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\(\frac{19}{37}+\left(1-\frac{19}{37}\right)\)
\(=\frac{19}{37}+1-\frac{19}{37}\)
\(=\left(\frac{19}{37}-\frac{19}{37}\right)+1\)
\(=0+1=1\)
\(\frac{4}{1\cdot3\cdot5}+\frac{4}{3\cdot5\cdot7}+\frac{4}{5\cdot7\cdot9}+\frac{4}{7\cdot9\cdot11}+\frac{4}{9\cdot11\cdot13}\)
\(=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{9.11}-\frac{1}{11.13}\)
\(=\frac{1}{1.3}-\frac{1}{11.13}\)
\(=\frac{1}{3}-\frac{1}{143}\)
\(=\frac{140}{429}\)
M = \(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\)
M = \(\frac{2}{1}-\frac{2}{3}+\frac{2}{3}-\frac{2}{5}+\frac{2}{5}-\frac{2}{7}+\frac{2}{7}-\frac{2}{9}+\frac{2}{9}-\frac{2}{11}\)
M = \(\frac{2}{1}-\frac{2}{11}\)
M = \(\frac{20}{11}\)
= \(\frac{1}{2}\cdot\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-...-\frac{1}{2x+1}+\frac{1}{2x+1}-\frac{1}{2x+3}\right)\)
= \(\frac{1}{2}\cdot\left(\frac{1}{3}-\frac{1}{2x+3}\right)=\frac{1}{6}-\frac{1}{4x+6}\)
Đặt \(A=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{\left(2x+1\right)\left(2x+3\right)}\)
\(\Rightarrow2A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{\left(2x+1\right)}-\frac{1}{2x+3}\)
\(\Rightarrow2A=\frac{1}{3}-\frac{1}{2x+3}\)
\(\Rightarrow A=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{2x+3}\right)\)
\(\Rightarrow A=\frac{1}{6}-\frac{1}{4x+6}\)
\(\frac{3}{5.7}+\frac{3}{7.9}+\frac{3}{9.11}+...+\frac{3}{59.61}\)
\(=\)\(\frac{3}{2}\left(\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+...+\frac{2}{59.61}\right)\)
\(=\)\(\frac{3}{2}\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{59}-\frac{1}{61}\right)\)
\(=\)\(\frac{3}{2}\left(\frac{1}{5}-\frac{1}{61}\right)\)
\(=\)\(\frac{3}{2}.\frac{56}{305}\)
\(=\)\(\frac{84}{305}\)
Chúc bạn học tốt ~
1) \(A=\frac{4}{5.7}+\frac{4}{7.9}+...+\frac{4}{59.61}\)
\(A=2.\left(\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{59.61}\right)\)
\(A=2.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+..+\frac{1}{59}-\frac{1}{61}\right)\)
tiếp theo bạn tính kết quả trong ngoặc rồi nhân với 2 là ra kết quả của phần 1
phần 2 tách 3^2 = 3.3 sau đó lấy thừa số chung là 3,tiếp theo làm như phần 1 là ra kết quả
d)\(\frac{2.3+4.6+14.21}{3.5+6.10+21.35}=\frac{2.3+2.2.6+2.7.21}{3.5+3.2.10+3.7.35}=\frac{2.3+2.12+2.147}{3.5+3.20+3.245}=\frac{2\left(3+12+147\right)}{3\left(5+20+245\right)}\)
\(=\frac{2.162}{3.270}=\frac{54}{135}=\frac{2}{5}\)
\(a.\frac{-2019.2018+1}{\left(-2017\right).\left(-2019\right)+2018}\)
\(=\frac{2019.\left(-2018\right)+1}{2019.2017+2018}\)
\(=\frac{2019.\left(-2018\right)+1}{2019.2018-1}\)
\(=-\frac{2018}{2018}\)
\(=-1\)
1/2A=2/5.7+2/7.9+...+2/2017.2019
1/2A=1/5-1/7+1/7-1/9+...+1/2017-1/2019
1/2A=1/5-1/2019
1/2A=2014/10095
A=4028/10095
\(\approx7.314\)