Question

17–4x[6–3x (m:4–15)]+95 = 100


*627+4x[63–17– n x3)] = 835


160-477: [5–4:(8- 0 :1)1= 1


20-[(20+ p :5)x10-10]:10= 0


5+[19x(q: 4)x5+50–1000]: 2 =5

Va rog frumos sa rezolvati prin metosa mersului invers..... Dau coroana, sa faceti si verificare sa va dea bine ca altfel nu dau coroana :) ​

Answer 1

17–4x[6–3x (m:4–15)]+95 = 100

17-4×[6–3x (m:4–15)]=5

4×[6–3x (m:4–15)]=12

6–3x (m:4–15)=3

3x (m:4–15)=3

m:4–15=1

m:4=16

m=4

627+4x[63–17– n x3)] = 835

4x[63–(17– n x3)]=208

63–(17– n x3)=52

(17– n x3)=11

n×3=6

n=2

160-477: [5–4:(8- 0 :1)]= 1. asta nu e bun ce ai scris aici

20-[(20+ p :5)x10-10]:10= 0

[(20+ p :5)x10-10]:10=20

(20+ p :5)x10-10=200

(20+ p :5)x10=210

20+ p :5=21

p:5=1

p=5

5+[19x(q: 4)x5+50–1000]: 2 =5

[19x(q: 4)x5+50–1000]: 2=0

19x(q: 4)x5+50–1000=0

19x(q: 4)x5+50=1000

19x(q: 4)x5=950

19x(q: 4)=190

q: 4=10

q=40