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Giusp m với. Tìm số A biết. 121:A dư 10, 61 : A dư 10. giải thích cách làm theo lớp 6 b nhé.Thank
Ta có: \(A=\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\)
\(\Leftrightarrow A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\)
\(\Rightarrow2A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\)
\(\Rightarrow2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\)
\(\Rightarrow2A=1-\frac{1}{11}=\frac{10}{11}\)
\(\Rightarrow A=\frac{10}{11}:2=\frac{5}{11}\)
Vậy \(A=\frac{5}{11}\)
A = \(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\)
A = \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\)
A = \(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\)
A = \(1-\frac{1}{11}\)
A = \(\frac{10}{11}\)
Bài làm:
Ta có:
\(B=-66\cdot\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{11}\right)+124.\left(-37\right)+63.\left(-124\right)\)
\(B=\left(-66\right).\frac{1}{2}+66.\frac{1}{3}-66.\frac{1}{11}-124.\left(37+63\right)\)
\(B=-33+22-6-124.100\)
\(B=17-12400\)
\(B=-12383\)
\(B=-66.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{11}\right)+124.\left(-37\right)+63.\left(-124\right)\)
\(=-66.\frac{1}{2}-\left(-66\right).\frac{1}{3}+\left(-66\right).\frac{1}{11}+\left(-124\right).37+63.\left(-124\right)\)
\(=-33+22-6+\left(-124\right).\left(37+63\right)\)
\(=-11-6+\left(-124\right).100\)
\(=-17-12400\)
\(=-12417\)
=2/10+3/10+4/10+......+13/10
=\(\frac{2+3+4+......+13}{10}\)
=90/10=9
k cho mình nha
\(1)\)\(\frac{3}{4}\cdot2+\frac{5}{2}\cdot\frac{1}{3}=\frac{3}{2}+\frac{5}{6}=\frac{9+5}{6}=\frac{14}{6}=\frac{7}{3}\)
\(2)\)\(\frac{5}{2}+\frac{3}{11}\cdot\frac{7}{26}\left(19-6\right)=\frac{5}{2}+\frac{3\cdot7}{11\cdot2}=\frac{5}{2}+\frac{21}{22}==\frac{38}{11}\)