Question

19. Determinati valorile lui „a" din:
a) [(324 + 205 : 5 - a) : 6] : 3 + 127 - 307;
b) [676:13:(31.20 - 31-a)].45-23.91= 247;
c) {[(205:5+362: 2) -32-6+a]: 221:13} - 140:5 = 635;
d) [(30+226: 2+a:4):29+20-100): 5-11-6-5-67 = 0;
e) 31.3.{[(41.10+95: a) +11.12]-63.2}: 421-93 = 0;
A) {495: [30:(3-8 -9.a)] + 226:2-324:18} - 25.4 = 94.
20. Calculati:
a) (84: 2+5.7-9:3+3): [(36:18+6.11) -19.3] +3;
b) [(2000-1236):4+2432:2] - 469.3;
c) [(3.3+3:3+3.3:3):13+(33:3+3-33:11+11):22;
d) 2.(1020:85+205): [33.18 -(50.10+23:4)] +1343:17 - 262;
e) 25.2 +135:3. {[1030:5-60:2+(2004.0+96:3): 8]:18} +99.Dau coroana și 45 puncte ​

Answer 1

Răspuns:

a)

[ ( 324 + 205 : 5 -x ) : 6 ] * 3 + 127 = 307[(324+205:5−x):6]∗3+127=307

[ ( 324 + 41 -x ) : 6 ] * 3 = 307-127[(324+41−x):6]∗3=307−127

[ ( 365 -x ) : 6 ] * 3 = 180[(365−x):6]∗3=180

[ ( 365 -x ) : 6 ] = 180:3[(365−x):6]=180:3

( 365 -x ) : 6=60(365−x):6=60

365 -x = 60*6365−x=60∗6

365 -x =360365−x=360

x=5x=5

b)

[676:13:(31*20-31-a)]*45-23*91=247[676:13:(31∗20−31−a)]∗45−23∗91=247

[52:(620-31-a)]*45-2093=247[52:(620−31−a)]∗45−2093=247

[52:(589-a)]*45=247+2093[52:(589−a)]∗45=247+2093

\frac{52}{589-a} *45=2340

589−a

52

∗45=2340

\frac{2340}{589-a} =2340

589−a

2340

=2340

589-a= \frac{2340}{2340}589−a=

2340

2340

589-a=1=\ \textgreater \ a=589-1=\ \textgreater \ a=588589−a=1= \textgreater a=589−1= \textgreater a=588

c)

\{[(41+362:2)-32*6+a]*221:13\}-140:5=635{[(41+362:2)−32∗6+a]∗221:13}−140:5=635

\{[(41+181)-32*6+a]*221:13\}-140:5=635{[(41+181)−32∗6+a]∗221:13}−140:5=635

\{[222-32*6+a]*221:13\}-140:5=635{[222−32∗6+a]∗221:13}−140:5=635

\{[190*6+a]*221:13\}-140:5=635{[190∗6+a]∗221:13}−140:5=635

\{[1140+a]*221:13\}-140:5=635{[1140+a]∗221:13}−140:5=635

\{[1140+a]*17\}-140:5=635{[1140+a]∗17}−140:5=635

\{1140+a}*17-28=635

\{1140+a\}*11=635{1140+a}∗11=635

1140+a=635-111140+a=635−11

1140+a=6241140+a=624

a=1140-624a=1140−624

a=516a=516

d)

[(30+226:2+a:4):29+20*100]:5-11*6-5*67=0[(30+226:2+a:4):29+20∗100]:5−11∗6−5∗67=0

[(30+113+a:4):29+2000]:5-66-335=0[(30+113+a:4):29+2000]:5−66−335=0

[(143+a:4):29+2000]:5=401[(143+a:4):29+2000]:5=401

(143+a:4):29+2000=2005(143+a:4):29+2000=2005

(143+a:4):29=5(143+a:4):29=5

143+a:4=145143+a:4=145

a:4=145-143a:4=145−143

a:4=2a:4=2

a=4*2a=4∗2

a=8a=8

Explicație pas cu pas:

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