$\sqrt{\frac{x-8}{1388}}+\sqrt{\frac{x-7}{1389}}+\sqrt{\frac{x-6}{1390}}=\sqrt{\frac{x-1388}{8}}+\sqrt{\frac{x-1389}{7}}+\sqrt{\frac{x-1390}{6}}$ How many solutions the following equation has in real numbers
?$$\sqrt{\frac{x-8}{1388}}+\sqrt{\frac{x-7}{1389}}+\sqrt{\frac{x-6}{1390}}=\sqrt{\frac{x-1388}{8}}+\sqrt{\frac{x-1389}{7}}+\sqrt{\frac{x-1390}{6}}$$
$1)1\quad\quad\quad\quad\quad\quad\quad2)2\quad\quad\quad\quad\quad\quad\quad3
 )3\quad\quad\quad\quad\quad\quad\quad4)4\quad\quad\quad\quad\quad5)\text{zero}$

Obviously we have $x\ge1390$. starting from $x=1390$ we can see LHS is very close to $1+1+1=3$ and RHS is close to zero. for higher values of $x$ we can see right side of the equation grows much faster than left side. so it seems the equation has not answer in real numbers.
I'm not sure how to proceed mathematically. I can rewrite it as follow:
$$\left(\sqrt{\frac{x-8}{1388}}-\sqrt{\frac{x-1388}{8}}\right)+\left(\sqrt{\frac{x-7}{1389}}-\sqrt{\frac{x-1389}{7}}\right)+\left(\sqrt{\frac{x-6}{1390}}-\sqrt{\frac{x-1390}{6}}\right)=0$$
I putted fractions with similar numbers in the same parentheses so it looks better now, but don't know how to continue.
Write $x=1396+t$. Then
$$\sqrt{1+\frac t{1388}}\:+\sqrt{1+\frac t{1389}}\:+\sqrt{1+\frac t{1390}}$$
$$=\sqrt{1+\frac t8}\:+\sqrt{1+\frac t7}\:+\sqrt{1+\frac t6}.$$
Clearly $t=0$ is one solution. Now, $t$ cannot be positive, because that would make the terms on the LHS each smaller than the terms on the RHS. Similarly, $t$ cannot be negative, for the opposite reason. Therefore $t=0$ or $x=1396$ is the unique solution.
The easiest but absurd way to do this is by squaring and squaring and squaring and..... To write out the equation of polynomial in $x$
$$\sqrt{\frac{x-8}{1388}}+\sqrt{\frac{x-7}{1389}}+\sqrt{\frac{x-6}{1390}}-\sqrt{\frac{x-1388}{8}}-\sqrt{\frac{x-1389}{7}}-\sqrt{\frac{x-1390}{6}} = 0$$
With the help of a cas, then we would have to observe the polynomial in $x$ to see how many real roots it has
$$( 166271366055421398492452275393603941230766862954455137380647248140804636783475714327462432993480052899307496588105274122048146156780410637021361 \cdot x^6 - 1420140036782905319437561361639895711607700448706531408943829995316272533767501640030686352288245638342331131237991828557619781758027571151457168696 \cdot x^5 + 5053362485747062046204356842284861702806221919004319273920353418806545439134535332974351841718762407835957590697696948539773086891685978542298184835440 \cdot x^4 - 9589047098047637215433435077525811816830831747049585209634085291233160531678815568240751343497601637440011806189213827409633807381969946623370920034699520 \cdot x^3 + 10233876003972506215993189306355354430848302521410622385212634701164151435860483711557613487835971006731199476091003849691449303854798610976036031554449411840 \cdot x^2 - 5824396568911083719726988788037339915692680410889570986360670278090638276801032058096056401851613752306491371113890184533047015490065483972816831803846901282816 \cdot x + 1381013855402990033245846225695823841737429380881009697772601168066108621988351698959635167325016829640478323081178371168331795718011254016938645598239579277398016 )\cdot (x-1396)= 0 $$
The polynomial is $7$th degree, and has $4$ real roots
EDIT:
But squaring an equation creates an environment for false root.... so all roots are tested, so that only $x-1396 = 0$ is a real root that satisfies the equation