Question

buna ajutati ma va rog cu aceasta...Dau 60 de puncte punctele 6 7 8​

Answer 1

Răspuns:

6.

16+[a-(38:2-a:7)-10]=19

16+[a-(19-1/7 ×a)-10]=19

16+(a-19+1/7 ×a-10)=19

16+(8/7 ×a-29)=19

16+8/7 ×a-29=19

-13+8/7 ×a=19

-91+8a=133

8a=133+91

8a=224

a=224:8

a=28

7.

42:[c+(c×2-3)-25]=3

42:[c+(2c-3)-25]=3

42:(c+2c-3-25)=3

42:(3c-28)=3

42/3c-28 =3

42=3(3c-28)

14=3c-28

-3c= -28-14

-3c= -42 |:(-3)

c=14

8.

a-5×4+a:12=136

a-20+ 1/12 ×a=136

13/12 ×a -20=136

13a-240=1632

13a= 1632+240

13a=1872

a= 1872:13

a= 144

Sper ca te-am ajutat :)

Answer 2

Exercițiu 6

$16+(a -( 38\div 2 - a\div 7) - 10)=19$

$(a - (19 - a \div 7) - 10) = 19 - 16$

$a - (19 - a \div 7) = 3 + 10$

$a - 19 + a \div 7 = 13$

$a + a \div 7 = 13 + 19$

$a + a \div 7 = 32$

$7a + a = 32 \cdot 7$

$8a = 224$

$a = 224 \div 8$

$\boxed{a = 28}$

Exercițiul 7

$42 \div (c + (c \cdot 2 - 3) - 25) = 3$

$42 = 3 \cdot(c + (2c - 3) - 25)$

$42 = 3 \cdot(c + 2c - 3 - 25)$

$42 = 3 \cdot(3c - 28)$

$42 = 9c - 84$

$42 + 84 = 9c$

$126 = 9c$

$c = 126 \div 9$

$\boxed{c = 14}$

Exercițiul 8

$a - 5 \times 4 + a \div12 = 136$

$a - 20 + a \div 12 = 136$

$a + a \div 12 = 136 + 20$

$12a + a = 156 \cdot12$

$13a = 1872$

$a = 1872 \div 13$

$\boxed{a = 144}$