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222-9+11+12:2*14+14 = ? ( )

Toán Lớp 6: Bài 5 : Tìm a ∈ Z để : ( 6a + 1 ) ⋮ ( 3a – 1 )

Toán Lớp 6: Bài 5 : Tìm a ∈ Z để :
( 6a + 1 ) ⋮ ( 3a – 1 )

Comments ( 2 )

  1. Lời giải và giải thích chi tiết:
    (6a + 1) \vdots (3a – 1)
    ⇒ 2(3a – 1) + 3 \vdots (3a – 1)
    ⇒ 3 \vdots (3a – 1)
    ⇒ 3a – 1 ∈ Ư_((3)) = {-3 ; -1 ; 1 ; 3}
    ⇒ a = {0}
    Vậy  a = 0

  2. Answer
    $\text{Ta có:}$ 6a + 1 \vdots 3a – 1
    $\text{Mà:}$ 3a – 1 \vdots 3a – 1
    => 2 . (3a – 1) \vdots 3a – 1
    => 6a – 2 \vdots 3a – 1
    $\text{Lập hiệu}$
    (6a + 1) – (6a – 2)
    = 6a + 1 – 6a + 2
    = (6a – 6a) + (1 + 2)
    = 3 \vdots 3a – 1
    => 3a – 1 \in Ư(3} = {+-1 ; +-3}
    $\text{Lập bảng giá trị:}$
    $\begin{array}{|c|c|c|}\hline 3a-1&1&3&-1&-3\\\hline a&\dfrac{2}{3}&\dfrac{4}{3}&0&\dfrac{-2}{3}\\\hline \text{Thử lại}&\text{KTM}&\text{KTM}&\text{TM}&\text{KTM}\\\hline\end{array}$
    $\text{Vậy}$ a = 0
     

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222-9+11+12:2*14+14 = ? ( )

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