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

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Răspunsuri

2014-03-19T15:47:57+02:00
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