Question

1 supra 2:{(1-1 supra 3-1 supra 4):[1 supra 5+4×(1 supra 6+1 supra 3)]+1 supra 12}=

E urgent dau coroana

Answer 1

Răspuns:

$\frac{1}{2}$:{(1-$\frac{1}{3}$ - $\frac{1}{4}$):[$\frac{1}{5}$+4×($\frac{1}{6}$+$\frac{1}{3}$)]+$\frac{1}{12}$}

    numitorul comun 12, amplifici $\frac{1}{4}$ cu 3, $\frac{1}{3}$ cu 4 si 1 cu 12

    la a doua paranteza rotunda amplifici $\frac{1}{3}$ cu 2,numitorul comun fiind 6

$\frac{1}{2}$:[$\frac{5}{12}$:($\frac{1}{5}$+4×$\frac{1}{2}$)+$\frac{1}{12}$]

simplifici 4 cu 2 prin 2

$\frac{1}{2}$:[$\frac{5}{12}$:($\frac{1}{5}$+2)+$\frac{1}{12}$]

$\frac{1}{2}$:($\frac{5}{12}$:$\frac{11}{5}$+$\frac{1}{12}$)

$\frac{1}{2}$:($\frac{5}{12}$×$\frac{5}{11}$+$\frac{1}{12}$)

$\frac{1}{2}$:($\frac{25}{132}$+$\frac{1}{12}$)

amplifica 12 cu 11

$\frac{1}{2}$:$\frac{3}{11}$

$\frac{1}{2}$×$\frac{11}{3}$

$\frac{11}{6}$