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

Toán Lớp 6: Bài17:Tìm x a)2x + 3 chia hết cho x b)8x + 4 chia hết cho 2x – 1 c)x^2 – x + 5x + 1 chia hết cho x – 1

Home/ Toán Lớp 6: Bài17:Tìm x a)2x + 3 chia hết cho x b)8x + 4 chia hết cho 2x – 1 c)x^2 – x + 5x + 1 chia hết cho x – 1

Toán Lớp 6: Bài17:Tìm x a)2x + 3 chia hết cho x b)8x + 4 chia hết cho 2x – 1 c)x^2 – x + 5x + 1 chia hết cho x – 1

Tháng Mười Một 6, 2022 Toán học 2 Comments 0 views

Toán Lớp 6: Bài17:Tìm x
a)2x + 3 chia hết cho x
b)8x + 4 chia hết cho 2x – 1
c)x^2 – x + 5x + 1 chia hết cho x – 1

Comments ( 2 )

  1. diemchau
    diemchau
    6 Tháng Mười Một, 2022 at 10:55 sáng
    Reply
    $\text{a, 2x+3 $\vdots$ x}$
    $\text{⇔    3  $\vdots$  x}$
    $\text{Hay x $\in$ Ư(3)={±1;±3}}$
    $\text{Vậy x $\in$ {±1;±3}}$
    $\text{b, 8x + 4 $\vdots$ 2x – 1}$
    $\text{⇔8x – 4 + 8 $\vdots$ 2x – 1}$
    $\text{⇔      8  $\vdots$ 2x – 1}$
    $\text{Hay 2x-1 $\in$ Ư(8)={±1;±2;±4;±8}}$
    $\text{Ta có bảng sau :}$
    \begin{array}{|c|c|c|}\hline \text{2x-1}&\text{-8}&\text{-4}&\text{-2}&\text{-1}&\text{1}&\text{2}&\text{4}&\text{8}\\\hline \text{x}&\text{$\frac{-7}{2}$ }&\text{$\frac{-3}{2}$ }&\text{$\frac{-1}{2}$ }&\text{0}&\text{1}&\text{$\frac{3}{2}$ }&\text{$\frac{5}{2}$ }&\text{$\frac{9}{2}$ }\\\hline \text{Kết quả}&\text{(L)}&\text{(L)}&\text{(L)}&\text{(TM)}&\text{(TM)}&\text{(L)}&\text{(L)}&\text{(L)}\\\hline\end{array}
    $\text{    Vậy x $\in$ {0;1}}$
    $\text{c, $x^{2}$ – x + 5x + 1 $\vdots$ x-1}$
    $\text{⇔$x^{2}$ -2x +6x + 1 $\vdots$ x-1}$
    $\text{⇔$(x-1)^{2}$+6x $\vdots$ x-1}$
    $\text{⇔  6x -6 +6 $\vdots$ x-1}$
    $\text{⇔    6 $\vdots$ x-1}$
    $\text{Hay x-1 $\in$ Ư(6)={±1;±2;±3;±6}}$
    $\text{Ta có bảng sau :}$
    \begin{array}{|c|c|c|}\hline \text{x-1}&\text{-6}&\text{-3}&\text{-2}&\text{-1}&\text{1}&\text{2}&\text{3}&\text{6}\\\hline \text{x}&\text{-5}&\text{-2}&\text{-1}&\text{0}&\text{2}&\text{3}&\text{4}&\text{7}\\\hline \text{Kết quả}&\text{(TM)}&\text{(TM)}&\text{(TM)}&\text{(TM)}&\text{(TM)}&\text{(TM)}&\text{(TM)}&\text{(TM)}\\\hline\end{array}
    $\text{  Vậy x $\in$ {-5;-2;-1;0;2;3;4;7}}$
    CHO MK CTLHN NHA:(

  2. hoquyly
    hoquyly
    6 Tháng Mười Một, 2022 at 10:54 sáng
    Reply
    $\\$
    Bổ sung : x ∈ ZZ
    a,
    2x + 3 \vdots x
    Vì x \vdots x
    -> 2x \vdots x
    -> 3 \vdots x
    -> x ∈ Ư (3) = {1;-1;3;-3}
    -> x ∈ {1;-1;3;-3} (Tm)
    Vậy x ∈ {1;-1;3;-3} để 2x+3 \vdots x
    b,
    8x + 4 \vdots 2x-1
    -> 8x -4 + 8 \vdots 2x-1
    -> 4 (2x-1) + 8 \vdots 2x-1
    Vì 2x-1 \vdots 2x-1
    -> 4 (2x-1) \vdots 2x-1
    -> 8 \vdots 2x-1
    ->2x-1 ∈ Ư (8)={1;-1;2;-2;4;-4;8;-8}
    -> 2x ∈ {2;0;3;-1; 5;-3; 9;-7}
    -> x ∈ {1;0;3/2; (-1)/2; 5/2; (-3)/2; 9/2;(-7)/2}
    Vì x ∈ ZZ
    -> x ∈ {1;0}
    Vậy x ∈ {1;0} để 8x+4 \vdots 2x-1
    c,
    x^2 -x + 5x+1 \vdots x-1
    -> x^2 – x + 5x-5 + 6 \vdots x-1
    -> (x^2 – x) + (5x-5) + 6 \vdots x-1
    -> x (x-1) + 5 (x-1) + 6 \vdots x-1
    -> (x-1) (x+5) + 6 \vdots x-1
    Vì x-1 \vdots x-1
    -> (x-1) (x+5) \vdots x-1
    -> 6 \vdots x-1
    ->x-1 ∈ Ư (6)={1;-1;2;-2;3;-3;6;-6}
    -> x∈{2;0;3;-1;4;-2;7;-5} (Tm)
    Vậy x ∈ {2;0;3;-1;4;-2;7;-5} để x^2 -x+5x+1 \vdots x-1
     

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

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