Cho \(A=\frac{1\cdot3\cdot5\cdot...\cdot995\cdot997}{4\cdot6\cdot8\cdot...\cdot998\cdot1000};B=\frac{2\cdot4\cdot6\cdot...\cdot996\cdot998}{5\cdot7\cdot9\cdot...\cdot999\cdot1001}\)So sánh A và B
Những câu hỏi liên quan
Tính nhanh:
\(987654321\cdot\frac{1+1\cdot2+2\cdot3+3\cdot4+4\cdot5+5\cdot6+6\cdot7+7\cdot8+8\cdot9+9\cdot...\cdot999+999\cdot1000+1000}{1000+1000\cdot1001+1001\cdot1002+1002\cdot1003+1003\cdot1004+1004\cdot1005+1005\cdot...\cdot9999+9999\cdot10000+10000}\)
\(\sqrt[2]{4\cdot9\frac{8}{8}+\frac{48\cdot11+5}{1\cdot\frac{814}{5+\frac{6145}{1\cdot\frac{821}{614}}}}}2548-\frac{8452}{14\cdot\frac{58}{96\cdot\frac{41}{\frac{24}{1\cdot\frac{975545}{1421+\frac{84874}{\frac{1+2+3+4+5+6+7+8+9\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8\cdot9}{2\cdot\frac{2}{1}}}}}}}}\)
tính nhanh:
\(\frac{1\cdot2\cdot3+2\cdot4\cdot6+3\cdot6\cdot9+4\cdot8\cdot12}{1\cdot3\cdot4+4\cdot6\cdot8+6\cdot9\cdot12+8\cdot12\cdot16}\)
dấu \(\cdot\)là dấu nhân nha!
=\(\frac{6\left(1+8+27+64\right)}{12\left(1+16+54+128\right)}\)
=\(\frac{6.100}{12.199}\)
=\(\frac{50}{199}\)
Tk mình với nha mọi người!!!!!
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\(\frac{1x2x3+2x4x6+3x6x9+4x8x12}{1x3x4+4x6x8+6x9x12+8x12x16}\)
\(\frac{6x\left(1+8+27+64\right)}{12x\left(1+16+54+128\right)}=\frac{6x100}{12x199}=\frac{50}{199}\)
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\(41\sqrt[9^1]{8\sqrt[2]{\frac{12}{2.85\frac{1\cdot2+3\cdot4+5\cdot6+7\cdot8+9\sqrt[4]{16}}{2\cdot\frac{12}{2}\sqrt{4^2}-7^2}}}4\cdot5\cdot6\cdot7\cdot8\cdot9}\)
Ô phép tính khủng. Cái này do bạn chế ra à !
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Bài 1:Tìm x
\(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{\left(2\cdot x+1\right)\cdot\left(2\cdot x+3\right)}=\frac{9}{19}\)
Bài 2: Tính nhanh
\(\frac{2}{2\cdot4}+\frac{2}{4\cdot6}+\frac{2}{6\cdot8}+...+\frac{2}{2016\cdot2018}\)
ai giúp mình với gấp lắm không có bài là bị phạt đó
Bài 1 :
\(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{\left(2x+1\right)\left(2x+3\right)}=\frac{9}{19}\)
\(\Leftrightarrow1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2x+1}-\frac{1}{2x+3}=\frac{9}{19}\)
\(\Leftrightarrow1-\frac{1}{2x+3}=\frac{9}{19}\)
\(\Leftrightarrow\frac{1}{2x+3}=1-\frac{9}{19}\)
\(\Leftrightarrow\frac{1}{2x+3}=\frac{10}{19}\)
\(\Leftrightarrow10.\left(2x+3\right)=19\Leftrightarrow2x+3=\frac{19}{10}\)
\(\Leftrightarrow2x=\frac{19}{10}-3\Leftrightarrow2x=-\frac{11}{10}\)
\(\Leftrightarrow x=-\frac{11}{20}=-0,55\)
Bài 2 :
\(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{2016.2018}\)
\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+....+\frac{1}{2016}-\frac{1}{2018}\)
\(=\frac{1}{2}-\frac{1}{2018}=\frac{504}{1009}\)
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a, \(\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+....+\frac{1}{24\cdot25}\)
b, \(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+.....+\frac{2}{99\cdot101}\)
c, \(5\frac{2}{7}\cdot\frac{8}{11}+5\frac{2}{7}\cdot\frac{5}{11}-5\frac{2}{7}\cdot\frac{2}{11}\)
giup minh voi mai nop roi!!!!
#)Giải :
a)\(\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{24.25}\)
\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{24}-\frac{1}{25}\)
\(=\frac{1}{5}-\frac{1}{25}\)
\(=\frac{4}{25}\)
b)\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
\(=1-\frac{1}{101}\)
\(=\frac{100}{101}\)
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a) \(\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{24.25}\)
= \(\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{24}-\frac{1}{25}\)
= \(\frac{1}{5}-\frac{1}{25}\)
= \(\frac{4}{25}\)
b) \(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}\)
= \(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\)
= \(1-\frac{1}{101}\)
= \(\frac{100}{101}\)
c) \(5\frac{2}{7}.\frac{8}{11}+5\frac{2}{7}.\frac{5}{11}-5\frac{2}{7}.\frac{2}{11}\)
= \(5\frac{2}{7}.\left(\frac{8}{11}+\frac{5}{11}-\frac{2}{11}\right)\)
= \(5\frac{2}{7}\)
= \(\frac{37}{7}\)
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a)\(\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{24.25}\)
\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{24}-\frac{1}{25}\)
\(=\frac{1}{5}-\frac{1}{25}\)
\(=\frac{5}{25}-\frac{1}{25}=\frac{4}{25}\)
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Xem thêm câu trả lời
Cho \(A=1\cdot3\cdot5\cdot7\cdot...\cdot49\)
\(B=\frac{1\cdot2\cdot3\cdot...\cdot48\cdot49\cdot50}{2\cdot4\cdot6\cdot...\cdot48\cdot50}\)
\(C=\frac{26}{2}\cdot\frac{27}{2}\cdot...\cdot\frac{50}{2}\)
So sánh A,B và C
TRẢ LỜI ĐI CÓ DC KO ĐỂ MK CÒN GIẢI
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\(A=\frac{1\cdot2}{2\cdot2}\cdot\frac{2\cdot3}{3\cdot3}\cdot\frac{3\cdot4}{4\cdot4}\cdot\frac{4\cdot5}{5\cdot5}\cdot.................\cdot\frac{2012\cdot2013}{2013\cdot2013}\)với
\(B=\frac{2012\cdot2013-2012\cdot2012}{2012\cdot2011+2012\cdot2}\)
A=\(\frac{1}{2}\).\(\frac{2}{3}\)....\(\frac{2012}{2013}\)=\(\frac{1}{2013}\)
B=\(\frac{2012}{2012.2013}\)=\(\frac{1}{2013}\)
vậy A=B
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\(A=\frac{1.2}{2.2}.\frac{2.3}{3.3}.\frac{3.4}{4.4}.\frac{4.5}{5.5}.....\frac{2012.2013}{2013.2013}=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}....\frac{2012}{2013}=\frac{1.2.3.4.5....2012}{2.3.4.5....2013}=\frac{1}{2013}\)
\(B=\frac{2012.2013-2012.2012}{2012.2011+2012.2}=\frac{2012.\left(2013-2012\right)}{2012.\left(2011+2\right)}=\frac{2012}{2012.2013}=\frac{1}{2013}\)
\(\Rightarrow A=B\)
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Xem thêm câu trả lời
\(A=2^0\cdot2^1\cdot2^2\cdot2^3\cdot2^4\cdot...\cdot2^{100}\)
\(B=6^0\cdot6^1\cdot6^2\cdot6^3\cdot6^4\cdot...\cdot6^{600}\)
\(C=7^0\cdot7^1\cdot7^2\cdot7^3\cdot7^4\cdot...\cdot7^{700}\)
\(D=8^1\cdot8^2\cdot8^3\cdot8^4\cdot8^5\cdot...\cdot8^{800}\)
A = 20 . 21 . 22 . 23. 24....2100
= 1 . 21 . 22 . 23 . 24 .... 2100
= 1 . 21 + 2 + 3 + .... + 100
Ta có : Số số hạng của dãy số 1 + 2 + 3 + .... + 100 là :
(100 - 1) : 1 + 1 = 100 ( số hạng )
Tổng của dãy số 1 + 2 + 3 + ... + 100 là :
(100 + 1) . 100 : 2 = 5050
Thay vào, ta được :
A = 1 . 25050 = 25050
Vậy A = 25050
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\(A=2^0.2^1.2^2.2^3.....2^{100}=2^1.2^2.2^3......2^{100}=2^{1+2+3+....+100}=2^{\left(1+100\right).\left(100-1+1\right):2}=2^{5050}\)
\(B=6^0.6^1.6^2.6^3.6^4......6^{600}=6^{1+2+3+4+...+600}=6^{\left(1+600\right).\left(600-1+1\right):2}=6^{180300}\)
\(C=7^0.7^1.7^2.7^3.7^4.....7^{700}=7^{0+1+2+3+4+...+700}=7^{\left(700+0\right).\left(700-0+1\right):2}=7^{245000}\)
\(D=8^1.8^2.8^3......8^{800}=8^{1+2+3+....+800}=8^{\left(800+1\right).\left(800-1+1\right):2}=8^{320400}\)
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