^{x+2}+17=0.

Положив $t=3^x>0$ , получим

t^2-18t+17=0 \Rightarrow t\in\{1,17\}.

Отсюда $x=0$ или $x=\log_3 17$ . Но $x\ge \log_2 5$ , поэтому остаётся

\boxed{x=\log_3 17}.

6.

5^{2x-10-3\sqrt{x-2}}-4\cdot5^{x-5}<5^{1+3\sqrt{x-2}},\qquad x\ge2.

Пусть $s=\sqrt{x-2}\ge0$ , $a=(x-5)-\tfrac32 s$ , $u=5^{a}>0,\; v=5^{\frac32 s}>0$ . Тогда

5^{2x-10-3s}=u^2,\qquad 5^{x-5}=uv,\qquad 5^{1+3s}=5v^2.

Неравенство перепишется:

u^2-4uv<5v^2 \;\;\Leftrightarrow\;\; (u-5v)(u+v)<0.

Так как $u+v>0$ , имеем $u-5v<0\Rightarrow u<5v$ . Это равносильно

5^{(x-5)-\tfrac32 s}<5\cdot5^{\tfrac32 s}\;\;\Leftrightarrow\;\; 5^{x-5-3s}<5\;\;\Leftrightarrow\;\; x-3s<6.

Подставим $x=s^2+2$ :

s^2-3s-4<0 \;\;\Leftrightarrow\;\; (s-4)(s+1)<0 \;\;\Rightarrow\;\; 0\le s<4.

Значит $x=s^2+2\in[2,18)$ .

Ответ:

\boxed{\text{№5: } x=\log_3 17;\qquad \text{№6: } x\in[2,18).}