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(x – 1)/4 – (3x – 5)/2 = x
segue que
(x – 1)/4 – 2(3x – 5)/4 = x
(x – 1)/4 – (6x – 10)/4 = x
(x – 1 – 6x + 10)/4 = x
x – 1 – 6x + 10 = 4x
–1 + 10 = 4x +6x – x
9 = 9x
9/9 = x
1 = x
x = 1
Portanto
S = {1}
(x – 1)/4 – (3x – 5)/2 = x
segue que
(x – 1)/4 – 2(3x – 5)/4 = x
(x – 1)/4 – (6x – 10)/4 = x
(x – 1 – 6x + 10)/4 = x
x – 1 – 6x + 10 = 4x
–1 + 10 = 4x +6x – x
9 = 9x
9/9 = x
1 = x
x = 1
Portanto
S = {1}
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