Question

Ajutati-ma va rog, toate trei exercitii, dau coronita

Answer 1

3)a)  (√2-1)⁴=(√2-1)²(√2-1)²=(2-2√2+1)(2-2√2+1)=(3-2√2)(3-2√2)=
9-6√2-6√2+8=17-12√2

b) (√3-√2)⁴=(√3-√2)²(√3-√2)²=(3-2√6+2)(3-2√6+2)=(5-2√6)(5-2√6)=
=25-20√6+24=49-20√6

4) =intai rationalizam si aducem la acelasi numitor ce avem in paranteza
deci rationalizam si rezulta
{[√5(√3+√2)/5]-[√5(√3+√2)/(3-2)]}⁻²-5√6/2·6=
{(√15+√10)/5-(√15+√10)/1}⁻²-5√6/12=
aducem paranteza la acelasi numitor si rezulta
[(√15+√10-5√15-5√10)/5]⁻²-5√6/12=
[(-4√15-4√10)/5]⁻²-5√6/12=
[-5/4(√15+√10)]²-5√6/12=
[-5(√15-√10)/4(15-10)]²-5√6/12=
[(-5√15-5√10)/20]²-5√6/12=
(75+50√150+50)/400-5√6/12=
(125+5·50√6)/400-5√6/12=
(125+250√6)/400-5√6/12=aducem la acelasi numitor care este 1200
(3·125+3·250√6-100·5√6)/1200=
(375+750√6-500√6)/1200=
(375+250√6)/1200=125(3+2√6)/1200=simplif cu 5=
=25(3+2√6)/240=simplif.cu 5=5(3+2√6)/48

5) Mg=?
a=[(√2/3)-(3√2/2)]²=[(2√2-3·3√2)/6]²=[(2√2-9√2)/6]²=(-7√2/6)²=
a=98/36=simplif cu 2=49/18

b=(2·3√2/√5):(9√2/3√5)=6√2/√5:3√2/√5=6√2/√5·√5/3√2=6√2/3√2=2
b=2
Mg=√(ab)=√(49/18·2/1)=simplif cu 2=√49/9=7/3
Mg=2,(3)