Question

Rezolcați ecuațiile : f)12+[104÷26+2×(3×5-x)-12÷2]÷2=15​

Answer 1

12+[4+2*(15-x)-6]:2=15

12+(4+30-2x-6):2=15

12+(28-2x):2=15

12+14-x=15

26-x=15

x=26-15

x=11

Answer 2

$f)12+[104÷26+2×(3×5-x)-12÷2]÷2=15$

$12+[4 + 2 \times (15 - x) - 6] \div 2 = 15$

$12 + (4 + 30 - 2x - 6) \div 2 = 15$

$12+ (28 - 2x) \div 2 = 15$

$12 + \frac{28 - 2x}{2} = 15$

$12 + \frac{2(14 - x)}{2} = 15$

$12 + 14 - x = 15$

$26 - x = 15$

$- x = 15 - 26$

$- x = - 11$

$x = 11$