Question

109-{[468+(255:5x3-2)x2]:10x4-199} prima
{[(2xa-79)x2-78]x5+340}:250=4 a doua
Va rog sa ma ajutati dau coroana

Answer 1

109-{[ 468 +(51x3-2)x2]:10x4-199}=
109-{[ 468 +(153-2)x2]:10x4-199}=

109-{[ 468 +302]:10x4-199}=

109-{770:10x4-199}=

109-{77x4-199}=

109-{308-199}=

109-109=0

A doua

{[(2xa-79)x2-78]x5+340}:250=4
{[4a-158-78]x5+340}:250=4
{[4a-236]x5+340}:250=4
{20a-1180+340}:250=4
{20a-1520}:250=4
20a-1520=4*250
20a-1520=1000
20a=1000+1520
20a=2520
a= 2520:20
a= 126

Answer 2

109-{[468+(255:5x3-2)x2]:10x4-199} =0
109-{[468+(51x3-2)x2]:10x4-199} =
109-{[468+(153-2)x2]:10x4199} =
109-{[468+151x2]:10x4-199} =
109-{[468+302]:10x4-199} =
109-{770:10x4-199}=
109-{77x4-199}=
109-{308-199}=
109-109=
0


{[(2xa-79)x2-78]x5+340}:250=4
{[(2xa-79)x2-78]x5+340}=4 x250
[(2xa-79)x2-78]x5+340=1000
[(2xa-79)x2-78]x5=1000-340
[(2xa-79)x2-78]x5=660
(2xa-79)x2-78=660:5
(2xa-79)x2-78=132
(2xa-79)x2=132+78
(2xa-79)x2=210
2xa-79=210:2
2xa-79=105
2xa=105+79
2xa=184
a=184:2
a=92