Question

CUM REZOLVI IN DOUA MODUIERI? 4X(3+2+4) SI 5X(9-3-5) FIE NR A,B SI C DACA:a=9x7-2x4-5x10;98-(7x7+3x7+5x3);c=(13+7x4)-39 calculeaza 3xa+3xc-b;b+4xa-2xc;(a+c)x9-b ofer coroana

Answer 1

la prima intrebare ai asa : I 4*9=36 sau II 4*3+4*2+4*4=12+8+16=20+16+36 ..... la cealalta ai I 5*1= 5 ( faci diferenta in paranteza si inmultesti dupa cu 5 din fata ) sau a II 5*9-5*3-5*5=45-15-25=30-25=5

A doua problema b= 98-(7*7+.....)? Nu inteleg :))

Answer 2

4 * 9 = 36  sau 4 * 3 + 4 * 2 + 4 * 4 = 12 + 8 + 16 = 36
5 * 1 = 5  sau 5 * 9 + 5 * ( - 3) + 5 * (- 5) = 45 - 15 - 25 = 5

a = 63 - 8 - 50 = 5
b = 98 - (49 + 21 + 15) = 98 - 85 = 13
c = (13 + 28) - 39 = 41 - 39 = 2

3a + 3c - b = 3 * 5 + 3 * 2 - 13 = 15 + 6 - 13 = 8
b + 4a - 2c = 13 + 4 * 5 - 2 * 2 = 13 + 20 - 4 = 29
(a + c) * 9 - b = (5 + 2) * 9 - 23 =63 - 13 = 50