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PenhaFerreira1
@PenhaFerreira1
December 2019
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A )Determine a. de modo que x=1 seja raíz da equação
(6a-1)x² - (2a + 5) x + 3 =
B) 3 /x- 2 = 2/x-3
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eliviamaia
(6a-1)x² -(2a+5)x +3=0 para x=1
(6a-1)1² -(2a+5)1 +3=0
6a-1 - 2a-5 +3=0
4a -3 =0
4a=3
a=3/4
b)
3 2
----- = ----- multiplica em cruz
x-2 x-3
3(x-3)=2(x-2) aplica distributiva
3x-9=2x-4
3x-2x=-4+9
x=5
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Thanks 1
PenhaFerreira1
muito obrigada pela atenção
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(6a-1)1² -(2a+5)1 +3=0
6a-1 - 2a-5 +3=0
4a -3 =0
4a=3
a=3/4
b)
3 2
----- = ----- multiplica em cruz
x-2 x-3
3(x-3)=2(x-2) aplica distributiva
3x-9=2x-4
3x-2x=-4+9
x=5