Let \(t=\sqrt{x}\). Then \(t\ge 0\) and \(x=t^2\).

Substitute into the equation:

\(t^2-t=6\).

Move everything to one side:

\(t^2-t-6=0\).

Factorize:

\((t-3)(t+2)=0\).

Hence \(t=3\) or \(t=-2\).

But \(t=\sqrt{x}\ge 0\), so \(t=-2\) is impossible.

Therefore, \(t=3\), and \(x=3^2=9\).

Answer: \(9\).

Both sides of the equation are non-negative, so we can square both sides:

\(x^2-10=3x\).

Move everything to one side:

\(x^2-3x-10=0\).

Factorize:

\((x-5)(x+2)=0\).

We get the candidates \(x=5\) and \(x=-2\).

Check the domain: the radicand \(3x\ge 0\), so \(x\ge 0\).

Therefore, \(x=-2\) is impossible, and only \(x=5\) remains.

Answer: \(5\).

The right-hand side is \(x\), while the left-hand side is the principal square root, so \(x\ge 0\).

Square both sides:

\(2x+8=x^2\).

Move everything to one side:

\(x^2-2x-8=0\).

Factorize:

\((x-4)(x+2)=0\).

We get \(x=4\) or \(x=-2\).

Since \(x\ge 0\), the value \(-2\) is impossible.

Check: \(\sqrt{2\cdot4+8}=\sqrt{16}=4\).

Answer: \(4\).

A product is equal to zero if at least one of its factors is zero.

We also need the square root to exist: \(2x-2\ge 0\Rightarrow x\ge 1\).

1) \(\sqrt{2x-2}=0\Rightarrow 2x-2=0\Rightarrow x=1\).

2) \(x^2-25=0\Rightarrow x=\pm5\).

Taking into account \(x\ge 1\), only \(x=5\) is valid.

So the roots are \(1\) and \(5\).

According to the instruction, we write their sum:

\(1+5=6\).

Answer: \(6\).

First find the domain:

\(2x-1\ge 0\Rightarrow x\ge \dfrac12\).

Since both sides are non-negative, square the inequality:

\(2x-1\le 9\).

Hence:

\(2x\le 10\Rightarrow x\le 5\).

So, \(\dfrac12\le x\le 5\).

The integer values are \(1,2,3,4,5\).

Their number is \(5\).

Answer: \(5\).