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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
\(a.\) \(3\frac{3}{4}+\left(4\frac{2}{4}-3\frac{1}{2}\right):\frac{3}{4}\)
\(=\frac{15}{4}\left(\frac{18}{4}-\frac{7}{2}\right):\frac{3}{4}\)
\(=\frac{15}{4}+\frac{4}{4}:\frac{3}{4}=\frac{15}{4}+\frac{4}{3}\)
\(=\frac{15}{4}+\frac{4}{3}=\frac{61}{12}\)
\(b.\) \(7\cdot\frac{2}{3}-\frac{2}{5}:\frac{1}{2}-\frac{2}{3}\)
\(=\frac{7}{3}-\frac{2}{5}\cdot2-\frac{2}{3}\)
\(=\frac{7}{3}-\frac{4}{5}-\frac{2}{3}=\frac{35-12-10}{15}\)
\(=\frac{13}{15}\)
\(c.\) \(68,7-100:20+70,8\)
\(=68,7-5+70,8\)
\(=63,7+70,8\)
\(=134,5\)
\(d.\)\(\left(5915+445:5\right)-76\cdot25\)
\(=\left(5915+89\right)-1900\)
\(=6004-1900=4104\)
a)=\(\frac{15}{4}+\left(\frac{18}{4}-\frac{7}{2}\right)\)/ \(\frac{3}{4}\)
=\(\left(\frac{33}{4}-\frac{14}{4}\right)\)/ \(\frac{3}{4}\)
= \(\frac{19}{4}\cdot\frac{4}{3}\)
=\(\frac{19}{3}\)
b) = \(\frac{2}{3}\cdot\left(7-1\right)-\frac{2}{5}\cdot\frac{2}{1}\)
= \(\frac{2}{3}\cdot6-\frac{4}{5}\)
= 1-\(\frac{4}{5}\)
= \(\frac{1}{5}\)
c) = 68.7 - 5 + 70.8
= 63.7 + 70.8
=134.5
d) = (5915 - 89) -76*25
= 5826 - 1900
= 3926
\(\frac{2}{3}.\frac{4}{5}+\frac{1}{3}.\frac{4}{5}=\frac{4}{5}\left(\frac{2}{3}+\frac{1}{3}\right)=\frac{4}{5}.\frac{3}{3}=\frac{4}{5}.1=\frac{4}{5}\)
\(\frac{1}{2}:\frac{3}{4}+\frac{1}{6}:\frac{3}{4}=\frac{3}{4}:\left(\frac{1}{2}+\frac{1}{6}\right)=\frac{3}{4}:\frac{2}{3}=\frac{9}{8}\)
\(\frac{2}{3}.\frac{4}{5}-\frac{1}{3}.\frac{4}{5}=\frac{4}{5}\left(\frac{2}{3}-\frac{1}{3}\right)=\frac{4}{5}.\frac{1}{3}=\frac{4}{15}\)
\(\frac{1}{2}:\frac{3}{4}-\frac{1}{6}:\frac{3}{4}=\frac{3}{4}:\left(\frac{1}{2}-\frac{1}{6}\right)=\frac{3}{4}:\frac{1}{3}=\frac{9}{4}\)
\(\frac{2}{3}.\frac{4}{5}+\frac{1}{3}.\frac{4}{5}=\left(\frac{2}{3}+\frac{1}{3}\right).\frac{4}{5}=1.\frac{4}{5}=\frac{4}{5}\)
\(\frac{1}{2}:\frac{3}{4}+\frac{1}{6}:\frac{3}{4}=\frac{1}{2}.\frac{4}{3}+\frac{1}{6}.\frac{4}{3}=\left(\frac{1}{2}+\frac{1}{6}\right).\frac{4}{3}=\frac{2}{3}.\frac{4}{3}=\frac{8}{9}\)
c,d tương tự
\(a.\)\(1\frac{2}{3}:\frac{2}{3}-\frac{3}{4}\cdot\frac{2}{3}+5\frac{3}{7}\)
\(=\frac{5}{3}:\frac{2}{3}-\frac{3}{4}\cdot\frac{2}{3}+\frac{38}{7}\)
\(=\frac{5}{3}\cdot\frac{3}{2}-\frac{3}{4}\cdot\frac{2}{3}+\frac{38}{7}\)
\(=\frac{5}{2}-\frac{1}{2}+\frac{38}{7}\)
\(=\frac{4}{2}+\frac{38}{7}\)
\(=2+\frac{38}{7}\)
\(=\frac{14}{7}+\frac{38}{7}\)
\(=\frac{52}{7}\)
\(b.1\frac{1}{3}-1\frac{1}{4}:1\frac{1}{2}+2\frac{3}{4}\cdot3\frac{2}{3}\)
\(=\frac{4}{3}-\frac{5}{4}:\frac{3}{2}+\frac{11}{4}\cdot\frac{11}{3}\)
\(=\frac{4}{3}-\frac{5}{4}\cdot\frac{2}{3}+\frac{11}{4}\cdot\frac{11}{3}\)
\(=\frac{4}{3}-\frac{5}{6}+\frac{121}{12}\)
\(=\frac{16}{12}-\frac{10}{12}+\frac{121}{12}\)
\(=\frac{6}{12}+\frac{121}{12}\)
\(=\frac{127}{12}\)
\(c.7\cdot\frac{2}{3}-\frac{2}{5}:\frac{1}{2}-\frac{2}{3}\)
\(=7\cdot\frac{2}{3}-\frac{2}{5}\cdot\frac{2}{1}-\frac{2}{3}\)
\(=7\cdot\frac{2}{3}-\frac{4}{5}-\frac{2}{3}\)
\(=\frac{14}{3}-\frac{4}{5}-\frac{2}{3}\)
\(=\frac{70}{15}-\frac{12}{15}-\frac{10}{15}\)
\(=\frac{58}{15}-\frac{10}{15}\)
\(=\frac{48}{15}=\frac{16}{5}\)
\(\frac{5}{3}:\frac{2}{3}-\frac{3}{4}\cdot\frac{2}{3}+\frac{38}{7}\)
\(\frac{5}{2}-\frac{1}{2}+\frac{38}{7}\)
\(2+\frac{38}{7}\)
\(\frac{52}{7}\)
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