
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời. a) \(B=3+3^2+3^3+...+3^{120}\) \(B=3\cdot1+3\cdot3+3\cdot3^2+...+3\cdot3^{119}\) \(B=3\cdot\left(1+3+3^2+...+3^{119}\right)\) Suy ra B chia hết cho 3 (đpcm) b) \(B=3+3^2+3^3+...+3^{120}\) \(B=\left(3+3^2\right)+\left(3^3+3^4\right)+\left(3^5+3^6\right)+...+\left(3^{119}+3^{120}\right)\) \(B=\left(1\cdot3+3\cdot3\right)+\left(1\cdot3^3+3\cdot3^3\right)+\left(1\cdot3^5+3\cdot3^5\right)+...+\left(1\cdot3^{119}+3\cdot3^{119}\right)\) \(B=3\cdot\left(1+3\right)+3^3\cdot\left(1+3\right)+3^5\cdot\left(1+3\right)+...+3^{119}\cdot\left(1+3\right)\) \(B=3\cdot4+3^3\cdot4+3^5\cdot4+...+3^{119}\cdot4\) \(B=4\cdot\left(3+3^3+3^5+...+3^{119}\right)\) Suy ra B chia hết cho 4 (đpcm) c) \(B=3+3^2+3^3+...+3^{120}\) \(B=\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+\left(3^7+3^8+3^9\right)+...+\left(3^{118}+3^{119}+3^{120}\right)\) \(B=\left(1\cdot3+3\cdot3+3^2\cdot3\right)+\left(1\cdot3^4+3\cdot3^4+3^2\cdot3^4\right)+...+\left(1\cdot3^{118}+3\cdot3^{118}+3^2\cdot3^{118}\right)\) \(B=3\cdot\left(1+3+9\right)+3^4\cdot\left(1+3+9\right)+3^7\cdot\left(1+3+9\right)+...+3^{118}\cdot\left(1+3+9\right)\) \(B=3\cdot13+3^4\cdot13+3^7\cdot13+...+3^{118}\cdot13\) \(B=13\cdot\left(3+3^4+3^7+...+3^{118}\right)\) Suy ra B chia hết cho 13 (đpcm) (-4;-3;-2;-1;0;1;2;3;4) Ko có dấu ngoặc nhọn nên mik xài ngoặc tròn nha TL ; A = { x E N / 0 ;1 ; 2 ; 3 ; 4 ; 5 } B = { x E N / 0 ; 1 ; 2 ; 3 } C = { x E N / 0 ; 1 } D = { x E N / 0 ; x ; y } Chúc bạn học tốt nhé ! \(A=\left\{36;48;60;72\right\}\) \(B=\left\{0;15;30;45;60;75;90\right\}\) \(C=\left\{12;18\right\}\) \(D=\left\{1;3;9\right\}\) A={36;48;60;72} B={0;15;30;45;60;75;90} C= {18;12} D={1;3;9} hok tốt nha!! \(a,\frac{6}{10}=\frac{3}{5};\frac{6}{16}=\frac{3}{8};-\frac{15}{20}=\frac{3}{4};-\frac{10}{30}=\frac{-1}{3}\) \(b,\frac{42}{28}=\frac{3}{2};\frac{54}{-21}=-\frac{10}{7};-\frac{27}{33}=-\frac{9}{11};\frac{25}{14}=\frac{25}{14}\) \(c,\frac{125}{1000}=\frac{1}{8};\frac{198}{126}=\frac{11}{7};\frac{3}{243}=\frac{1}{81};\frac{103}{2090}=\frac{103}{2090}\) \(d,\frac{2.3}{9.14}=\frac{28}{3}\)
