Question

"a" si "b" de la ex 30, va rog

Answer 1

a)[1/(√2+1)+1/(√3+√2)+1/(√4+√3)] ×[1/(√2-1)-1/(√3-√2)+1/(√4-√3)]=
rationalizam numitorii
=[(√2-1)/(√2+1)(√2-1)+(√3-√2)/(√3+√2)(√3-√2)+(√4-√3)/(√4+√3)(√4-√3)] × ×[(√2+1)/(√2-1)(√2+1)-(√3+√2)/(√3-√2)(√3+√2)+(√4+√3)/(√4-√3)(√4+√3)]=
=[(√2-1)/(2-1)+(√3-√2)/(3-2)+(√4-√3)/(4-3)] × [(√2+1)/(2-1)-(√3+√2)/(3-2)+
+(√4+√3)/(4-3)]=
=[(√2-1)/1+(√3-√2)/1+(√4-√3)/1] × [(√2+1)/1-(√3+√2)/1+(√4+√3)/1]=
=(√2-1+√3-√2+√4-√3) × (√2+1-√3-√2+√4+√3)=
=(√4-1) × (√4+1)=4-1=3



b)
[1/(√4+√3)+1/(√5+√4)+1/(√6+√5)+1/(√7+6)]×[1/(√7-√3)]=
=[(√4-√3)/(√4+√3)(√4-√3)+(√5-√4)/(√5+√4)(√5-√4)+(√6-√5)/(√6+√5)(√6-√5)+ +(√7-√6)/(√7+√6)(√7-√6)]×[1/(√7-√3)]=
=[(√4-√3)/(4-3)+(√5-√4)/(5-4)+(√6-√5)/(6-5)+(√7-√6)/(7-6)]×[1/(√7-√3)]=
=[(√4-√3)/1+(√5-√4)/1+(√6-√5)/1+(√7-√6)/1]×[1/(√7-√3)]=
=(√4-√3+√5-√4+√6-√5+√7-√6)×[1/(√7-√3)]=
=(√7-√3)×1/(√7-√3)=1