Question

Trebuie sa calculam rationalizand mai intai numitorii
Sunt exercitii de clasa a 8 a va rog cn e bun la matematica sa ma ajute!
Multumesc Anticipat! :)

Answer 1

Ca sa rationalizezi numitori trebuie sa amplifici cu radicalul de aflat la numitor.
De exemplu √7/3√3 il amplifici cu √3 si iti da √7×√3/3×3=√21/9.
Asa le poti rezolva pe toate.

Answer 2

c) √7/3√3=(3√3·√7)/(3√3)²=3√21/9·3=simplif.cu3=√21/9asa se calculeaza, restul o sa le fac direct si calculezi tu
√3/2√7=√21/4·7=√21/28
2/√21=2√21/21
√21/9+√21/28-2√21/21= aducem la acelasi numitor=
=(28√21+9√21-2·12√21)/252=13√21/252
d)(1+2√3)/2√2=[2√2(1+2√3)]/4·2=(2√2+4√6)/8=2(√2+2√6)/8=(√2+2√6)/4
(1-√3)/3√2=[3√2(1-√3)/9·2=(3√2-3√6)/18=3(√2-√6)/18=(√2-√6)6
1/6√2=6√2/36·2=√2/12
(√2+2√6)/4-(√2-√6)/6-√2/12=(3√2+6√6-2√2+2√6-√2)/12=8√6/12=4√6/6=2√6/3
e)
(√3-√5)/√27=(√3-√5)/3√3=(3·3-13√15)/9·3=(9-3√15)/27=3(3-√15)/27=(3-√15)/9
(√3+√5)/√12=(√3+√5)/2√3=(6+2√15)4·3=2(3+√15)/12=(3+√15)/6
5√5/√108=5√5/6√3=30√15/36·3=5√15/18
(3-√15)/9-(3+√15)/6+5√15/18=(6-2√15-9-3√15+5√15)/18=(-3)/18=-1/6
f)
(√10+1)/20=(√10+1)/2√5=(2√50+2√5)/4·5=(10√2+2√5)/20=2(5√2+√5)/20=(5√2+√5)/10
(1-√10)/√50=(1-√10)/5√2=(5√2-5√20)/25·2=(5√2-10√5)/50=5(√2-2√5)/50=
(√2-2√5)/10
(√5-√2)/√10=(√50-√20)/10=(5√2-2√5)/10
(5√2+√5)/10+(√2-2√5)/10-(5√2-2√5)/10=(5√2+√5+√2-2√5-5√2+2√5)/10=
=(√2+√5)/10