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Đặt \(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{100}\)
\(A=\frac{99}{100}\)
1. \(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{40.43}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}\)
\(=1-\frac{1}{43}\)
\(=\frac{42}{43}\)
2. Đặt \(A=\frac{2}{2}+\frac{2}{6}+\frac{2}{12}+...+\frac{2}{90}\)
\(=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\right)\)
\(=2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=2.\left(1-\frac{1}{10}\right)\)
\(=2.\frac{9}{10}\)
\(=\frac{9}{5}\)
Ủng hộ mk nha !!! ^_^
1) 3/1×4 + 3/4×7 + 3/7×10 + ... + 3/40×43
= 1 - 1/4 + 1/4 - 1/7 + 1/7 - 1/10 + ... + 1/40 - 1/43
= 1 - 1/43
= 42/43
2) 2/2 + 2/6 + 2/12 + ... + 2/90
= 2 × (1/2 + 1/6 + 1/12 + ... + 1/90)
= 2 × (1/1×2 + 1/2×3 + 1/3×4 + ... + 1/9×10)
= 2 × (1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/9 - 1/10)
= 2 × (1 - 1/10)
= 2 × 9/10
= 9/5
\(2\frac{4}{3}+\frac{4}{12}=\frac{2.3+4}{3}+\frac{4}{12}=\frac{10}{3}+\frac{4}{12}=\frac{40}{12}+\frac{4}{12}=\frac{44}{12}=\frac{11}{3}\)
\(2\frac{4}{3}=\frac{10}{3}\)
\(\frac{10}{3}+\frac{4}{12}=\frac{120}{36}+\frac{12}{36}\)
\(=\frac{132}{36}=\frac{22}{6}=\frac{11}{3}\)
\(12\frac{4}{5}\)=\(\frac{64}{5}\).
Ta có :
\(\frac{64}{5}+\frac{4}{5}=\frac{68}{5}\)
\(12\frac{4}{5}+\frac{4}{5}=\frac{64}{5}+\frac{4}{5}=\frac{68}{5}\)
\(9\frac{5}{34}-12+5\)
\(9\frac{5}{34}=\frac{311}{34}\)
\(\Rightarrow9\frac{5}{34}-12+5\)
\(=\frac{311}{34}-12+5\)
\(=-\frac{97}{34}+5\)
\(=\frac{73}{34}\)
2\(\frac{2}{12}\)+ X = \(\frac{5}{7}\)+ \(\frac{8}{12}\)
\(\frac{13}{6}\)+ X=\(\frac{29}{21}\)
X= \(\frac{29}{21}\)- \(\frac{13}{6}\)
X=\(\frac{-11}{14}\)
\(|2^2_{12}\)\(+x=\frac{5}{7}+\frac{8}{12}\)
\(2+x=\frac{5}{7}+\frac{1}{2}\)
\(2+x=\frac{17}{14}\)
\(x=\frac{17}{14}-2\)
\(x=\frac{-11}{14}\)