cho em hỏi 1+1x3+2
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NT
2
XO
16 tháng 8 2019
2) \(\frac{3}{1\times3}+\frac{3}{3\times5}+\frac{3}{5\times7}+...+\frac{3}{99\times101}+\frac{3}{101\times103}\)
\(=\frac{3}{2}\times\left(\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{101\times103}\right)\)
\(=\frac{3}{2}\times\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{101}-\frac{1}{103}\right)\)
\(=\frac{3}{2}\times\left(1-\frac{1}{103}\right)\)
\(=\frac{3}{2}\times\frac{101}{103}\)
\(=\frac{303}{206}\)
DB
3
VN
Võ Ngọc Phương
VIP
8 tháng 8 2023
a) \(\dfrac{1}{1\times3}+\dfrac{1}{3\times5}+\dfrac{1}{5\times7}+...+\dfrac{1}{x\times\left(x+3\right)}=\dfrac{99}{200}\)
Ta có: \(\left(1-\dfrac{1}{3}\right)\times\dfrac{1}{2}+\left(\dfrac{1}{3}-\dfrac{1}{5}\right)\times\dfrac{1}{2}+\left(\dfrac{1}{5}-\dfrac{1}{7}\right)\times\dfrac{1}{2}+...+\left(\dfrac{1}{x}-\dfrac{1}{x+3}\right).\dfrac{1}{2}=\dfrac{99}{200}\)
\(\dfrac{1}{2}\times\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{x}-\dfrac{1}{x+3}\right)=\dfrac{99}{200}\)
\(\dfrac{1}{2}\times\left(1-\dfrac{1}{x+3}\right)=\dfrac{99}{200}\)
\(1-\dfrac{1}{x+3}=\dfrac{99}{200}:\dfrac{1}{2}\)
\(1-\dfrac{1}{x+3}=\dfrac{99}{100}\)
\(\dfrac{1}{x+1}=1-\dfrac{99}{100}\)
\(\dfrac{1}{x+1}=\dfrac{1}{100}\)
\(\Rightarrow x+1=100\)
\(x=100-1\)
\(x=99\)
17 tháng 12 2017
a, Đặt :
\(A=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+..............+\dfrac{1}{19.21}\)
\(\Leftrightarrow2A=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+............+\dfrac{2}{19.21}\)
\(\Leftrightarrow2A=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+..........+\dfrac{1}{19}-\dfrac{1}{21}\)
\(\Leftrightarrow2A=1-\dfrac{1}{21}\)
\(\Leftrightarrow2A=\dfrac{20}{21}\)
\(\Leftrightarrow A=\dfrac{10}{21}\)
17 tháng 12 2017
b, \(A=\dfrac{1}{1.3}+\dfrac{1}{3.5}+...........+\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}\)
\(\Leftrightarrow2A=\dfrac{2}{1.3}+\dfrac{2}{3.5}+............+\dfrac{2}{\left(2n-1\right)\left(2n+1\right)}\)
\(\Leftrightarrow2A=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+........+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\)
\(\Leftrightarrow2A=1-\dfrac{1}{2n+1}\)
\(\Leftrightarrow2A=\dfrac{2n}{2n+1}\)
\(\Leftrightarrow A=\dfrac{n}{2n+1}\)
HT
27 tháng 10 2020
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 \(A=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+...+\frac{1}{213\cdot215}\)
\(A\cdot2=\frac{2}{1\cdot3}+\frac{2}{3.5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{213\cdot215}\)
\(A\cdot2=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{213}-\frac{1}{215}\)
\(A\cdot2=\frac{1}{1}-\frac{1}{215}\)
\(A\cdot2=\frac{214}{215}\)
\(A=\frac{214}{215}:2\)
\(A=\frac{107}{215}\)
31 tháng 8 2016
\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{11.13}+\frac{2}{13.15}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}\)
\(=1-\frac{1}{15}\)
\(=\frac{14}{15}\)
NH
1
TL :
1 + 1 x 3 + 2
= 2 x 5
= 10
HT
=8 nha ! 1+1=2x3=6+2=8 ! Chúc em học tốt ! TIM TIM !