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

Toán Lớp 6: Bài 5:tìm n ∈ N biết 1, 7 chia hết cho n-2 2, n+2 chia hết cho n-4 3, 2n+5 chia hết cho n+1 4, n+4 chia hết cho 2n+3 5, 2n+9 chia hết c

Tháng Một 22, 2023 Toán học 2 Comments 0 views

Toán Lớp 6: Bài 5:tìm n ∈ N biết
1, 7 chia hết cho n-2
2, n+2 chia hết cho n-4
3, 2n+5 chia hết cho n+1
4, n+4 chia hết cho 2n+3
5, 2n+9 chia hết cho 4n+3
6, n²+n+4 chia hết cho n+1

Comments ( 2 )

  1. tuminh25
    tuminh25
    22 Tháng Một, 2023 at 12:36 sáng
    Reply
    Đáp án:
     
    Giải thích các bước giải:
     

    toan-lop-6-bai-5-tim-n-n-biet-1-7-chia-het-cho-n-2-2-n-2-chia-het-cho-n-4-3-2n-5-chia-het-cho-n

  2. thuminhthong
    thuminhthong
    22 Tháng Một, 2023 at 12:36 sáng
    Reply
    Giải đáp:
    $1)n \in \{1;3;9\}\\ 2) n \in \{1;2;3;5;6;7;10\}\\ 3) n \in \{0;2\}\\ 4) n \in \{1\}\\ 5) n \in \{0;3\}\\ 6) n \in \{0;1;3\}$
    Lời giải và giải thích chi tiết:
    $1)\\ 7 \ \vdots \ (n-2)\\ \Rightarrow (n-2) \in Ư(7)\\ \Leftrightarrow (n-2) \in \{\pm 1; \pm 7\}\\ \Rightarrow n \in \{-5;1;3;9\}\\ n \in \mathbb{N}\\ \Rightarrow n \in \{1;3;9\}\\ 2)\\ (n+2) \ \vdots \ (n-4)\\ \Rightarrow \dfrac{n+2}{n-4} \in \mathbb{Z}\\ =\dfrac{n-4+6}{n-4} \in \mathbb{Z}\\ =1+\dfrac{6}{n-4} \in \mathbb{Z}\\ \Rightarrow \dfrac{6}{n-4} \in \mathbb{Z}\\ \Rightarrow (n-4) \in Ư (6)\\ \Leftrightarrow (n-4) \in \{\pm 1 ; \pm 2 ;\pm 3; \pm 6\}\\ \Rightarrow n \in \{-2;1;2;3;5;6;7;10\}\\ n \in \mathbb{N} \Rightarrow n \in \{1;2;3;5;6;7;10\}\\ 3)\\ (2n+5) \ \vdots \ (n+1)\\ \Rightarrow \dfrac{2n+5}{n+1} \in \mathbb{Z}\\ =\dfrac{2n+2+3}{n+1} \in \mathbb{Z}\\ =2+\dfrac{3}{n+1} \in \mathbb{Z}\\ \Rightarrow \dfrac{3}{n+1} \in \mathbb{Z}\\ \Rightarrow (n+1) \in Ư (3)\\ \Leftrightarrow (n+1) \in \{\pm 1; \pm 3\}\\ \Rightarrow n \in \{-4;-2;0;2\}\\ n \in \mathbb{N} \Rightarrow n \in \{0;2\}\\ 4)\\ (n+4) \ \vdots \ (2n+3)\\ \Rightarrow 2(n+4) \ \vdots \ (2n+3)\\ \Rightarrow \dfrac{2(n+4)}{2n+3} \in \mathbb{Z}\\ =\dfrac{2n+3+5}{2n+3} \in \mathbb{Z}\\ =1+\dfrac{5}{2n+3} \in \mathbb{Z}\\ \Rightarrow \dfrac{5}{2n+3} \in \mathbb{Z}\\ \Rightarrow (2n+3) \in Ư (5)\\ \Leftrightarrow (2n+3) \in \{\pm 1; \pm 5\}\\ \Rightarrow n \in \{-4;-2;-1;1\}\\ n \in \mathbb{N} \Rightarrow n \in \{1\}\\ \text{Thử lại thấy n=1 thoả mãn}\\ 5)\\ (2n+9 ) \ \vdots \ (4n+3)\\ \Rightarrow 2(2n+9) \ \vdots \ (4n+3)\\ \Rightarrow \dfrac{2(2n+9)}{4n+3} \in \mathbb{Z}\\ =\dfrac{4n+3+15}{4n+3} \in \mathbb{Z}\\ =1+\dfrac{15}{4n+3} \in \mathbb{Z}\\ \Rightarrow \dfrac{15}{4n+3} \in \mathbb{Z}\\ \Rightarrow (4n+3) \in Ư (15)\\ \Leftrightarrow (4n+3) \in \{\pm 1; \pm 3; \pm 5\; \pm 15\}\\ \Rightarrow n \in \left\{-\dfrac{9}{2};-2;-\dfrac{3}{2};-1;-\dfrac{1}{2};0;\dfrac{1}{2};3\right\}\\ n \in \mathbb{N} \Rightarrow n \in \{0;3\}\\ \text{Thử lại thấy $n \in \{0;3\}$ thoả mãn}\\ 6)\\ (n^2+n+4) \ \vdots \ (n+1)\\ \Rightarrow \dfrac{n^2+n+4}{n+1} \in \mathbb{Z}\\ =\dfrac{n(n+1)+4}{n+1} \in \mathbb{Z}\\ =n+\dfrac{4}{n+1} \in \mathbb{Z}\\ \Rightarrow \dfrac{4}{n+1} \in \mathbb{Z}\\ \Rightarrow (n+1) \in Ư (4)\\ \Leftrightarrow (n+1) \in \{\pm 1; \pm 2; \pm 4\}\\ \Rightarrow n \in \{-5;-3;-2;0;1;3\}\\ n \in \mathbb{N} \Rightarrow n \in \{0;1;3\}$

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

Thanh Hường

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