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1 tháng 5 2018
Đặt \(A=\frac{4}{12}+\frac{4}{20}+\frac{4}{30}+...+\frac{4}{306}\)
\(\Rightarrow A=4\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{306}\right)\)
\(\Rightarrow A=4\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{17}-\frac{1}{18}\right)\)
\(\Rightarrow A=4\left(\frac{1}{3}-\frac{1}{18}\right)\)
\(\Rightarrow A=4.\frac{5}{18}\)
\(\Rightarrow A=\frac{10}{9}\)
1 tháng 5 2018
Đặt A= \(\frac{4}{12}+\frac{4}{20}+\frac{4}{30}+....+\frac{4}{306}\)
\(=4\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{306}\right)\)
\(=4\left(\frac{1}{3.4}+\frac{1}{3.5}+\frac{1}{5.6}+......+\frac{1}{17.18}\right)\)
\(=4\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{17}-\frac{1}{18}\right)\)
\(=4.\left(\frac{1}{3}-\frac{1}{18}\right)\)
Vậy A = \(\frac{10}{9}\)
30 tháng 3 2018
Bài 1:
\(\frac{4}{12}+\frac{4}{20}+\frac{4}{30}+...+\frac{4}{306}\)
\(=4\cdot\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{306}\right)\)
\(=4\cdot\left(\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+...+\frac{1}{17\cdot18}\right)\)
\(=4\cdot\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{17}-\frac{1}{18}\right)\)
\(=4\cdot\left(\frac{1}{3}-\frac{1}{18}\right)\)
\(=4\cdot\left(\frac{6}{18}-\frac{1}{18}\right)\)
\(=4\cdot\frac{5}{18}\)
\(=\frac{10}{9}\)
Bài 2 :
\(\left(3x-4\right)-\left(6x+7\right)=8\)
\(3x-4-6x-7=8\)
\(\left(3x-6x\right)-\left(4+7\right)=8\)
\(-3x-11=8\)
\(-3x=8+11\)
\(-3x=19\)
\(x=19:\left(-3\right)\)
\(x=\frac{-19}{3}\)
Vậy \(x=\frac{-19}{3}\)
b ) \(\left(\frac{4}{5}x+3\right):\left(-4\right)=\frac{1}{2}\)
\(\frac{4}{5}x+3=\frac{1}{2}\cdot\left(-4\right)\)
\(\frac{4}{5}x+3=-2\)
\(\frac{4}{5}x=\left(-2\right)-3\)
\(\frac{4}{5}x=-5\)
\(x=\left(-5\right):\frac{4}{5}\)
\(x=\left(-5\right)\cdot\frac{4}{5}\)
\(x=-4\)
Vậy \(x=-4\)
k nha !
30 tháng 3 2018
\(\frac{4}{12}\)+\(\frac{4}{20}\)+...+\(\frac{4}{306}\)=\(\frac{4}{3.4}\)+\(\frac{4}{4.5}\)+...+\(\frac{4}{17.18}\)=4(\(\frac{1}{3}\)-\(\frac{1}{4}\)+\(\frac{1}{4}\)-\(\frac{1}{5}\)+...+\(\frac{1}{17}\)-\(\frac{1}{18}\))
=4(\(\frac{1}{3}\)-\(\frac{1}{8}\))=4.\(\frac{5}{24}\)=\(\frac{5}{6}\)
17 tháng 2 2020
\(M=\frac{2.6.10+4.12.20+...+20.60.100}{1.2.3+2.4.6+...+10.20.30}=\frac{2.6.10.1^3+2.6.10.2^3+...+2.6.10.10^3}{1.2.3.1^3+1.2.3.2^3+...+1.2.3.10^3}\)
\(=\frac{2.6.10.\left(1^3+2^3+...+10^3\right)}{1.2.3.\left(1^3+2^3+...+10^3\right)}=\frac{2.6.10}{1.2.3}=20\)
vậy M=20
ND
3
28 tháng 7 2015
\(A=\frac{4}{2}+\frac{4}{6}+\frac{4}{12}+\frac{4}{20}+\frac{4}{30}+\frac{4}{42}\)
\(A=\frac{4}{1.2}+\frac{4}{2.3}+\frac{4}{3.4}+\frac{4}{4.5}+\frac{4}{5.6}+\frac{4}{6.7}\)
\(A=4\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)
\(A=4\left(1-\frac{1}{7}\right)\)
\(A=4.\frac{6}{7}\)
\(A=\frac{24}{7}\)
28 tháng 7 2015
\(A=\frac{4}{2}+\frac{4}{6}+\frac{4}{12}+\frac{4}{20}+\frac{4}{30}+\frac{4}{42}=4\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)\)
\(=4\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)\)
\(=4\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)
\(=4\left(1-\frac{1}{7}\right)=\frac{6}{7}.4=\frac{24}{7}\)
Đặt \(A=\frac{4}{12}+\frac{4}{20}+\frac{4}{30}+...+\frac{4}{306}\)
\(=4.\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{306}\right)\)
\(=4.\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{17.18}\right)\)
\(=4.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{17}-\frac{1}{18}\right)\)
\(=4.\left(\frac{1}{3}-\frac{1}{18}\right)\)
\(=4.\frac{5}{18}=\frac{10}{9}\)
Vậy A=\(\frac{10}{9}\)
\(\frac{4}{12}+\frac{4}{20}+\frac{4}{30}+.......+\frac{4}{306}\)
\(=\frac{4}{3.4}+\frac{4}{4.5}+\frac{4}{5.6}+.....+\frac{4}{17.18}\)
\(=4\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+.......+\frac{1}{17.18}\right)\)
\(=4\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+......+\frac{1}{17}-\frac{1}{18}\right)\)
\(=4\left(\frac{1}{3}-\frac{1}{18}\right)\)
\(=4.\frac{5}{18}\)
\(=\frac{10}{9}\)