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Calculati suma inverselor a, b, c, unde a = √3 + √2, b = √4 + √3, c = √5 + √4

Răspuns :

inversul lui a este 1/a
1/a+1/b+1/c=1/(√3+√2)+1/(√4+√3)+1/(√5+√4)=
=1/(√3+√2)+1/(2+√3)+1/(√5+2)

Obs: 1/(√3+√2)=rationalizam=(√3-√2)/(√3+√2)(√3-√2)=(√3-√2)/(9-4)=(√3-√2)/5
1/(2+√3)=(2-√3)/(4-3)=(2-√3)
1/(√5+2)=(√5-2)/(5-2)=(√5-2)/3

avem 1/(√3+√2)+1/(2+√3)+1/(√5+2)=(√3-√2)/5+(2-√3)+(√5-2)/3=aducem la acelasi numitor=[3(√3-√2)+15(2-√3)+5(√5-2)]/15=
=(3√3-3√2+30-15√3+5√5-10)/15=(-12√3-3√2+5√5+20)/15=
=-3(4√3+√2)/15+5(√5+4)/15=(√5+4)/3-(4√+√2)/3