If $\gcd(a,b) = 1$ and $a,b\mid x$ then $ab\mid x$. If $\gcd(a,b) = 1$ and $a,b\mid x$ then $ab\mid x$.
My attempt at answering the question:
\begin{align*}
x &\equiv 0 \pmod{a}\\\
&\Longrightarrow x\text{ is divisible by $a$}\\\
&\Longrightarrow x = ma\text{ for some integer $m$}\\\
\ \\\
x &\equiv 0 \pmod{b}\\\
&\Longrightarrow  x\text{ is divisible by $b$}\\\
&\Longrightarrow x = mb\text{ for some integer $m$}\\\
\ \\\
x^2 &= (ma)(mb)\\\
x^2 &= (m^2)(ab)\\\
x &= \sqrt{m^2ab}\\\
x &= m\sqrt{a}\sqrt{b}
\end{align*}
Let $m$ be $k\sqrt{a}\sqrt{b}$. Then
\begin{align*}
x &= kab\\\
&\Longrightarrow x \equiv 0 \pmod{ab}
\end{align*}
Is this correct, if not can someone point me in the right direction?
 A: $\gcd(a,b)=1$ so there are integers $s$ and $t$ such that $sa+tb=1$, whence $sax+tbx=x$. But $a$ divides $x$ so $x=ax'$ for some integer $x'$. As $b$ divides $x$, $x=bx''$ for some integer $x''$. Substituting on the left-hand side of the preceding equation yields $sabx''+tabx'=x$. Factor out the common $ab$ on the left to get $ab(ax''+tx'')=x$, so that $ab$ divides $x$.
A: No. This is not correct.
1. Integers $m_1$ and $m_2$ in $x=m_1 a$ and $x=m_2 b$ can be different.
2. In the beginning of your proof all numbers are integers. However when you write "let m be k*sqrt(a)*sqrt(b)", you don't know, that k is integer too.  
Right direction: use, the fact, that each number $y$ can be uniquely represented as $p_1^{\alpha_1} p_2^{\alpha_2}\dots p_n^{\alpha_n}$, where $p_i$ are primes and $\alpha_i$ are integers.
A: Since $\rm\ \color{#C00}{\gcd(a,b) = 1},\,$ by Euclid's Lemma, $\rm\ \ \color{#c00}{a\mid b}(x/b)\ \Rightarrow\ a\mid x/b\ \Rightarrow\ ab\mid x$
Alternatively $\rm\ \ b,a\:|\:x\ \Rightarrow\ ab\: |\: ax,bx\ \Rightarrow\ ab\ |\ gcd(ax,bx)\ =\ x\ \color{#c00}{gcd(a,b)}\ =\ x $
This is the special case $\rm\ gcd(a,b) = 1\ $ of $\rm\ gcd(a,b)\ lcm(a,b)\ =\ ab\ $ which has a similar proof which has an elegant view in terms of cofactor reflection.
See also the LCM Universal Property $\ a,b\mid x\iff {\rm lcm}(a,b)\mid x$
By induction this generalizes to
$\qquad$ if  $\,a_i\,$ are pair-coprime then $\, a_1,\cdots a_k\mid x\Rightarrow a_1\cdots a_k \mid x$
i.e. lcm = product for pair coprimes (see also here)
A: You are given that $a$ divides $x$. Therefore, you can write $x = ma$ where m is an integer. You are also given that $b$ divides $x$. This implies that $b$ divides $ma$. But $b$ and $a$ are coprime. Therefore $b$ must divide $m$. So you can write $m = kb$ where $k$ is another integer. Therfore, you have $x = kab$.
