Question

Determinat x din egalitatea 0,1 (x)+0.2 (x)+.....+0.9(x)=4.7

Answer 1

Ne ajutam de formula lui Gauss

1+2+3+...+9=9x(9+1):2=9x10:2=90:2=45

0,1 (x)+0.2 (x)+.....+0.9(x)=4.7
x/10+2x/10+...+9x/10=4,7
x/10(1+2+...+9)=4,7
x(1+2+...+9)=4,7x10
x*45=47
x=47:45
x=1,0(4)