p= 2/3x5+2/5x7+2/7x9+...+2/103x105
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Ta có:
\(\frac{2}{5.7}+\frac{2}{7.9}+........+\frac{2}{103.105}\)
\(=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+..........+\frac{1}{103}-\frac{1}{105}\)
\(=\frac{1}{5}-\frac{1}{105}=\frac{21}{105}-\frac{1}{105}=\frac{20}{105}=\frac{4}{21}\)
M=2/3.5+2/5.7+/.............+2/97/99
=1/3-1/5+1/5-1/7+......+1/97-1/99
=1/3-1/99
=32/99
OK
xin 5 tích vì mất 5 phút
Ta có : \(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{11.13}\)
\(=\frac{1}{3}+\frac{1}{5}-\frac{1}{5}+......+\frac{1}{11}-\frac{1}{13}\)
\(=\frac{1}{3}-\frac{1}{13}\)
\(=\frac{10}{39}\)
\(\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{11\times13}\)
\(=\frac{1}{3\times5}+\frac{1}{5\times7}+...+\frac{1}{11\times13}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{13}\)
\(=\frac{1}{3}-\frac{1}{13}=\frac{10}{39}\)
\(\dfrac{2}{1\times3}+\dfrac{2}{3\times5}+\dfrac{2}{5\times7}+\dfrac{2}{7\times9}+\dfrac{2}{9\times11}\)
\(=2\times\dfrac{1}{2}\times\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}\right)\)
\(=1-\dfrac{1}{11}\)
\(=\dfrac{11}{11}-\dfrac{1}{11}\)
\(=\dfrac{10}{11}\)
sửa đề câu a và câu b nhá , mik nghĩ đề như này :
\(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{213\cdot215}\)
\(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{213}-\frac{1}{215}\)
= \(\frac{1}{1}-\frac{1}{215}\)
\(=\frac{214}{215}\)
b, đặt \(A=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+...+\frac{1}{213\cdot215}\)
\(A\cdot2=\frac{2}{1\cdot3}+\frac{2}{3.5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{213\cdot215}\)
\(A\cdot2=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{213}-\frac{1}{215}\)
\(A\cdot2=\frac{1}{1}-\frac{1}{215}\)
\(A\cdot2=\frac{214}{215}\)
\(A=\frac{214}{215}:2\)
\(A=\frac{107}{215}\)
\(B=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{99.101}\)
\(\Rightarrow B=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
\(\Rightarrow B=1-\frac{1}{101}=\frac{101}{101}-\frac{1}{101}=\frac{100}{101}\)
_Học tốt_
\(\frac{2}{1\times3}+\frac{2}{3\times5}+...+\frac{2}{19\times21}=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{19}-\frac{1}{21}\)
\(=1-\frac{1}{21}=\frac{20}{21}\)
đúng cái nhé
\(=1-\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{19}-\dfrac{1}{21}\right)\)
\(=1-\dfrac{6}{21}=\dfrac{15}{21}=\dfrac{5}{7}\)
\(P=\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{103\cdot105}\)
\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{103}-\dfrac{1}{105}\)
\(=\dfrac{1}{3}-\dfrac{1}{105}=\dfrac{34}{105}\)
\(P=\dfrac{2}{3\times5}+\dfrac{2}{5\times7}+\dfrac{2}{7\times9}+...+\dfrac{2}{103\times105}\)
\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{103}-\dfrac{1}{105}\)
\(=\dfrac{1}{3}-\dfrac{1}{105}=\dfrac{34}{105}\)
Công thức: \(\dfrac{a}{n\left(n+a\right)}=\dfrac{1}{n}-\dfrac{1}{n+a}\)