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

Toán Lớp 6: 1 cmr a=9^3 + 3^12 + 3^10 chia hết cho 37 cmr b=2^7 + 4^3 + 8^2 chia hết cho 4 Tìm a;b biết: a)4a5b chia hết cho 2;3;5;9 b)a25b chia hế

Toán Lớp 6: 1
cmr a=9^3 + 3^12 + 3^10 chia hết cho 37
cmr b=2^7 + 4^3 + 8^2 chia hết cho 4
Tìm a;b biết:
a)4a5b chia hết cho 2;3;5;9
b)a25b chia hết cho 2;5;9
c)7ab5b chia hết cho 5 và 9
d)2a3a5 chia hết cho 3
e)3a6a chia hết cho 5 thannks

Comments ( 1 )

  1. Giải đáp:
    $2)\\ a) \left\{\begin{array}{l} b =0 \\ a= 9 \end{array} \right. \\ b) \left\{\begin{array}{l} b=0\\ a=2 \end{array} \right. \\ c) \left[\begin{array}{l}  \left\{\begin{array}{l}  b=0 \\ a=6  \end{array} \right.  \\ \left\{\begin{array}{l}  b =5 \\ a=5  \end{array} \right.  \end{array} \right. \\ d) a \in \{1;4;7\}\\ e)  a \in \{0;5\}$
    Lời giải và giải thích chi tiết:
    $1)\\ a=9^3 + 3^12 + 3^10\\ =(3^2)^3 + 3^12 + 3^10\\ =3^{2.3} + 3^12 + 3^10\\ =3^6 + 3^12 + 3^10\\ =3^3(3^3 + 3^9 + 3^7)\\ =27(3^3 + 3^9 + 3^7) \ \vdots \ 27\\ b=2^7 + 4^3 + 8^2\\ =2^7 + (2^2)^3 + (2^3)^2\\ =2^7 + 2^{2.3} + 2^{3.2}\\ =2^7 + 2^6 + 2^6\\ =2^2(2^5 + 2^4 + 2^4)\\ =4(2^5 + 2^4 + 2^4) \ \vdots \ 4\\ 2)\\ a)\left\{\begin{array}{l} \overline{4a5b} \ \vdots \ 2 \\\overline{4a5b} \ \vdots \ 3 \\ \overline{4a5b} \ \vdots \ 5\\ \overline{4a5b} \ \vdots \ 9 \end{array} \right. \\ \Leftrightarrow \left\{\begin{array}{l} b \ \vdots \ 2 \\ (4+a+5+b) \ \vdots \ 3 \\ b \ \vdots \ 5\\ (4+a+5+b) \ \vdots \ 9 \end{array} \right. \\ \Leftrightarrow \left\{\begin{array}{l} b \in\{0;2;4;6;8\} \\ (9+a+b) \ \vdots \ 3 \\ b \in \{0;5\}\\ (9+a+b) \ \vdots \ 9 \end{array} \right. \\ \Leftrightarrow \left\{\begin{array}{l} b =0 \\ a+b \ \vdots \ 3 \\ a+b \ \vdots \ 9 \end{array} \right. \\ \Leftrightarrow \left\{\begin{array}{l} b =0 \\ a \ \vdots \ 3 \\ a \ \vdots \ 9 \end{array} \right. \\ \Leftrightarrow \left\{\begin{array}{l} b =0 \\ a= 9 \end{array} \right. \\ b)\left\{\begin{array}{l} \overline{a25b} \ \vdots \ 2 \\ \overline{a25b} \ \vdots \ 5\\ \overline{a25b} \ \vdots \ 9 \end{array} \right. \\ \Leftrightarrow \left\{\begin{array}{l} b \ \vdots \ 2 \\ b \ \vdots \ 5\\ (a+2+5+b) \ \vdots \ 9 \end{array} \right. \\ \Leftrightarrow \left\{\begin{array}{l} b=0\\ (a+7+b) \ \vdots \ 9 \end{array} \right. \\ \Leftrightarrow \left\{\begin{array}{l} b=0\\ a+7 \ \vdots \ 9 \end{array} \right. \\ \Leftrightarrow \left\{\begin{array}{l} b=0\\ a=2 \end{array} \right. \\ c)\left\{\begin{array}{l} \overline{7ab5b} \ \vdots \ 5\\ \overline{7ab5b} \ \vdots \ 9 \end{array} \right. \\ \Leftrightarrow \left\{\begin{array}{l} b \ \vdots \ 5\\ (7+a+b+5+b) \ \vdots \ 9 \end{array} \right. \\ \Leftrightarrow \left\{\begin{array}{l} b \in \{0;5\} \\ (12+a+2b) \ \vdots \ 9 \end{array} \right. \\ \Leftrightarrow \left[\begin{array}{l} \left\{\begin{array}{l} b=0 \\ (12+a+2b) \ \vdots \ 9 \end{array} \right. \\ \left\{\begin{array}{l} b =5 \\ (12+a+2b) \ \vdots \ 9 \end{array} \right. \end{array} \right. \\ \Leftrightarrow \left[\begin{array}{l} \left\{\begin{array}{l} b=0 \\ (12+a) \ \vdots \ 9 \end{array} \right. \\ \left\{\begin{array}{l} b =5 \\ (22+a) \ \vdots \ 9 \end{array} \right. \end{array} \right. \\ \Leftrightarrow \left[\begin{array}{l} \left\{\begin{array}{l} b=0 \\ a=6 \end{array} \right. \\ \left\{\begin{array}{l} b =5 \\ a=5  \end{array} \right. \end{array} \right. \\ d)\overline{2a3a5} \ \vdots \ 3\\ \Leftrightarrow (2+a+3+a+5) \ \vdots \ 3\\ \Leftrightarrow (10+2a) \ \vdots \ 3\\ \Leftrightarrow a \in \{1;4;7\}\\ e)\overline{3a6a} \ \vdots \ 5\\ \Leftrightarrow a \ \vdots \ 5\\ \Leftrightarrow a \in \{0;5\}$

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