Question

Sa se afle x din egalitatea;((203+202×203)÷(203×204-203))×x=111×90÷(111+222+333+....+999)

Answer 1

[(203+202×203)÷(203×204-203)]×x=111×90÷(111+222+333+....+999) =

[ 203×(1+202)): 203×(204-1)]×x = 111×90 : [111×(1+2+3+...+9)]

[(203 × 203) : (203× 203)]×x = 9 990 : (111× 9 × 10 : 2)

1 × x = 9 990 : (111 × 9 × 5)

x = 9 990 : 4 995

x = 2

Answer 2

203²:203²  * x=111*90/(111 * (1+2+...+9))

x=90/45
 x=2
 as simple as that!



Obs
am tinut cont ca 203+202*203=203(1+202)=203*203=203²
si ca 204*203-203=203(204-1)=203*203=203²
sica 1+2+..+9=9*10/2=45b (suma Gauss)