\(\frac{13.2-13.17}{42.21+4211}\) Và \(\frac{48.13-48.19}{18.31-18.7}\)
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\(x+\frac{3}{5.9}+\frac{3}{9.13}+\frac{3}{13.17}+...+\frac{4}{41.45}=-\frac{37}{45}\)
\(\Leftrightarrow x+3\left(\frac{1}{5.9}+\frac{1}{9.13}+\frac{1}{13.17}+...+\frac{1}{41.45}\right)=-\frac{37}{45}\)
\(\Leftrightarrow x+\frac{3}{4}\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+...+\frac{1}{41}-\frac{1}{45}\right)=-\frac{37}{45}\)
\(\Leftrightarrow x+\frac{3}{4}\left(\frac{1}{5}-\frac{1}{45}\right)=-\frac{37}{45}\)
\(\Leftrightarrow x+\frac{3}{4}.\frac{8}{45}=-\frac{37}{45}\)
\(\Leftrightarrow x+\frac{2}{15}=-\frac{37}{45}\)
\(\Leftrightarrow x=-\frac{43}{45}\)
\(x+\frac{2}{5.9}+\frac{2}{9.13}+\frac{2}{13.17}+...+\frac{2}{41.45}=\frac{-37}{45}\)
\(\Leftrightarrow x+\left[\frac{2}{4}\cdot\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+...+\frac{1}{41}-\frac{1}{45}\right)\right]=\frac{-37}{45}\)
\(\Leftrightarrow x+\left[\frac{1}{2}\cdot\left(\frac{1}{5}-\frac{1}{45}\right)\right]=\frac{-37}{45}\)
\(\Leftrightarrow x+\left[\frac{1}{2}\cdot\frac{8}{45}\right]=\frac{-37}{45}\)
\(\Leftrightarrow x+\frac{4}{45}=\frac{-37}{45}\)
\(\Leftrightarrow x=-\frac{41}{45}\)
\(=\left(\dfrac{7}{18}+\dfrac{3}{34}\right):\left[\dfrac{14}{4}\left(\dfrac{1}{9}-\dfrac{1}{13}+\dfrac{1}{14}-\dfrac{1}{18}+\dfrac{1}{13}-\dfrac{1}{17}\right)\right]\)
\(=\dfrac{73}{153}:\left[\dfrac{14}{4}\cdot\dfrac{73}{1071}\right]\)
\(=\dfrac{73}{153}:\dfrac{73}{306}=2\)
theo đề bài ta có:
\(x+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}=\frac{-37}{45}\)
\(x+\left(\frac{1}{5}-\frac{1}{45}\right)=\frac{-37}{45}\)
\(x+\frac{8}{45}=\frac{-37}{45}\)
\(x=\frac{-37}{45}-\frac{8}{45}\)
\(x=\frac{-45}{45}=1\)
đặt A=4/5.9+4/9.13+4/13.17+...+4/41.45
=1/5-1/9+1/9-1/13+1/13-1/17+...+1/41-1/45
=1/5-1/45
=8/45
suy ra x+8/45=-37/45
suy ra x=-1
\(P=\left(\frac{7.9}{9.18}+\frac{7.3}{2.7.17}\right):\left(\frac{14}{117}+\frac{1}{18}+\frac{14}{221}\right)\)
\(P=\left(\frac{7}{18}+\frac{3}{34}\right):\left(..\right)\)
Tới đây bạn tự làm nốt.
\(A=\frac{4}{10.13}+\frac{4}{13.14}+...+\frac{4}{100.103}\)
\(A=4.\left(\frac{1}{10.13}+\frac{1}{13.16}+...+\frac{1}{100.103}\right)\)
\(A=4.\frac{1}{3}.\left(\frac{3}{10.13}+\frac{3}{13.16}+...+\frac{3}{100.103}\right)\)
\(A=\frac{4}{3}.\left(\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+...+\frac{1}{100}-\frac{1}{103}\right)\)
\(A=\frac{4}{3}.\left(\frac{1}{10}-\frac{1}{103}\right)\)
\(A=\frac{4}{3}.\frac{93}{1030}\)
\(A=\frac{62}{515}\)
so sanh hả bạn
đề bài là gì vậy bạn ???????????????