Given: 2√5+√3+1√3+√2−3√5+√2
Rationalising the denominators,
=2(√5−√3)(√5+√3)(√5−√3)+√3−√2(√3+√2)(√3−√2)
−3(√5−√2)(√5+√2)(√5−√2)
=2(√5−√3)(√5)2−(√3)2+√3−√2(√3)2−(√2)2−3(√5−√2)(√5)2−(√2)2
[ using identity, a2−b2=(a−b)(a+b) ]
=2(√5−√3)5−3+√3−√23−2−3(√5−√2)5−2
=2(√5−√3)2+√3−√21−3(√5−√2)3
=√5−√3+√3−√2−(√5−√2)
=√5−√3+√3−√2−√5+√2
=0