$\displaystyle x\sqrt{45}+x\sqrt{72}=\sqrt{18}$

Notice how all of the quantities in square roots are divisible by 9

$\displaystyle x\sqrt{9\times 5}+x\sqrt{9\times 8}=\sqrt{9\times 2}$

$\displaystyle x\sqrt{9}\sqrt{5}+x\sqrt{9}\sqrt{4\times 2}=\sqrt{9}\sqrt{2}$

$\displaystyle 3x\sqrt{5}+3x\sqrt{4}\sqrt{2}=3\sqrt{2}$

$\displaystyle 3x\sqrt{5}+6x\sqrt{2}=3\sqrt{2}$

$\displaystyle x(3\sqrt{5}+6\sqrt{2})=3\sqrt{2}$

$\displaystyle x=\frac{3\sqrt{2}}{3\sqrt{5}+6\sqrt{2}}$

Simplifying, this becomes

$\displaystyle x=\frac{\sqrt{2}}{\sqrt{5}+2\sqrt{2}}$

$\displaystyle x\sqrt{72}+x\sqrt{108}=12$

Note that all of the square root terms share a common factor of 36, which itself is a square of 6:

$\displaystyle x\sqrt{36\times 2}+x\sqrt{36\times 3}=12$

$\displaystyle 6x\sqrt{2}+6x\sqrt{3}=12$

Factoring $\displaystyle 6x$ from both terms on the left side of the equation:

$\displaystyle 6x(\sqrt{2}+\sqrt{3})=12$

$\displaystyle 6x=\frac{12}{\sqrt{2}+\sqrt{3}}$

$\displaystyle x=\frac{2}{\sqrt{2}+\sqrt{3}}$

$\displaystyle 2x\sqrt{12}+3x\sqrt{28}=18$

Note that both $\displaystyle \sqrt{12}$ and $\displaystyle \sqrt{28}$ have a common factor of $\displaystyle 4$ and $\displaystyle 4$ is a perfect square:

$\displaystyle 2x\sqrt{4\times 3}+3x\sqrt{4\times 7}=18$

$\displaystyle 2x\cdot2 \sqrt3 +3x\cdot2\sqrt7=18$

$\displaystyle 4x\sqrt{3}+6x\sqrt{7}=18$

From here, we can factor $\displaystyle 2x$ out of both terms on the lefthand side 

$\displaystyle 2x(2\sqrt{3}+3\sqrt{7})=18$

$\displaystyle 2x(2\sqrt{3}+3\sqrt{7})=18$

$\displaystyle x=\frac{18}{2(2\sqrt{3}+3\sqrt{7})}$

$\displaystyle x=\frac{9}{2\sqrt{3}+3\sqrt{7}}$

In order to solve for $\displaystyle x$, first note that all of the square root terms on the left side of the equation have a common factor of 9 and 9 is a perfect square:

$\displaystyle x\sqrt{27} + \sqrt{63}=9$

$\displaystyle x\sqrt{9\times 3} + \sqrt{9\times 7}=9$

$\displaystyle 3x\sqrt{3} + 3\sqrt{7}=9$

Simplifying, this becomes:

$\displaystyle 3(x\sqrt{3} + \sqrt{7})=9$

$\displaystyle x\sqrt{3} + \sqrt{7}=3$

$\displaystyle x\sqrt{3} =3 -\sqrt{7}$

$\displaystyle x =\frac{3 -\sqrt{7}}{\sqrt{3}}$

To begin with, factor out the contents of the radicals.  This will make answering much easier:

$\displaystyle \sqrt{210}+\sqrt{55}=\sqrt{2*3*5*7}+\sqrt{5*11}$

They both have a common factor $\displaystyle 5$.  This means that you could rewrite your equation like this:

$\displaystyle \sqrt{5}\sqrt{2*3*7}+\sqrt{5}\sqrt{11}$

This is the same as:

$\displaystyle \sqrt{5}\sqrt{42}+\sqrt{5}\sqrt{11}$

These have a common $\displaystyle \sqrt{5}$.  Therefore, factor that out:

$\displaystyle \sqrt{5}\sqrt{42}+\sqrt{5}\sqrt{11}=\sqrt{5}(\sqrt{42}+\sqrt{11})$

These three roots all have a $\displaystyle 5$ in common; therefore, you can rewrite them:

$\displaystyle \sqrt{15}-\sqrt{20}+\sqrt{35}=\sqrt{3*5}-\sqrt{4*5}+\sqrt{7*5}$

Now, this could be rewritten:

$\displaystyle \sqrt{3*5}-\sqrt{4*5}+\sqrt{7*5}=\sqrt{5}\sqrt{3}-\sqrt{5}\sqrt{4}+\sqrt{5}\sqrt{7}$

Now, note that $\displaystyle \sqrt{4}=2$

Therefore, you can simplify again:

$\displaystyle \sqrt{5}\sqrt{3}-\sqrt{5}\sqrt{4}+\sqrt{5}\sqrt{7}=\sqrt{5}\sqrt{3}-2\sqrt{5}+\sqrt{5}\sqrt{7}$

Now, that looks messy! Still, if you look carefully, you see that all of your factors have $\displaystyle \sqrt{5}$; therefore, factor that out:

$\displaystyle \sqrt{5}\sqrt{3}-2\sqrt{5}+\sqrt{5}\sqrt{7}=\sqrt{5}(\sqrt{3}-2+\sqrt{7})$

This is the same as:

$\displaystyle \sqrt{5}(\sqrt{3}+\sqrt{7}-2)$

Examining the terms underneath the radicals, we find that $\displaystyle 128$ and $\displaystyle 192$ have a common factor of $\displaystyle 64$. $\displaystyle 64$ itself is a perfect square, being the product of $\displaystyle 8$ and $\displaystyle 8$. Hence, we recognize that the radicals can be re-written in the following manner:

$\displaystyle \sqrt{128}=\sqrt{64} \sqrt{2}$, and $\displaystyle \sqrt{192}=\sqrt{64} \sqrt{3}$.

The equation can then be expressed in terms of these factored radicals as shown:

$\displaystyle 2x\sqrt{128}-3x\sqrt{192}=16$ 

$\displaystyle \Rightarrow 2x(\sqrt64 \sqrt2)-3x(\sqrt{64} \sqrt3)=16$  

$\displaystyle \Rightarrow 2x(8)\sqrt2-3x(8)\sqrt3=16$

$\displaystyle \Rightarrow 16x\sqrt2-24x\sqrt3=16$

Factoring the common term $\displaystyle 8x$ from the lefthand side of this equation yields

$\displaystyle 8x (2\sqrt2-3x\sqrt3)=16$

Divide both sides by the expression in the parentheses:

$\displaystyle 8x=\frac{16}{2\sqrt2-3x\sqrt3}$

Divide both sides by $\displaystyle 8$ to yield $\displaystyle x$ by itself on the lefthand side:

$\displaystyle x=\frac{16}{8(2\sqrt2-3x\sqrt3)}$

Simplify the fraction on the righthand side by dividing the numerator and denominator by $\displaystyle 8$:

$\displaystyle x=\frac{2}{2\sqrt{2}-3\sqrt{3}}$

This is the solution for the unknown variable $\displaystyle x$ that we have been required to find.

When multiplying square roots, you are allowed to multiply the numbers inside the square root. Then simplify if necessary.

$\displaystyle \sqrt{7}*\sqrt{14}=\sqrt{7*14}=\sqrt{98}=\sqrt{49}*\sqrt{2}=7\sqrt{2}$

When multiplying square roots, you are allowed to multiply the numbers inside the square root. Then simplify if necessary.

$\displaystyle \sqrt{10}*\sqrt{15}=\sqrt{10*15}=\sqrt{150}=\sqrt{25}*\sqrt{6}=5\sqrt{6}$

When multiplying square roots, you are allowed to multiply the numbers inside the square root. Then simplify if necessary.

$\displaystyle \sqrt{24}*3\sqrt{8}=3\sqrt{24*8}=3\sqrt{192}=3\sqrt{64}*\sqrt{3}=24\sqrt{3}$