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d) Ta có: \(x+\dfrac{4}{5\cdot9}+\dfrac{4}{9\cdot13}+...+\dfrac{4}{41\cdot45}=\dfrac{-37}{45}\)
\(\Leftrightarrow x+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{41}-\dfrac{1}{45}=\dfrac{-37}{45}\)
\(\Leftrightarrow x+\dfrac{1}{5}-\dfrac{1}{45}=\dfrac{-37}{45}\)
\(\Leftrightarrow x=\dfrac{-37}{45}+\dfrac{1}{45}-\dfrac{1}{5}=\dfrac{-36}{45}-\dfrac{1}{5}=\dfrac{-4}{5}-\dfrac{1}{5}=-1\)
Vậy: x=-1
Đặt A=\(\dfrac{2}{3.5}.\dfrac{2}{7.9}.....\dfrac{2}{99.101}\)
A=\(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\)
A=\(\dfrac{1}{3}-\dfrac{1}{101}=\dfrac{98}{303}\)
Ta có: \(P=\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+\dfrac{2}{9\cdot11}+\dfrac{2}{11\cdot13}+\dfrac{2}{13\cdot15}\)
\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{13}-\dfrac{1}{15}\)
\(=\dfrac{1}{3}-\dfrac{1}{15}\)
\(=\dfrac{4}{15}\)
\(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+\dfrac{2}{11.13}+\dfrac{2}{13.15}\)
\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+.....+\dfrac{1}{13}-\dfrac{1}{15}\)
(do \(\dfrac{n}{a.\left(a+n\right)}=\dfrac{1}{a}-\dfrac{1}{a+n}\) với \(a\in N\)*)
\(=\dfrac{1}{3}-\dfrac{1}{15}=\dfrac{4}{15}\)
Chúc bạn học tốt!!!
\(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+\dfrac{2}{11.13}+\dfrac{2}{13.15}\)
= \(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{15}\)
= \(\dfrac{1}{3}-\dfrac{1}{15}\)
= \(\dfrac{4}{15}\)
Ta có: \(N=\dfrac{5}{3.5}-\dfrac{5}{5.7}-\dfrac{5}{7.9}-\dfrac{5}{9.11}-\dfrac{5}{11.13}-\dfrac{5}{13.15}
\)
\(\Rightarrow N=5\left(\dfrac{1}{3.5}-\dfrac{1}{5.7}-\dfrac{1}{7.9}-\dfrac{1}{9.11}-\dfrac{1}{11.13}-\dfrac{1}{13.15}\right)\)
\(\Rightarrow N=\dfrac{5}{2}\left(\dfrac{1}{3}+\dfrac{1}{5}-\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{7}+\dfrac{1}{9}-\dfrac{1}{9}+\dfrac{1}{11}-\dfrac{1}{11}+\dfrac{1}{13}-\dfrac{1}{13}+\dfrac{1}{15}\right)\)
\(\Rightarrow N=\dfrac{5}{2}\left(\dfrac{1}{3}+\dfrac{1}{15}\right)=\dfrac{5}{2}\left(\dfrac{5+1}{15}\right)=\dfrac{5}{2}.\dfrac{6}{15}\)
\(\Rightarrow N=\dfrac{3}{3}=1\)
\(=2x+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{9}-\frac{1}{11}=-\frac{2016}{2017}\)
\(=2x+\frac{1}{3}-\frac{1}{11}=-\frac{2016}{2017}\)
\(2x+\frac{8}{33}=-\frac{2016}{2017}\)
\(2x=\frac{-2016}{2017}-\frac{8}{33}\)
\(2x=\frac{-2024}{2017}\)
\(x=-\frac{1012}{2017}\)
\(2x+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}=\frac{-2016}{2017}\)
\(2x+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}=\frac{-2016}{2017}\)
\(2x+\frac{1}{3}-\frac{1}{11}=\frac{-2016}{2017}\)
\(2x+\frac{8}{33}=\frac{-2016}{2017}\)
\(2x=\frac{-2016}{2017}-\frac{8}{33}\)
Số dư dài quá. Đến đây bạn tự làm tiếp nhé
1) P = 2/3.5 + 2/5.7 + 2/7.9 + 2/9.11 + 2/11.13 + 2/13.15
P= (1/3-1/5) + (1/5-1/7) + (1/7-1/9) + (1/9-1/11) + (1/11-1/13) + (1/13-1/15)
P=1/3-1/15= 4/15
2) a/ 0,2:1+3/5+80%
= 2/10:8/5+8/10
= 2/10.5/8+8/10
= 1/8 + 4/5 = 5/40 + 32/40 = 37/40
b/ 0,5:5/4-2+1/5
= 5/10:5/4-11/5
= 5/10.4/5-11/5
=2/5-11/5 = -9/5
Tính :
\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}+\frac{1}{13.15}\Rightarrow\frac{1}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}+\frac{2}{13.15}\right)=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}\right)=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{15}\right)=\frac{1}{2}\left(\frac{5}{15}-\frac{1}{15}\right)=\frac{1}{2}.\frac{4}{15}=\frac{2}{15}\)
\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}+\frac{1}{13.15}\)
=\(\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}\right)\)
=\(\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{15}\right)=\frac{1}{2}.\frac{4}{15}=\frac{2}{15}\)
\(\text{Ta có:}\) \(\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\right).x=\frac{2}{3}\)
\(\Leftrightarrow2.\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\right).x=\frac{2}{3}.2\)
\(\Leftrightarrow\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right).x=\frac{4}{3}\)
\(\Leftrightarrow\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+.....+\frac{1}{9}-\frac{1}{11}\right).x=\frac{4}{3}\)
\(\Leftrightarrow\left(1-\frac{1}{11}\right)x=\frac{4}{3}\)
\(\Leftrightarrow\frac{10}{11}x=\frac{4}{3}\)
\(\Leftrightarrow x=\frac{4}{3}:\frac{10}{11}=\frac{22}{15}\)
1. Ta có: \(\left|x\right|=7\Rightarrow\left[{}\begin{matrix}x=7\\x=-7\end{matrix}\right.\)
Vậy \(x\in\left\{\pm7\right\}\)
2. \(M=\dfrac{1}{2}.\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+\dfrac{2}{11.13}\right)\)
\(\Rightarrow M=\dfrac{1}{2}.\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13}\right)\)
\(\Rightarrow M=\dfrac{1}{2}.\left(\dfrac{1}{3}-\dfrac{1}{13}\right)\)
\(\Rightarrow M=\dfrac{1}{2}.\left(\dfrac{13}{39}-\dfrac{3}{39}\right)\)
\(\Rightarrow M=\dfrac{1}{2}.\dfrac{10}{39}=\dfrac{1.10}{2.39}=\dfrac{5}{39}\)
Tick mk vs! Thank nhiều!
1. Theo đb ta có: |x|=7
=> Có 2 TH:\(\left\{{}\begin{matrix}x=7\\x=-7\end{matrix}\right.\) \(\in Z\)
Vậy x=7 \(\veebar\) x= -7 ( x\(\in\) Z) thì |x|=7
2. \(\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.11}+\dfrac{1}{11.13}\)
Đặt A= \(\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.11}+\dfrac{1}{11.13}\)
Ta thấy: \(\dfrac{1}{3}-\dfrac{1}{5}=\dfrac{2}{3.5}\)
\(\dfrac{1}{5}-\dfrac{1}{7}=\dfrac{2}{5.7}\)
... \(\dfrac{1}{11}-\dfrac{1}{13}=\dfrac{2}{11.13}\)
=> 2D=2(\(\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.11}+\dfrac{1}{11.13}\))
<=> 2D= \(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+\dfrac{2}{11.13}\)
<=>2D=\(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13}\)
<=> 2D= \(\dfrac{1}{3}-\dfrac{1}{13}\)
<=>2D= \(\dfrac{13}{39}-\dfrac{3}{39}\)
<=>2D=\(\dfrac{10}{39}\)
=> D= \(\dfrac{10}{39}:2\)
<=> D= \(\dfrac{10}{39}.\dfrac{1}{2}\)
<=> D=\(\dfrac{5}{39}\)
Vậy D= \(\dfrac{5}{39}\)
_ Chc bn hk tốt_
\(\frac{x}{3.5}+\frac{x}{5.7}+\frac{x}{7.9}+...+\frac{x}{13.15}=\frac{4}{45}\)
\(\Leftrightarrow\frac{x}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{13.15}\right)=\frac{4}{45}\)
\(\Leftrightarrow\frac{x}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{13}-\frac{1}{15}\right)=\frac{4}{45}\)
\(\Leftrightarrow\frac{x}{2}.\left(\frac{1}{3}-\frac{1}{15}\right)=\frac{4}{45}\)
\(\Leftrightarrow\frac{x}{2}.\frac{4}{15}=\frac{4}{45}\)
\(\Leftrightarrow\frac{x}{2}=\frac{4}{45}:\frac{4}{15}\)
\(\Leftrightarrow\frac{x}{2}=\frac{1}{3}\)
\(\Leftrightarrow x=\frac{1}{3}.2\)
\(\Leftrightarrow x=\frac{2}{3}\)
Vậy x = \(\frac{2}{3}\)
_Chúc bạn học tốt_