Question

ajutor dau coroana rpd pls

ex 10​

Answer 1

Răspuns:

Mg(a,b) = rad(a*b)

a) x = (rad(2)-1)(3+rad(5))(rad(7)-rad(6))(4 + rad(7))

y = (rad(2)+1)(3-rad(5))(rad(7)+rad(6))(4-rad(7))

Mg(x, y) =

= rad(xy)

Reordonam factorii in mod convenabil. Ne folosim de formula (a+b)(a-b) = a^2-b^2

(rad(2)-1)(rad(2)+1) = 2-1 = 1

(3+rad(5))(3-rad(5)) = 9-5 = 4

(rad(7)-rad(6))(rad(7)+rad(6)) = 7-6 = 1

(4+rad(7))(4-rad(7)) = 16-7 = 9

Mg(x, y) = rad(9*4) = 3*2 = 6

De acum scriu r() in loc de rad().

b) x = r(7-4r(3))

Inauntrul radicalului se observa un binom, in care nr vor fi a si b.

a^2 + b^2 = 7

2ab = 4r(3)

ab = 2r(3)

Prin incercari se afla a=2, b=r(3)

x = r((2-r(3))^2)

x = |2 - r(3)|

2 > r(3) =>

x = 2 - r(3)

Analog pt y, a.î. avem y = 2 + r(3)

Mg(x, y) =

= r(xy) =

= r((2-r(3))(2+r(3))) =

= r(4 - 3) =

= r(1) =

= 1

c) x = r((3+2r(3))^2)

x = |3 + 2r(3)|

x = 3 + 2r(3)

y se bazeaza pe aceeasi teorie ca la punctul b.

a^2 + b^2 = 21

2ab = 12r(3)

ab = 6r(3)

a = 3, b = 2r(3)

y = r((2r(3)-3)^2)

y = |2r(3) - 3|

2r(3) > 3

y = 2r(3) - 3

x = 2r(3)+3

xy = 12 - 9 = 3

Mg(x,y) = rad(xy) = rad(3)