Question

Ma poate ajuta cineva cu ex 3 si ex 4 doar a si b va rog? :)

Answer 1

ex 3

a) 3 ⁻²⁰⁰¹:3⁻¹⁹⁹⁹=3⁽⁻²⁰⁰¹⁻⁽⁻¹⁹⁹⁹⁾=3⁻²⁰⁰¹⁺¹⁹⁹⁹=3⁻²=(1/3)²=1²/3²=1/9

b)(1/10000)⁴ ori 100⁻⁵ ori 10²⁶=10000⁻⁴ ori (10²)⁻⁵ ori 10²⁶=(10⁴)⁻⁴ ori 10⁻¹⁰ ori 10²⁶=10⁻¹⁶ ori 10⁻¹⁰ ori 10²⁶=10⁽⁻¹⁶⁺⁽⁻¹⁰⁾⁺²⁶⁾=10⁽⁻¹⁶⁻¹⁰⁺²⁶⁾=10⁽⁻²⁶⁺²⁶⁾=10°=1

ex 4

a)(2/5√3-1/√12+3/√75):(2√3)⁻¹=

=(2/√75-1/√12+3/√75):(√12)⁻¹=

=(2√75/75-1√12/12+3√75/75):(1/√12)=

=(5√75/75-1√12/12):(1√12/12)=

=(5×5√3/75-2√3/12):√12/12=

=(25√3/75-2√3/12):√12/12=

pe 25√3/75 o rationalizam cu 4= 100√3/300

pe 2√3/12 o rationalizam cu 25=50√3/300

=(100√3/300-50√3/300):√12/12=

=50√3/300 ori 12/√12=

50√3/300 se simplifica cu 50=√3/6

=√3/6 ori 12√12/12=

12√12/12 se simplifica cu 12=√12

=√3/6 ori √12=√3/6 ori2√3=2√3×√3/6=2×3/6=6/6=1

b) rezolvam din paranteza asta

√3×(2/√3-1)=√3×2/√3-√3×1=2-√3

rezolvam de aici

1/√2×(√6-2)=1/√2×√6-1/√2-2=√6/√2-1√2/2-2=√3-1√2/2-2×2/2=√3-1√2/2-4/2=√3-(√2-4)/2

acum exercitiul devine

=[(2-√3)+(√3-√2-4)/2]×√6⁻²=

=[(2-√3)+(√3-√2-4)/2]×1²/√6²=

=[(2-√3)+(√3-√2-4)/2]×1/6=

=(2-√3)×1/6+(√3-√2-4)/2×1/6=

=(2-√3)/6+(√3-√22+4)/12

de aici nu mai stiu