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Bài 1:\(A=1-\frac{1}{2}+1-\frac{1}{6}+.......+1-\frac{1}{9900}\)
\(=1-\frac{1}{1.2}+1-\frac{1}{2.3}+........+1-\frac{1}{99.100}\)
\(=99-\left(\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{99.100}\right)=99-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{99}-\frac{1}{100}\right)\)
\(=99-\left(1-\frac{1}{100}\right)=99-\frac{99}{100}=\frac{9801}{100}\)
Bài 2:\(A=\frac{1}{299}.\left(\frac{299}{1.300}+\frac{299}{2.301}+.........+\frac{299}{101.400}\right)\)
\(=\frac{1}{299}.\left(1-\frac{1}{300}+\frac{1}{2}-\frac{1}{301}+.........+\frac{1}{101}-\frac{1}{400}\right)\)
\(=\frac{1}{299}.\left(1+\frac{1}{2}+......+\frac{1}{101}-\frac{1}{300}-\frac{1}{301}-.......-\frac{1}{400}\right)\)
\(=\frac{1}{299}.\left[\left(1+\frac{1}{2}+.......+\frac{1}{101}\right)-\left(\frac{1}{300}+\frac{1}{301}+......+\frac{1}{400}\right)\right]\)(đpcm)
1/
\(=\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{6}\right)+...+\left(1-\frac{1}{9900}\right)\)
\(=\left(1+1+...+1\right)\left(50so\right)-\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{9900}\right)\)
\(=50-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)\)
\(=50-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(=50-\left(1-\frac{1}{100}\right)=49+\frac{1}{100}=\frac{4901}{100}\)
2/
\(=\frac{1}{299}\left(\frac{299}{1.300}+\frac{299}{2.301}+...+\frac{299}{101.400}\right)\)
\(=\frac{1}{299}\left(1-\frac{1}{300}+\frac{1}{2}-\frac{1}{301}+...+\frac{1}{101}-\frac{1}{400}\right)\)
\(=\frac{1}{299}\left[\left(1+\frac{1}{2}+...+\frac{1}{101}\right)-\left(\frac{1}{300}+\frac{1}{301}+...+\frac{1}{400}\right)\right]\)
\(a,2,5\cdot x=\frac{11}{15}\)
\(\Rightarrow x=\frac{11}{15}:2,5\)
\(\Rightarrow x=\frac{11}{15}:\frac{25}{10}\)
\(\Rightarrow x=\frac{11}{15}\cdot\frac{10}{25}\)
\(\Rightarrow x=\frac{11}{3}\cdot\frac{2}{25}=\frac{22}{75}\)
\(b,x-15\%\cdot x=\frac{1}{3}\)
\(\Rightarrow x-\frac{15}{100}\cdot x=\frac{1}{3}\)
\(\Rightarrow x-\frac{3}{20}\cdot x=\frac{1}{3}\)
\(\Rightarrow\frac{20x}{20}-\frac{3x}{20}=\frac{1}{3}\)
\(\Rightarrow\frac{17x}{20}=\frac{1}{3}\)
\(\Rightarrow17x\cdot3=20\)
\(\Rightarrow17x=\frac{20}{3}\)
\(\Rightarrow x=\frac{20}{3}:17=\frac{20}{3}\cdot\frac{1}{17}=\frac{20}{51}\)
Câu c mk làm sau :v
c,\(\left[\frac{3x}{7}+1\right]:\left[-4\frac{1}{7}\right]=\frac{-3}{28}\)
\(\Rightarrow\left[\frac{3x}{7}+1\right]:\left[-\frac{29}{7}\right]=\frac{-3}{28}\)
\(\Rightarrow\frac{3x}{7}+1=-\frac{3}{28}\cdot-\frac{29}{7}=\frac{87}{196}\)
\(\Rightarrow\frac{3x}{7}=\frac{87}{196}-1\)
Tìm nốt :v
a) \(\frac{-3}{7}+\frac{15}{26}-\left(\frac{2}{13}-\frac{3}{7}\right)=\frac{-3}{7}+\frac{15}{26}-\frac{2}{13}+\frac{3}{7}=\left(\frac{-3}{7}+\frac{3}{7}\right)+\left(\frac{15}{26}-\frac{2}{13}\right)\)
\(=\frac{15-4}{26}=\frac{11}{26}\)
c) \(\frac{-11}{23}.\frac{6}{7}+\frac{8}{7}.\frac{-11}{23}-\frac{1}{23}=\frac{-11}{23}.\left(\frac{6}{7}+\frac{8}{7}\right)-\frac{1}{23}\)
\(=\frac{-11}{23}.2-\frac{1}{23}=\frac{-22-1}{23}=\frac{-23}{23}=-1\)
a. \(\frac{-3}{7}+\frac{15}{26}-\left(\frac{2}{13}-\frac{3}{7}\right)=\frac{-3}{7}+\frac{15}{26}-\frac{2}{13}+\frac{3}{7}=\frac{15}{13.2}-\frac{2}{13}=\frac{15}{13.2}-\frac{2.2}{13.2}=\frac{15-4}{26}=\frac{11}{26}\)
C. \(\frac{-11}{23}.\frac{6}{7}+\frac{8}{7}.\frac{-11}{23}-\frac{1}{23}=\frac{1}{23}.\left(-11.\frac{6}{7}-11.\frac{8}{7}-1\right)=\frac{1}{23}.\left(-22-1\right)=\frac{1}{23}.\left(-23\right)=-1\)
\(a,=\left(-\frac{3}{7}+\frac{3}{7}\right)+\frac{5}{13}=0+\frac{5}{13}=\frac{5}{13}\)
\(b,=-\frac{5}{21}+\left(-\frac{2}{21}\right)+\frac{8}{24}=-\frac{1}{3}+\frac{1}{3}=0\)
\(c,=\left[-\frac{6}{11}+\frac{\left(-5\right)}{11}\right]+2=-1+2=1\)
\(d,=\frac{-1+\left(-15\right)}{32}+\frac{1}{2}=-\frac{1}{2}+\frac{1}{2}=0\)
a) \(\frac{-3}{7}+\frac{5}{13}+\frac{3}{7}\)
\(=\left(\frac{-3}{7}+\frac{3}{7}\right)+\frac{5}{13}\)
\(=0+\frac{5}{13}=\frac{5}{13}\)
b) \(\frac{-5}{21}+\frac{-2}{21}+\frac{8}{24}\)
\(=\frac{-7}{21}+\frac{1}{3}\)
\(=\frac{-1}{3}+\frac{1}{3}=0\)
c) \(\frac{-5}{11}+\left(\frac{-6}{11}+2\right)\)
\(=\frac{-5}{11}+\frac{-6}{11}+2\)
\(=\left(-1\right)+2=1\)
d) \(\left(\frac{-1}{32}+\frac{1}{2}\right)+\frac{-15}{32}\)
\(=\frac{-1}{32}+\frac{1}{2}+\frac{-15}{32}\)
\(=\left(\frac{-1}{32}+\frac{-15}{32}\right)+\frac{1}{2}\)
\(=\frac{-16}{32}+\frac{1}{2}\)
\(=\frac{-1}{2}+\frac{1}{2}=0\)
\(\frac{11}{36}-\left(-\frac{7}{24}\right)\)
\(=\frac{11}{36}+\frac{7}{24}\)
\(=\frac{22}{72}+\frac{21}{72}\)
\(=\frac{43}{72}\)
= 43/72 nha