Solve the equation (1)/(x)x sqrt(x2 x5)((1)/(sqrt(x)))3 + (2)/(sqrt(x))x [2sqrt(x3)-((1)/(sqrt(x)))⁻³] + x(x3sqrt(x))^-(2)/(7) = 4.

The first term is $\frac{1}{x} \times \sqrt{\frac{x^2 \sqrt{x^5}}{(\frac{1}{\sqrt{x}})^3}}$.
First, convert all terms to powers of $x$:
$\sqrt{x^5} = x^{\frac{5}{2}}$
$\frac{1}{\sqrt{x}} = x^{-\frac{1}{2}}$
$(\frac{1}{\sqrt{x}})^3 = (x^{-\frac{1}{2}})^3 = x^{-\frac{3}{2}}$

Substitute these into the expression under the square root:

$$\frac{x^2 \cdot x^{\frac{5}{2}}}{x^{-\frac{3}{2}}} = \frac{x^{2+\frac{5}{2}}}{x^{-\frac{3}{2}}} = \frac{x^{\frac{4}{2}+\frac{5}{2}}}{x^{-\frac{3}{2}}} = \frac{x^{\frac{9}{2}}}{x^{-\frac{3}{2}}}$$

Now, apply the division rule for exponents ($a^m/a^n = a^{m-n}$):

$$x^{\frac{9}{2} - (-\frac{3}{2})} = x^{\frac{9}{2} + \frac{3}{2}} = x^{\frac{12}{2}} = x^6$$

So, the square root becomes $\sqrt{x^6} = (x^6)^{\frac{1}{2}} = x^3$.
Finally, multiply by $\frac{1}{x}$:

$$\frac{1}{x} \times x^3 = x^{-1} \times x^3 = x^{3-1} = x^2$$

The first term simplifies to $x^2$.

The second term is $\frac{2}{\sqrt{x}} \times \left[2\sqrt{x^3} - \left(\frac{1}{\sqrt{x}}\right)^{-3}\right]$.
Convert terms to powers of $x$:
$\frac{2}{\sqrt{x}} = 2x^{-\frac{1}{2}}$
$\sqrt{x^3} = x^{\frac{3}{2}}$
$\left(\frac{1}{\sqrt{x}}\right)^{-3} = (x^{-\frac{1}{2}})^{-3} = x^{\frac{3}{2}}$

Substitute these into the expression:

$$2x^{-\frac{1}{2}} \times \left[2x^{\frac{3}{2}} - x^{\frac{3}{2}}\right]$$

Simplify the expression inside the brackets:

$$2x^{\frac{3}{2}} - x^{\frac{3}{2}} = (2-1)x^{\frac{3}{2}} = x^{\frac{3}{2}}$$

Now, multiply:

$$2x^{-\frac{1}{2}} \times x^{\frac{3}{2}} = 2x^{-\frac{1}{2}+\frac{3}{2}} = 2x^{\frac{2}{2}} = 2x^1 = 2x$$

The second term simplifies to $2x$.

The third term is $x(x^3\sqrt{x})^{-\frac{2}{7}}$.
First, simplify the term inside the parenthesis:

$$x^3\sqrt{x} = x^3 \cdot x^{\frac{1}{2}} = x^{3+\frac{1}{2}} = x^{\frac{6}{2}+\frac{1}{2}} = x^{\frac{7}{2}}$$

Now, raise this to the power of $-\frac{2}{7}$:

$$(x^{\frac{7}{2}})^{-\frac{2}{7}} = x^{\frac{7}{2} \times (-\frac{2}{7})} = x^{-1}$$

Finally, multiply by $x$:

$$x \cdot x^{-1} = x^{1-1} = x^0 = 1$$

The third term simplifies to 1.

The original equation was $T_1 + T_2 + T_3 = 4$.
Substituting the simplified terms:

$$x^2 + 2x + 1 = 4$$

Rearrange the equation into a standard quadratic form:

$$x^2 + 2x + 1 - 4 = 0$$ $$x^2 + 2x - 3 = 0$$

Factor the quadratic equation:

$$(x+3)(x-1) = 0$$

This gives two possible solutions:
$x+3 = 0 \implies x = -3$
$x-1 = 0 \implies x = 1$

For the terms involving $\sqrt{x}$ in the original equation to be defined in real numbers, $x$ must be greater than 0.
If $x = -3$, terms like $\sqrt{x}$ (e.g., $\sqrt{-3}$) are not real numbers. Therefore, $x=-3$ is not a valid solution.
If $x = 1$, all terms are well-defined.
Let's verify $x=1$:
$1^2 + 2(1) + 1 = 1 + 2 + 1 = 4$. This matches the right side of the equation.

Thus, the only valid solution is $x=1$.