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\(A=\frac{3}{1}+\frac{3}{\frac{\left(2+1\right).2}{2}}+\frac{3}{\frac{\left(3+1\right).3}{2}}+....+\frac{3}{\frac{\left(100+1\right).100}{2}}\)
\(\Rightarrow A=\frac{3}{1}+\frac{6}{2.3}+\frac{6}{3.4}+...+\frac{6}{100.101}\)
\(\Rightarrow A=\frac{3}{1}+6.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...-\frac{1}{101}\right)\)
\(\Rightarrow A=\frac{3}{1}+6.\left(\frac{1}{2}-\frac{1}{101}\right)\)
\(\Rightarrow A=\frac{3}{1}+\frac{6.99}{202}=\frac{297}{101}+\frac{3}{1}=\frac{600}{101}\)
kết quả k bik có sai k
a)\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}......\frac{99}{100}\)
\(=\frac{1.2.3.4.....99}{2.3.4.5.6.....100}\)
\(=\frac{1}{100}\)
b) Tương tự như câu a
\(\frac{10}{18}+\frac{4}{9}+\frac{26}{10}+\frac{12}{5}+\frac{9}{15}\)
\(=\frac{5}{9}+\frac{4}{9}+\frac{13}{5}+\frac{12}{5}+\frac{3}{5}\)
\(=\left(\frac{5}{9}+\frac{4}{9}\right)+\left(\frac{13}{5}+\frac{12}{5}+\frac{3}{5}\right)\)
\(=1+\frac{28}{5}\)
\(=\frac{33}{5}\)
Ta có:
a) \(\frac{10}{18}+\frac{4}{9}+\frac{26}{10}+\frac{12}{5}+\frac{9}{15}=\frac{5}{9}+\frac{4}{9}+\frac{13}{5}+\frac{12}{5}+\frac{9}{15}=1+1+\frac{9}{15}=1\frac{9}{15}\)
b)\(\frac{10}{18}+\frac{4}{9}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}=\left(\frac{5}{9}+\frac{4}{9}\right)+\left(\frac{16}{128}+\frac{8}{128}+\frac{4}{128}+\frac{2}{128}+\frac{1}{128}\right)\)
\(=1+\frac{31}{128}=1\frac{31}{128}\)
Chỗ cuối mk nhầm, sửa lại nha :
\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{90}+\frac{1}{100}\)
\(=\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+\frac{1}{5\times6}+...+\frac{1}{9\times10}+\frac{1}{100}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{9}-\frac{1}{10}+\frac{1}{100}\)
\(=\frac{1}{2}-\frac{1}{10}+\frac{1}{100}\)
\(=\frac{50}{100}-\frac{10}{100}+\frac{1}{100}\)
\(=\frac{41}{100}\)
\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{90}+\frac{1}{100}\)
\(=\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+\frac{1}{5\times6}+...+\frac{1}{9\times10}+\frac{1}{100}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{9}-\frac{1}{10}+\frac{1}{100}\)
\(=\frac{1}{2}-\frac{1}{10}+\frac{1}{100}\)
\(=\frac{50}{100}-\frac{10}{100}-\frac{1}{100}\)
\(=\frac{39}{100}\)
Đúng thì k nha bn !!!!
Ta có:2/2(1+2)+2/2(1+2+3)+2/2(1+2+3+4)+...+2/2(1+2+3+...+100)
=2/6+2/12+2/20+...+2/5050
=2/2.3+2/3.4+2/4.5+...+2/100.101
=2.(1/2.3+1/3.4+1/4.5+...+1/100.101)
=2.(1-1/2+1/3-1/4+1/4-1/5+...+1/100-1/101)
=2.(1-1/101)
=2.100/101
=200/101
\(S=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{90}+\frac{1}{100}\)
\(S=\left(\frac{1}{6}+\frac{1}{12}\right)+\left(\frac{1}{20}+\frac{1}{100}\right)+\left(\frac{1}{30}+\frac{1}{90}\right)\)
\(S=\left(\frac{2}{12}+\frac{1}{12}\right)+\left(\frac{5}{100}+\frac{1}{100}\right)+\left(\frac{3}{90}+\frac{1}{90}\right)\)
\(S=\frac{3}{12}+\frac{6}{100}+\frac{4}{90}\)
\(S=\frac{1}{4}+\frac{3}{50}+\frac{2}{45}\)
\(S=\frac{319}{900}\)
S=\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.15}+..\)
\(S=SAI\)
HÌNH NHƯ SAI RỒI
HÌNH NHỨIA
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{90}\)
\(=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{9\cdot10}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(=1-\frac{1}{10}\)
\(=\frac{9}{10}\)
Chú ý : Dấu '' . '' là dấu nhân
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49/50
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