\(6,52+\frac{7}{4}\times9\)
kb với mk nha
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m = 1/3-1/5+1/5-1/7+1/7-1/9+...+1/97-1/99
m = 1/3-1/99=32/99
Sorry chị em ko làm đc câu b vì em mới học lớp 4
k em ha
a) \(M=\frac{2}{3\times5}+\frac{2}{5\times7}+\frac{2}{7\times9}+...+\frac{2}{97\times99}\)
\(\Rightarrow M=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\)
\(\Rightarrow M=\frac{1}{3}-\frac{1}{99}\)
\(\Rightarrow M=\frac{33}{99}-\frac{1}{99}=\frac{32}{99}\)
b) \(N=\frac{3}{5\times7}+\frac{3}{7\times9}+\frac{3}{9\times11}+...+\frac{3}{197\times199}\)
\(\Rightarrow N=3\times\left(\frac{1}{5\times7}+\frac{1}{7\times9}+\frac{1}{9\times11}+...+\frac{1}{197\times199}\right)\)
\(\Rightarrow N=3\times\left[2\times\left(\frac{1}{5\times7}+\frac{1}{7\times9}+\frac{1}{9\times11}+...+\frac{1}{197\times199}\right)\right]\)
\(\Rightarrow N=3\times\left(\frac{2}{5\times7}+\frac{2}{7\times9}+\frac{2}{9\times11}+...+\frac{2}{197\times199}\right)\)
\(\Rightarrow N=3\times\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{197}-\frac{1}{199}\right)\)
\(\Rightarrow N=3\times\left(\frac{1}{5}-\frac{1}{199}\right)\)
\(\Rightarrow N=3\times\frac{194}{995}=\frac{582}{995}\)
----Chúc em học giỏi !----
\(x+78\cdot9=9\cdot45\)
\(\Leftrightarrow x+702=405\)
\(\Leftrightarrow x=-297\)
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=\(\frac{-11}{14}\)
\(|2^2_{12}\)\(+x=\frac{5}{7}+\frac{8}{12}\)
\(2+x=\frac{5}{7}+\frac{1}{2}\)
\(2+x=\frac{17}{14}\)
\(x=\frac{17}{14}-2\)
\(x=\frac{-11}{14}\)
\(2\frac{4}{3}+\frac{4}{12}=\frac{2.3+4}{3}+\frac{4}{12}=\frac{10}{3}+\frac{4}{12}=\frac{40}{12}+\frac{4}{12}=\frac{44}{12}=\frac{11}{3}\)
\(2\frac{4}{3}=\frac{10}{3}\)
\(\frac{10}{3}+\frac{4}{12}=\frac{120}{36}+\frac{12}{36}\)
\(=\frac{132}{36}=\frac{22}{6}=\frac{11}{3}\)
\(12\frac{4}{5}\)=\(\frac{64}{5}\).
Ta có :
\(\frac{64}{5}+\frac{4}{5}=\frac{68}{5}\)
\(12\frac{4}{5}+\frac{4}{5}=\frac{64}{5}+\frac{4}{5}=\frac{68}{5}\)
163/25+63/4
2227/100
tk mk nha bn iu