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TA CÓ:\(1\cdot3\cdot....\cdot99=\frac{\left(1\cdot3\cdot...\cdot99\right)\left(2\cdot4\cdot...\cdot100\right)}{2\cdot4....\cdot100}=\frac{1\cdot2\cdot3\cdot....\cdot100}{2\cdot2\cdot2\cdot...\cdot2\left(50\right)\cdot1\cdot2\cdot3\cdot..\cdot50}\)
\(=\frac{51\cdot52\cdot...\cdot100}{2\cdot2\cdot2\cdot...\cdot2}=\frac{51}{2}\cdot\frac{52}{2}\cdot\frac{53}{2}\cdot...\cdot\frac{100}{2}\)(ĐPCM)
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 :
\(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{2016.2018}\)
\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+....+\frac{1}{2016}-\frac{1}{2018}\)
\(=\frac{1}{2}-\frac{1}{2018}=\frac{504}{1009}\)
A = \(\frac{3469-54}{6938-108}\)
= \(\frac{3469-54}{2\left(3469-54\right)}\)
= \(\frac{1}{2}\)
B = \(\frac{2468-98}{3702-147}\)
= \(\frac{2\left(1234-49\right)}{3\left(1234-49\right)}\)
= \(\frac{2}{3}\)
BCNN(2,3) = 6
\(\frac{1}{2}=\frac{1.3}{2.3}=\frac{3}{6}\)
\(\frac{2}{3}=\frac{2.2}{3.2}=\frac{4}{6}\)
\(A=\frac{3469-54}{6938-108}=\frac{3469-54}{3469.2-54.2}=\frac{3469-54}{2.\left(3469-54\right)}=\frac{1}{2}=\frac{1.3}{2.3}=\frac{3}{6}\)
\(B=\frac{2468-98}{3702-147}=\frac{1234.2-49.2}{1234.3-49.3}=\frac{2.\left(1234-49\right)}{3.\left(1234-49\right)}=\frac{2}{3}=\frac{2.2}{3.2}=\frac{4}{6}\)
đụ cha mi
mi trù ta thi rớt HK II mà ta giúp mày hả
mấy bài này cũng dễ ẹt nữa
đừng có mơ ta sẽ giúp mày
ha ha ha ha ha ha ha ha ha ha ha ha ha ha ha ha
\(B=\left(1+\frac{1}{1\cdot3}\right)\left(1+\frac{1}{2\cdot4}\right)\left(1+\frac{1}{3\cdot5}\right)...\left(1+\frac{1}{99\cdot101}\right)\)
\(B=\frac{2^2}{1\cdot3}\cdot\frac{3^2}{2\cdot4}\cdot\frac{4^2}{3\cdot5}\cdot\cdot\cdot\frac{100^2}{99\cdot101}\)
\(B=\frac{2^2\cdot3^2\cdot4^2\cdot\cdot\cdot100^2}{1\cdot3\cdot2\cdot4\cdot3\cdot5\cdot\cdot\cdot99\cdot101}\)
\(B=\frac{\left(2\cdot3\cdot4\cdot\cdot\cdot100\right)\cdot\left(2\cdot3\cdot4\cdot\cdot\cdot100\right)}{\left(1\cdot2\cdot3\cdot\cdot\cdot99\right)\cdot\left(3\cdot4\cdot5\cdot\cdot\cdot101\right)}\)
\(B=\frac{100\cdot2}{1\cdot101}\)
\(B=\frac{200}{101}\)
thiệt chớ bạn ra đề kiểu đó ma nào làm
=1.3.5....49/100