Question

Daca x+ y+ z =17calculati
5*x +5*y+5*z

13*x+13*y+13*z-1313

882-3*(7*x+7*y+7*)

2. Stiind ca a+b=6 determinati nr x din 4*a+3*x+4*b=
3. Identificand un factor comun in fiecare suma calculati.
10+20+30+.......+1230=

7+14+21+.......+315=

Answer 1

1 5(x +y +z)= 5*17=75
13(x+y+z)-1313=13*17-1313=-1092
882-3*(7*17)=882-3*119=882-357=525
2.4(a+b)+3x 4*6+3x+=24+3x (mai lipseste ceva_)
3 10(1+2+3+....+123)=5*124*123=76260
7(1+2+3+...+45)=7*45*23=7245

Answer 2

$1).x+y+z=17 \\ 5x+5y+5z=5(x+y+z)=5*17=85 \\ 13x+13y+13z-1313=13(x+y+z)-1313= \\ =13*17-1313=221-1313=-1092 \\ 882-3(7x+7y+7z)=882-3[7(x+y+z)]= \\ =882-3(7*17)=882-3*119=882-357=525$
$2).a+b=6 \\ 4a+3x+4b=66 \\ 4(a+b)+3x=66 \\ 4*6+3x=66 \\ 24+3x=66 \\ 3x=66-24 \\ 3x=42 \\ x= \frac{42}{3} \\ x=14$
$3a).10+20+30+...+1230=10(1+2+3+...+123)= \\ =10* \frac{123(123+1)}{2} =10* \frac{123*124}{2} =10* \frac{15252}{2} =10*7626=76260 \\ b).7+14+21+...+315=7(1+2+3+...+45)= \\ =7* \frac{45(45+1)}{2} =7* \frac{45*46}{2} =7* \frac{2070}{2} =7*1035=7245$