tinh:
\(\frac{\frac{4}{5}-\frac{4}{71}+\frac{4}{123}}{\frac{3}{5}-\frac{3}{71}+\frac{3}{123}}\)-\(\frac{\frac{1}{19}+\frac{1}{1315}-\frac{1}{287}}{\frac{3}{19}+\frac{3}{1315}-\frac{3}{287}}\)
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Tính bằng cách thuận tiện
1,\(\frac{13}{19}\left(\frac{19}{26}-\frac{71}{43}\right)+\frac{71}{43}\left(\frac{13}{19}-\frac{86}{71}\right)\)
2,\(\frac{5}{7}\left(\frac{1}{2}-\frac{1}{3}+\frac{4}{7}\right)+\left(\frac{1}{2}-\frac{1}{3}-\frac{4}{7}\right):\frac{7}{5}\)
\(\frac{\frac{4}{17}+\frac{4}{19}-\frac{4}{2111}}{\frac{5}{17}+\frac{5}{19}-\frac{5}{2111}}-\frac{\frac{1}{123}-\frac{1}{19}+\frac{1}{371}-\frac{1}{5}}{\frac{-5}{123}+\frac{5}{19}-\frac{5}{371}+1}\)
\(\frac{\frac{4}{17}+\frac{4}{19}-\frac{4}{2111}}{\frac{5}{17}+\frac{5}{19}-\frac{5}{2111}}-\frac{\frac{1}{123}-\frac{1}{19}+\frac{1}{371}-\frac{1}{5}}{-\frac{5}{123}+\frac{5}{19}-\frac{5}{371}+1}\)
\(=\frac{4.\left(\frac{1}{17}+\frac{1}{19}-\frac{1}{2111}\right)}{5.\left(\frac{1}{17}+\frac{1}{19}-\frac{1}{2111}\right)}+\frac{\frac{1}{123}-\frac{1}{19}+\frac{1}{371}-\frac{1}{5}}{5.\left(\frac{1}{123}-\frac{1}{19}+\frac{1}{371}-\frac{1}{5}\right)}=\frac{4}{5}+\frac{1}{5}=1\)
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Cho tam giác ABC có đường cao AD .Gọi E là trung điểm của AB .F đối xứng vs D qua E c/m AB = DF
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Tinh B=1\(\frac{6}{41}\frac{12+\frac{12}{19}-\frac{12}{37}-\frac{12}{53}}{3+\frac{1}{3}-\frac{3}{37}-\frac{3}{53}}:\frac{4+\frac{4}{17}+\frac{4}{19}+\frac{4}{2006}}{5+\frac{5}{17}+\frac{5}{19}+\frac{5}{2006}}.\frac{124242423}{2373735}\)
\(\frac{5}{23}x\frac{17}{6}+\frac{5}{23}x\frac{9}{26}\)
\(=\frac{5}{23}x\left(\frac{17}{26}+\frac{9}{26}\right)\)
\(=\frac{5}{23}x1=\frac{5}{23}\)
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1/ Chứng tỏ rằng frac{1}{1cdot2}+frac{1}{2cdot3}+frac{1}{3cdot4}+...+frac{1}{99cdot100}12/ Chứng tỏ rằng Bfrac{1}{4}+frac{1}{5}+frac{1}{6}+...+frac{1}{19}13/ Rút gọn biểu thức A1+frac{1}{2}+frac{1}{2^2}+frac{1}{2^3}+...+frac{1}{2^{2012}}4/ Tính nhanhfrac{frac{4}{2010}+frac{4}{2011}-frac{4}{2012}}{frac{5}{2010}+frac{5}{2011}-frac{5}{2012}}-frac{frac{1}{123}-frac{1}{19}+frac{1}{371}-frac{1}{5}}{-frac{5}{123}+frac{5}{19}-frac{5}{371}+1} GIÚP ĐƯỢC CÂU NÀO THÌ GIÚP NHÉ, MÌNH TICK CHO
Đọc tiếp
1/ Chứng tỏ rằng \(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}<1\)
2/ Chứng tỏ rằng \(B=\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{19}<1\)
3/ Rút gọn biểu thức \(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\)
4/ Tính nhanh\(\frac{\frac{4}{2010}+\frac{4}{2011}-\frac{4}{2012}}{\frac{5}{2010}+\frac{5}{2011}-\frac{5}{2012}}-\frac{\frac{1}{123}-\frac{1}{19}+\frac{1}{371}-\frac{1}{5}}{-\frac{5}{123}+\frac{5}{19}-\frac{5}{371}+1}\)
GIÚP ĐƯỢC CÂU NÀO THÌ GIÚP NHÉ, MÌNH TICK CHO
c)\(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{2012}}\)
\(2A=2\left(1+\frac{1}{2}+\frac{1}{2^2}+.....+\frac{1}{2^{2012}}\right)\)
\(2A=2+1+\frac{1}{2^2}+\frac{1}{2^3}+.....+\frac{1}{2^{2011}}\)
\(2A-A=\left(2+1+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{2011}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....\frac{1}{2^{2012}}\right)\)
\(A=2-\frac{1}{2^{2012}}\)
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1/
A=1/1-1/2+1/2-1/3+1/3-1/4+...+1/99-1/100
A=1/1-1/100
Vì 1/100>0
-->1/1-1/100<1
-->A<1
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a)\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{99}-\frac{1}{100}\)
\(\frac{1}{1}-\frac{1}{100}\)=\(\frac{99}{100}<1\)
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Xem thêm câu trả lời
\(\frac{\frac{4}{2010}+\frac{4}{2011}-\frac{4}{2012}}{\frac{5}{2010}+\frac{5}{2011}-\frac{5}{2012}}-\frac{\frac{1}{123}-\frac{1}{19}+\frac{1}{37}-\frac{1}{5}}{-\frac{5}{123}+\frac{5}{19}-\frac{5}{37}+1}\)= ?
\(A=\frac{1}{71}+\frac{2}{70}+\frac{3}{69}+\frac{4}{68}+...+\frac{70}{2}+\frac{71}{1};B=\frac{1}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{71}+\frac{1}{72}}AxB=?\)
B=\(-1\frac{1}{5}.\frac{4\left(3+\frac{1}{3}-\frac{3}{7}-\frac{3}{53}\right)}{3+\frac{1}{3}-\frac{3}{37}-\frac{3}{53}}:\frac{4+\frac{4}{17}+\frac{4}{19}+\frac{4}{2003}}{5+\frac{5}{17}+\frac{5}{19}+\frac{5}{2003}}\)
t tưởng mọi hôm bài này m làm thạo lắm mà bây h chịu ak
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B=\(-1\frac{1}{5}.\frac{4\left(3+\frac{1}{3}-\frac{3}{7}-\frac{3}{53}\right)}{3+\frac{1}{3}-\frac{3}{37}-\frac{3}{53}}:\frac{4+\frac{4}{17}+\frac{4}{19}+\frac{4}{2003}}{5+\frac{5}{17}+\frac{5}{19}+\frac{5}{2003}}\)
\(B=-1\frac{1}{5}\cdot\frac{4\left(3+\frac{1}{3}-\frac{3}{7}-\frac{3}{53}\right)}{3+\frac{1}{3}-\frac{3}{7}-\frac{3}{53}}\div\frac{4+\frac{4}{17}+\frac{4}{19}+\frac{4}{2003}}{5+\frac{5}{17}+\frac{5}{19}+\frac{5}{2003}}\)
\(B=\frac{-6}{5}\cdot4\div\frac{4\left(1+\frac{1}{17}+\frac{1}{19}+\frac{1}{2003}\right)}{5\left(1+\frac{1}{17}+\frac{1}{19}+\frac{1}{2003}\right)}\)
\(B=\frac{-24}{5}\div\frac{4}{5}\)
\(B=-6\)
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\(B=-1\frac{1}{5}.\frac{4.\frac{3}{7}}{\frac{3}{37}}:\frac{4+3.\frac{4}{1}}{5+3.\frac{5}{1}}\)
\(B=-\frac{6}{5}.\frac{148}{7}:\frac{4}{5}\)
\(B=-\frac{222}{7}\)
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\(-1\frac{1}{5}.\frac{4.\left(3+\frac{1}{3}-\frac{3}{7}-\frac{3}{53}\right)}{3+\frac{1}{3}-\frac{3}{37}-\frac{3}{53}}:\frac{4+\frac{4}{17}+\frac{4}{19}+\frac{4}{2003}}{5+\frac{5}{17}+\frac{5}{19}+\frac{5}{2003}}\)