Question

ratonalizati numitorii

Answer 1

Amplifici cu conjugatul la fiecare

La primul cu sqrt de 5 + sqrt de 2

Answer 2

Răspuns

E foarte mult.

Explicație pas cu pas:

Se inmulteste cu CONJUGATA (semnul invers). Invata!

(a-b)(a+b)=a²-b²  si se scapa de radical.

A.  3/(√5-√2)=3×(√5+√2)/ [(√5-√2)(√5+√2)] =

=3(√5+√2)/(5-2)=√5+√2

B. 25/(6+√6)=25×(6-√6) / [(6-√6)(6+√6)]=

=25×(6-√6) / (36-6)= 5×(6-√6)/6= 5-  5√6 /6

C. 16/(7√2 +4√6)=16×(7√2 -4√6) /[(7√2-4√6)(7√2+4√6)=

=16×(7√2-4√6) /(98-96)=8×(7√2-4√6)

D. 3√6 /(3√2+√6)=3√2 /(√3 +1)=3√2 ×(√3-1) /[(√3 -1)(√3 +1)]=

=(3√6 -3√2)/(3-1)=(3/2)(√6-√2)

E. 12/(3√2-2√3)=12×(3√2+2√3) / [(3√2-2√3)(3√2+2√3)]

=12×(3√2+2√3)/(18-12)=6√2 +4√3

F. 18/(2√3-3)=18×(2√3+3) /[(2√3+3)(2√3-3)=

=18×(2√3+3) /(12-9)=12√3+18

G. 3/(2√7 -3√3)=3×(2√7 +3√3) /[(2√7+3√3)(2√7-3√3)]=

=3×(2√7 +3√3) /(28-27)=6√7 +9√3

H. (2-√2) /(6-3√2)=(2-√2)/[3×(2-√2)]=1/3

I. (√2-1)/(√2+1)=(√2-1)²/[(√2-1)(√2+1)]=

=(√2-1)²/1= 2-2√2+1=3-2√2

J. (2√3 +√2)/(√3+√2)=(2√3+√2)(√3-√2)/[(√3-√2)(√3+√2)]=

=(2√9+2√6+√6+2) /(3-2)=8+3√6

K. (5+√5)/(10+2√5)=(5+√5) /[2×(5+√5)]=1/2=0.5

L. (√3-√2)/(√3+√2)=(√3-√2)²/ [(√3-√2)(√3+√2)]=

=(3-2√6+2)/1= 5-2√6.