Question

Va rog sa ma ajutati .

Answer 1

Răspuns:

1) ab cu bara deasupra+a+b=54

10a+b+a+b=54

11a+2b=54=>a=nr par

a, b<10  => a=2  =>b>10

a=4  =>44+2b=54 =>2b=10=> b=10:2=>b=5

ab=45 numarul cautat

45+4+5=54

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2) a+b=19

a:b=3r3=>a=3b+3

3b+3+b=19

4b=19-3

b=16:4=4

a=3·4+3=15

______________

3) a+b+c=502

b=3a=>a=b/3

c:b=7r2=>c=7b+2

b/3+b+7b+2=502 amplificam

b+3b+21b+6=1506

25b=1506-6

b=1500:25

b=60

a=60/3=20

c=7·60+2=422

Answer 2

Răspuns:

11*4+2*5 = 54

Explicație pas cu pas:

1.    10a+b+a+b=54  11a+2b=54   a=4   b=5      

2.   x+y=19   x/y=3+3/y  => y=x/3-1  

3x+x-3=19*3=57    => 4x=60   =>x=15   y=6

3.   x+y+z=502    y=3x      z/y = 7+2/y  => z=7y+2 =  21x+2

    x+3x+21x+2=502    25x=500   x=20; y=60;  z= 7*60+2=422