Question

gaseste valoarea lui x care face adevarata fiecare relatie :     
a)  (x ori 2 -40) :23=12 
b) [(x-416:4+2)x5+3]x2=1986
c) (72x4-32x9)x(1985x1986+1987x1988-1989x1990)+x=1000
d) 5 ori x-2000:50- (1291-1091)=40

Answer 1

a)$\frac{2x-40}{23} =12$
$2x-40=276$
$2x=316$
$x=158$

b)$[( \frac{x-416}{4} +2)*5+3]*2=1986$
$( \frac{x-416}{4} +2)*5+3=993$
$( \frac{x-416}{4} +2)*5=990$
$\frac{x-416}{4} +2=198$
$\frac{x-416}{4} =196$
$x-416=49$
$x=465$

c) 72*4-32*9=288-288=0 ⇒ (72*4-32-9)*(1985*1986*1987*1988-1989*1990)=0 obtinem x=1000

d) $1291-1091=200$
Obtinem  $\frac{5x-2000}{50} -200=40$
$\frac{5x-2000}{50} =240$
$2x-2000= 2000$
$2x=4000$
$x=2000$