$\sqrt{49a^4{b}^2}\backslash sqrt\{49a^4b^2\}$

This expression is not defined whenever there is a negative number under the root.

Both exponents of $aa$ and $bb$ are even, so $a^{4}a^4$ and $b^{2}b^2$ are always positive or zero. Thus, the expression under the root is also positive or zero and the root is always defined:

$D_{\max }=\mathrm{R}D\_\{\backslash max\}=ℝ$

Now, take the root.

$\sqrt{49a^4{b}^2}=7\cdot a^2\cdot \mid b\mid \backslash sqrt\{49a^4b^2\}=7\backslash cdot\; a^2\backslash cdot\backslash left|b\backslash right|$

You have to put dashes, since $bb$ could be negative.

$\sqrt{{(-b)}^2}\backslash sqrt\{\backslash left(-b\backslash right)^2\}$

The root is net defined, if the expression under the root is negative. Now, squares of any number (also of $-b-b$) are always positive of zero, so there are never any problems with the definition of the root.

$D_{\max }=\mathrm{R}D\_\{\backslash max\}=ℝ$

Square $-b-b$ .

${\sqrt{-b}}^2\backslash sqrt\{-b\}^\{\backslash ;2\}$

The number under the root must not be negative. $\Rightarrow \backslash ;\backslash ;\backslash Rightarrow\backslash ;\backslash ;$ Only negative numbers or 0 are allowed for $bb$.

$D_{\max }={\mathrm{R}}_0^{-}D\_\{\backslash max\}=ℝ\_0^-$

${\sqrt{-b}}^2\backslash sqrt\{-b\}^\{\backslash ;2\}$$=-b=-b$

So finally, power and root cancel.

$\sqrt{{(1-2x)}^2}\backslash sqrt\{\backslash left(1-2x\backslash right)^2\}$

The number under the root must not be negative. However, the term in the squared bracket is always positive or zero, so we never have any problem with taking the root.

${\mathrm{D}}_{\mathrm{m}\mathrm{a}\mathrm{x}}=\mathbb{R}\{\backslash mathrm\; D\}\_\backslash mathrm\{max\}=\backslash mathbb\{R\}$

$\sqrt{{(1-2x)}^2}=\mid 1-2x\mid \backslash sqrt\{\backslash left(1-2x\backslash right)^2\}=\backslash left|1-2x\backslash right|$

Take the root. Root and square cancel each other out, but since the term in the bracket can also be negative and could be made positive by squaring, you have to put an absolute value.

$\sqrt{{(x-y)}^2}\backslash sqrt\{\backslash left(x-y\backslash right)^2\}$

The number under the root must not be negative. However, squaring makes a term non-negative, so there is never a problem wit the root being defined.

${\mathrm{D}}_{\mathrm{m}\mathrm{a}\mathrm{x}}=\mathbb{R}\{\backslash mathrm\; D\}\_\backslash mathrm\{max\}=\backslash mathbb\{R\}$

$\sqrt{{(x-y)}^2}=\mid x-y\mid \backslash sqrt\{\backslash left(x-y\backslash right)^2\}=\backslash left|x-y\backslash right|$

Take the root. Root and square cancel each other out, but since the term in the bracket can also be negative and could be made positive by squaring, you have to put an absolute value.

$\sqrt{x^2+y^2}\backslash sqrt\{x^2+y^2\}$

The number under the root must not be negative. But since both squares are positive or zero, we never get any definition problems with the root.

${\mathrm{D}}_{\mathrm{m}\mathrm{a}\mathrm{x}}=\mathbb{R}\{\backslash mathrm\; D\}\_\backslash mathrm\{max\}=\backslash mathbb\{R\}$

$\sqrt{x^2+y^2}=\sqrt{x^2+y^2}\backslash sqrt\{x^2+y^2\}=\backslash sqrt\{x^2+y^2\}$

It is not possible to further simplify this term. It is already in the simplest possible form.

$\sqrt{x^2\cdot y^2}\backslash sqrt\{x^2\backslash cdot\; y^2\}$

Both $x^{2}x^2$ and $y^{2}y^2$ are never zero. The same holds for the product, so there are never any problems with definition of the square root.

$\to \mathbb{D}=\mathbb{R}\backslash rightarrow\; \backslash mathbb\{D\}=\; \backslash mathbb\{R\}$

Now, $\sqrt{x^2\cdot y^2}=\sqrt{{(x\cdot y)}^2}\backslash sqrt\{x^2\backslash cdot\; y^2\}=\backslash sqrt\{\backslash left(x\backslash cdot\; y\backslash right)^2\}$. Take the root. Root and square cancel each other out, but it is necessary to retain absolute values, since $xx$ or $yy$ might be negative and turned positive by the squaring.

$=\mid x\cdot y\mid =\backslash left|x\backslash cdot\; y\backslash right|$