We need to solve the following inequalities: (a) $10 - \sqrt{2x+7} \le 3$ (b) $3 \le \sqrt{2x+5} < 6$ (c) $\sqrt{2x+9} - \sqrt{9+x} > 0$ (d) $\sqrt{2} - \sqrt{x+6} \le -\sqrt{x}$ (e) $\sqrt{x-3} > \sqrt{x+4} - 1$ (f) $3 + \sqrt{2x-7} \le 6$

1. Problem Description

We need to solve the following inequalities:

(a) $10 - \sqrt{2x+7} \le 3$

(b) $3 \le \sqrt{2x+5} < 6$

(c) $\sqrt{2x+9} - \sqrt{9+x} > 0$

(d) $\sqrt{2} - \sqrt{x+6} \le -\sqrt{x}$

(e) $\sqrt{x-3} > \sqrt{x+4} - 1$

(f) $3 + \sqrt{2x-7} \le 6$

2. Solution Steps

(a) $10 - \sqrt{2x+7} \le 3$

Subtract 10 from both sides:

$-\sqrt{2x+7} \le -7$

Multiply by -1 (and flip the inequality sign):

$\sqrt{2x+7} \ge 7$

Square both sides:

$2x+7 \ge 49$

Subtract 7 from both sides:

$2x \ge 42$

Divide by 2:

$x \ge 21$

Also, we need $2x+7 \ge 0$, so $2x \ge -7$, which means $x \ge -3.5$.

Since $x \ge 21$, the condition $x \ge -3.5$ is satisfied.

(b) $3 \le \sqrt{2x+5} < 6$

Square all parts:

$9 \le 2x+5 < 36$

Subtract 5 from all parts:

$4 \le 2x < 31$

Divide by 2:

$2 \le x < 15.5$

Also, we need $2x+5 \ge 0$, so $2x \ge -5$, which means $x \ge -2.5$.

Since $2 \le x < 15.5$, the condition $x \ge -2.5$ is satisfied.

(c) $\sqrt{2x+9} - \sqrt{9+x} > 0$

$\sqrt{2x+9} > \sqrt{9+x}$

Square both sides:

$2x+9 > 9+x$

Subtract 9 from both sides:

$2x > x$

Subtract x from both sides:

$x > 0$

Also, we need $2x+9 \ge 0$, so $2x \ge -9$, which means $x \ge -4.5$.

We also need $9+x \ge 0$, so $x \ge -9$.

Since $x > 0$, the other two conditions are satisfied.

(d) $\sqrt{2} - \sqrt{x+6} \le -\sqrt{x}$

$\sqrt{2} + \sqrt{x} \le \sqrt{x+6}$

Square both sides:

$(\sqrt{2}+\sqrt{x})^2 \le x+6$

$2 + 2\sqrt{2x} + x \le x+6$

$2\sqrt{2x} \le 4$

$\sqrt{2x} \le 2$

Square both sides:

$2x \le 4$

$x \le 2$

Also, we need $x+6 \ge 0$, so $x \ge -6$.

We also need $x \ge 0$.

So we have $0 \le x \le 2$.

(e) $\sqrt{x-3} > \sqrt{x+4} - 1$

$\sqrt{x-3} + 1 > \sqrt{x+4}$

Square both sides:

$(\sqrt{x-3}+1)^2 > x+4$

$x-3 + 2\sqrt{x-3} + 1 > x+4$

$x-2 + 2\sqrt{x-3} > x+4$

$2\sqrt{x-3} > 6$

$\sqrt{x-3} > 3$

Square both sides:

$x-3 > 9$

$x > 12$

We need $x-3 \ge 0$, so $x \ge 3$.

We need $x+4 \ge 0$, so $x \ge -4$.

Since $x>12$, all conditions are satisfied.

(f) $3 + \sqrt{2x-7} \le 6$

$\sqrt{2x-7} \le 3$

Square both sides:

$2x-7 \le 9$

$2x \le 16$

$x \le 8$

We need $2x-7 \ge 0$, so $2x \ge 7$, which means $x \ge 3.5$.

So $3.5 \le x \le 8$.

3. Final Answer

(a) $x \ge 21$

(b) $2 \le x < 15.5$

(c) $x > 0$

(d) $0 \le x \le 2$

(e) $x > 12$

(f) $3.5 \le x \le 8$