\(\frac{1}{4x9}+\frac{1}{9x14}+\frac{1}{14x19}+\frac{1}{44x49}\)
Những câu hỏi liên quan
So sánh:
\(A=\frac{5}{4x9}+\frac{5}{9x14}+\frac{5}{14x19}+...+\frac{5}{1704x1709}+\frac{5}{1709x1714}\)
\(B=\frac{6}{7x9}+\frac{6}{9x11}+\frac{6}{11x13}+...+\frac{6}{1283x1285}\)
( Làm xong nhớ kb với mk nhé )
tớ nghĩ là A = B
nếu đúng thì tk hộ tớ với , tớ đang bị âm 998 điểm !
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dễ thôi nhưng tôi giúp câu A còn đâu bn tự nghĩ nhé
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A =\(\frac{5}{4x9}\)+....+\(\frac{5}{1709x1714}\)
=\(\frac{1}{5}\)
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Xem thêm câu trả lời
G = 3/4x9 =3/9x14+ 3/14x19+................+3/34+35
giúp nhanh nha
Tính \(\frac{1}{2x4}+\frac{1}{4x9}+\frac{1}{6x12}+....+\frac{1}{36x57}+\frac{1}{38x60}\)
tính:1/4x9 + 1/9x4 + 1/14x19 + ... + 1/1999x2004
\(=\dfrac{1}{5}\left(\dfrac{5}{4\cdot9}+\dfrac{5}{9\cdot14}+...+\dfrac{5}{1999\cdot2004}\right)\)
\(=\dfrac{1}{5}\left(\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{14}+...+\dfrac{1}{1999}-\dfrac{1}{2004}\right)\)
\(=\dfrac{1}{5}\left(\dfrac{1}{4}-\dfrac{1}{2004}\right)=\dfrac{1}{5}\cdot\dfrac{125}{501}=\dfrac{25}{501}\)
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Sửa \(\dfrac{1}{9.4}\rightarrow\dfrac{1}{9.14}\) nha
\(=\dfrac{1}{5}\left(\dfrac{5}{4.9}+\dfrac{5}{9.14}+...+\dfrac{5}{1999.2004}\right)\)
\(=\dfrac{1}{5}\left(\dfrac{1}{4}-\dfrac{1}{2004}\right)=\dfrac{1}{5}.\dfrac{125}{501}=\dfrac{25}{501}\)
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\(=2\left(\dfrac{1}{4.9}+\dfrac{1}{9.14}+.....+\dfrac{1}{1999.2004}\right)\)
\(=2\left(\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{14}+...+\dfrac{1}{1999}-\dfrac{1}{2004}\right)\)
\(=2\left(\dfrac{1}{4}-\dfrac{1}{2004}\right)=2\cdot\dfrac{125}{501}=\dfrac{250}{501}\)
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\(A=\frac{2^{12}x3^5-4^6x9^2}{\left(2^2x3\right)^6+8^4x3^5}-\frac{5^{10}x7^3-25^5x49^2}{\left(125x7\right)^3+5^9x14^3}\)
\(B=\frac{\left(\frac{-1}{2}\right)^3-\left(\frac{3}{4}\right)^3x\left(-2\right)^2}{2x\left(-1\right)^5+\left(\frac{3}{4}\right)^2-\frac{3}{8}}\)
\(C=2^2+3\left(\frac{1}{2}\right)^0-2^{-2}+\left[\left(-2^2\right):\frac{1}{2}\right]:8\)
Bài 3 .Tính:
a) A = 2/1x5 + 2/5x9 + 2/9x13 +....+2/81x85
b)B = 3/4x9 + 3/9x14 + 3/14x19 +....+ 3/34x39
c)C = 5/11x14 + 5/14x17 + 5/17x20 +....+ 5/47x50
Đây là một dang tính nhanh phân số bằng cách rút gọn.Mời các bạn thử ngẫm.
\(A=\frac{2}{1.5}+\frac{2}{5.9}+\frac{2}{9.13}+....+\frac{2}{81.85}\)
\(\Rightarrow2A=\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+....+\frac{4}{81.85}\)
\(\Rightarrow2A=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+....+\frac{1}{81}-\frac{1}{85}\)
\(\Rightarrow2A=1-\frac{1}{85}\)
\(\Rightarrow A=\frac{84}{85}:2=\frac{42}{85}\)
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tính A còn lại tự tính nha
a) A = 2/1x5 + 2/5x9 + 2/9x13 +....+2/81x85
\(\frac{2}{1x5}+\frac{2}{5x9}+\frac{2}{9x13}+...+\frac{2}{81x85}\)
\(A=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{81}-\frac{1}{85}\)
\(A=1-\frac{1}{85}\)
\(\Rightarrow A=\frac{84}{85}\)
k nha
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A = \(\frac{2}{1\cdot5}\) + \(\frac{2}{5\cdot9}\) + \(\frac{2}{9\cdot13}\) + ... + \(\frac{2}{81\cdot85}\)
= \(\frac{1}{2}\) * ( \(\frac{1}{1\cdot5}\) + \(\frac{1}{5\cdot9}\) + \(\frac{1}{9\cdot13}\) + ... + \(\frac{1}{81\cdot85}\) )
= \(\frac{1}{2}\) * ( 1 - \(\frac{1}{5}\) + \(\frac{1}{5}\) - \(\frac{1}{9}\) + \(\frac{1}{9}\) - \(\frac{1}{13}\) + ... + \(\frac{1}{81}\) - \(\frac{1}{85}\) )
= \(\frac{1}{2}\) * ( 1 - \(\frac{1}{85}\) )
= \(\frac{1}{2}\) * \(\frac{84}{85}\)
= \(\frac{42}{85}\)
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Xem thêm câu trả lời
Afrac{1+frac{1}{3}+frac{1}{5}+frac{1}{7}+......+frac{1}{999}}{frac{1}{1.999}+frac{1}{3.997}+frac{1}{5.995}+......+frac{1}{999.1}}Bfrac{1+left(1+2right)+left(1+2+3right)+left(1+2+3+4right)+......+left(1+2+3+...+98right)}{1.2+2.3+3.4+4.5+......+98.99}Cfrac{frac{1}{1.300}+frac{1}{2.301}+frac{1}{3.302}+......+frac{1}{100.400}}{frac{1}{1.102}+frac{1}{2.103}+frac{1}{3.104}+......+frac{1}{299.400}}Dfrac{frac{1}{99}+frac{2}{98}+frac{3}{97}+......+frac{99}{1}}{frac{1}{2}+frac{1}{3}+frac{1}{4}+......+frac...
Đọc tiếp
\(A=\frac{1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+......+\frac{1}{999}}{\frac{1}{1.999}+\frac{1}{3.997}+\frac{1}{5.995}+......+\frac{1}{999.1}}\)
\(B=\frac{1+\left(1+2\right)+\left(1+2+3\right)+\left(1+2+3+4\right)+......+\left(1+2+3+...+98\right)}{1.2+2.3+3.4+4.5+......+98.99}\)
\(C=\frac{\frac{1}{1.300}+\frac{1}{2.301}+\frac{1}{3.302}+......+\frac{1}{100.400}}{\frac{1}{1.102}+\frac{1}{2.103}+\frac{1}{3.104}+......+\frac{1}{299.400}}\)
\(D=\frac{\frac{1}{99}+\frac{2}{98}+\frac{3}{97}+......+\frac{99}{1}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+......+\frac{1}{100}}:\frac{92-\frac{1}{9}-\frac{2}{10}-\frac{3}{97}-......-\frac{92}{100}}{\frac{1}{45}+\frac{1}{50}+\frac{1}{55}+......+\frac{1}{500}}\)
Giup tui voi !!!!!!!!!!!!!!!!!!!!!!!!!!! Mai phai nop roi !!!!!!!!!!!!!!!!!!!
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19 tháng 8 2019 lúc 11:16
e,Afrac{1}{2}+frac{5}{6}+frac{11}{12}+frac{19}{20}+frac{29}{30}+frac{41}{42}left(1-frac{1}{2}right)+left(1-frac{1}{6}right)+left(1-frac{1}{12}right)+left(1-frac{1}{20}right)+left(1-frac{1}{20}right)+left(1-frac{1}{42}right)Rightarrow A1-frac{1}{2}+1-frac{1}{6}+1-frac{1}{12}+1-frac{1}{20}+1-frac{1}{30}+1-frac{1}{42}4-left(frac{1}{2}+frac{1}{6}+frac{1}{12}+frac{1}{20}+frac{1}{30}+frac{1}{42}right)Rightarrow A4-left(frac{1}{1.2}+frac{1}{2.3}+frac{1}{3.4}+frac{1}{4.5}+frac{1}{5.6}+frac{1}{6.7}right)...
Đọc tiếp
e,\(A=\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+\frac{29}{30}+\frac{41}{42}=\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{6}\right)+\left(1-\frac{1}{12}\right)+\left(1-\frac{1}{20}\right)+\left(1-\frac{1}{20}\right)+\left(1-\frac{1}{42}\right)\)
\(\Rightarrow A=1-\frac{1}{2}+1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+1-\frac{1}{30}+1-\frac{1}{42}=4-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)\)
\(\Rightarrow A=4-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)=4-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)
\(\Rightarrow A=4-\left(\frac{1}{1}-\frac{1}{7}\right)=4-\frac{6}{7}=3\frac{1}{7}\)
BN mún hỏi j vậy, đây k phải câu hỏi, mà có thì phải là toán lớp 6
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frac{frac{1}{2}+frac{1}{3}+......+frac{1}{2013}}{frac{2012}{1}+frac{2012}{2}+frac{2011}{3}+...+frac{1}{2013}}frac{frac{1}{2}+frac{1}{3}+..+frac{1}{2013}}{frac{2012}{1}+2+frac{2012}{2}+1+frac{2011}{3}+1+...+frac{1}{2013}+1-2014}frac{frac{1}{2}+frac{1}{3}+...+frac{1}{2013}}{frac{2014}{1}+frac{2014}{2}+...+frac{2014}{2013}-2014}frac{frac{1}{2}+frac{1}{3}+...+frac{1}{2013}}{2014left(1+frac{1}{2}+frac{1}{3}+...+frac{1}{2013}-1right)}frac{1}{2014}
Đọc tiếp
\(\frac{\frac{1}{2}+\frac{1}{3}+......+\frac{1}{2013}}{\frac{2012}{1}+\frac{2012}{2}+\frac{2011}{3}+...+\frac{1}{2013}}\)
=\(\frac{\frac{1}{2}+\frac{1}{3}+..+\frac{1}{2013}}{\frac{2012}{1}+2+\frac{2012}{2}+1+\frac{2011}{3}+1+...+\frac{1}{2013}+1-2014}\)
=\(\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}}{\frac{2014}{1}+\frac{2014}{2}+...+\frac{2014}{2013}-2014}\)
=\(\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}}{2014\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}-1\right)}\)
=\(\frac{1}{2014}\)
