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Bài 1 :
\(S=1.3+3.5+5.7+...+99.101=3+15+35+...9999\)
Ta thấy :
\(3=2^2-1\)
\(15=4^2-1\)
\(35=6^2-1\)
.....
\(9999=100^2-1\)
\(\Rightarrow S=2^2+4^2+...+100^2-\left(1\right).\left(\left(100-2\right):2+1\right)\)
\(\Rightarrow S=\dfrac{100.\left(100+1\right)\left(2.100+1\right)}{6}-51\)
\(\Rightarrow S=\dfrac{100.101.201}{6}-51=338299\)
8.9.14+6.7.12+19.4.18
=72.14+72.7+72.19
=72(14+7+19)
=72.40
=2880
\(S=\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+...+\dfrac{1}{97.99}\)
\(2S=\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{97.99}\)
\(2S=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{97}-\dfrac{1}{99}\)
\(2S=\dfrac{1}{3}-\dfrac{1}{99}\)
\(2S=\dfrac{32}{99}\)
\(S=\dfrac{32}{99}:2\)
\(S=\dfrac{16}{99}\)
Cảm ơn bạn rất nhiều! Bạn đã cứu mình rùi! Xin trân thành cảm ơn!
Ta có : \(S=\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)
\(\Rightarrow2S=2\left(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\right)\)
\(\Rightarrow2S=\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\)
\(\Rightarrow2S=\left(\frac{1}{3}-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{7}\right)+...+\left(\frac{1}{97}-\frac{1}{99}\right)\)
\(\Rightarrow2S=\left(\frac{1}{3}-\frac{1}{99}\right)+\left[\left(\frac{1}{5}+...+\frac{1}{97}\right)-\left(\frac{1}{5}+\frac{1}{7}+...+\frac{1}{97}\right)\right]\)
\(\Rightarrow2S=\left(\frac{33}{99}-\frac{1}{99}\right)+0\)
\(\Rightarrow2S=\frac{32}{99}\)
\(\Rightarrow S=\frac{32}{99}\div2\)
\(\Rightarrow S=\frac{16}{99}\)
\(S=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{97.99}\)
\(S=\frac{1}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\right)\)
\(S=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(S=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{99}\right)\)
\(S=\frac{1}{2}.\frac{32}{99}=\frac{16}{99}\)
\(8.6+288\left(x-3\right)^2=50\)
\(=>48+228\left(x-3\right)^2=50\)
\(=>288\left(x-3\right)^2=2\)
\(=>\left(x-3\right)^2=\frac{1}{144}\)
\(=>x-3=\frac{1}{12}\)
\(=>x=\frac{37}{12}\)
12.T=2.6.12+6.10.12+10.14.12+...+102.106.12=
=2.6.(10+2)+6.10.(14-2)+10.14.(18-6)+...+102.106.(110-98)=
=2.2.6+2.6.10-2.6.10+6.10.14-6.10.14+10.14.18-...-98.102.106+102.106.110=
=2.2.6+102.106.110
\(\Rightarrow T=\dfrac{2.2.6+102.106.110}{12}=99112\)
\(A=\frac{1}{6.10}+\frac{1}{10.14}+\frac{1}{14.18}+...+\frac{1}{402.406}\)
4\(A=\frac{1}{6}-\frac{1}{10}+\frac{1}{10}-\frac{1}{14}+\frac{1}{14}-\frac{1}{18}+...+\frac{1}{402}-\frac{1}{406}\)
4\(A=\frac{1}{6}-\frac{1}{406}\)
4\(A=\frac{100}{609}\)
\(\Rightarrow A=\frac{100}{609}:4\)\(=\frac{25}{609}\)
=1/6-1/10+1/10-1/14+1/14-1/18+...........+1/402-1/406
=1/6-1/406
Sửa đề: \(D=\dfrac{3}{2\cdot6}+\dfrac{3}{6\cdot10}+\dfrac{3}{10\cdot14}+...+\dfrac{3}{26\cdot30}\)
Ta có: \(D=\dfrac{3}{2\cdot6}+\dfrac{3}{6\cdot10}+\dfrac{3}{10\cdot14}+...+\dfrac{3}{26\cdot30}\)
\(=\dfrac{3}{4}\left(\dfrac{4}{2\cdot6}+\dfrac{4}{6\cdot10}+\dfrac{4}{10\cdot14}+...+\dfrac{4}{26\cdot30}\right)\)
\(=\dfrac{3}{4}\left(\dfrac{1}{2}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{14}+...+\dfrac{1}{26}-\dfrac{1}{30}\right)\)
\(=\dfrac{3}{4}\left(\dfrac{1}{2}-\dfrac{1}{30}\right)\)
\(=\dfrac{3}{4}\cdot\dfrac{28}{60}\)
\(=\dfrac{21}{60}=\dfrac{7}{20}\)
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