$\sqrt{49x-49}-\sqrt{36x-36}=3\sqrt{2}_{}$ $ĐK:_{}$ $x_{}$ $\geq1$
$⇔7\sqrt{x-1}-\sqrt{36.(x-1)}=3\sqrt{2}_{}$
$⇔7\sqrt{x-1}-\sqrt{36}.\sqrt{x-1}=3\sqrt{2}_{}$
$⇔7\sqrt{x-1}-6\sqrt{x-1}=3\sqrt{2}_{}$
$⇔\sqrt{x-1}=3\sqrt{2}_{}$
$⇔x-1=(3\sqrt{2})^2_{}$
$⇔x-1=18_{}$
$⇔x=19_{}$ $(TMĐK)_{}$
Vậy phương trình trên có nghiệm x = 19