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NG
5
11 tháng 1 2018
2/1.3+2/3.5+...+2/x(x+2)= 40/41
1-1/3+1/3-1/5+...+1/x-1/(x+2)=40/41
1-1/(x+2)=40/41
1/(x+2)=1-40/41=1/41
x+2=41
x=41-2=39
LD
6
13 tháng 8 2019
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x\left(x+2\right)}=\frac{20}{41}\)
\(\Leftrightarrow1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{20}{41}\)
\(\Leftrightarrow1-\frac{1}{x+2}=\frac{21}{41}\)
\(\Leftrightarrow\frac{1}{x+2}=1-\frac{21}{41}\)
\(\Leftrightarrow\frac{1}{x+2}=\frac{20}{41}\)
\(\Leftrightarrow20\left(x+2\right)=41\)
\(\Leftrightarrow x-2=\frac{41}{20}\)
\(\Leftrightarrow x=\frac{41}{20}+2\)
\(\Leftrightarrow x=\frac{81}{20}\)
S
13 tháng 8 2019
\(\frac{1}{1.3}+...+\frac{1}{a\left(a+2\right)}=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+....+\frac{2}{a\left(a+2\right)}\right)=\frac{1}{2}\left(1-\frac{1}{3}+....-\frac{1}{a+2}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{a+2}\right)=\frac{20}{41}\Rightarrow a+2=41\Leftrightarrow a=39\)
19 tháng 2 2017
a) Ta có
1/1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + ... + 1/x - 1/x + 2 = 5
rút gọn ta được : 1 - 1/x+2 = 5
<=> x + 2 - 1 / x+ 2
<=> x + 1 / x + 2 = 5
<=> x+ 1 = 5x + 10
<=> - 4x = 9
<=> x = -9/4
B) / x +4/ = 2^0 + 1 ^ 2013
=> /x + 4/ = 1 + 1
=> / x + 4 / = 2
TH 1 : x+ 4 = 2
=> x = 2 - 4 = -2
TH2 : x + 4 = -2
=> x = -2 - 4 = -6
=> x = { - 2 , -6 }
RL
6 tháng 4 2018
Ta có :
\(\dfrac{1}{2}\)(\(\dfrac{1}{1}\)-\(\dfrac{1}{3}\)+\(\dfrac{1}{3}\)-\(\dfrac{1}{5}\)+\(\dfrac{1}{5}\)-\(\dfrac{1}{7}\)+...+\(\dfrac{1}{x}\)-\(\dfrac{1}{x+2}\))=\(\dfrac{20}{41}\)
\(\dfrac{1}{2}\)(\(\dfrac{1}{3}\)-\(\dfrac{1}{x+2}\))=\(\dfrac{20}{41}\)
\(\dfrac{1}{3}\)-\(\dfrac{1}{x+2}\)=\(\dfrac{40}{41}\)
\(\dfrac{1}{x+2}\)=\(\dfrac{1}{3}\)-\(\dfrac{40}{41}\)
SC
3
26 tháng 4 2015
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x\left(x+2\right)}=\frac{20}{41}\)
\(\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{x\left(x+2\right)}\right)=\frac{20}{41}\)
\(\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}\right)=\frac{20}{41}\)
\(\frac{1}{2}\left(1-\frac{1}{x+2}\right)=\frac{20}{41}\)
\(\frac{1}{2}.\frac{x+1}{x+2}=\frac{20}{41}\)
\(\frac{x+1}{x+2}=\frac{20}{41}:\frac{1}{2}\)
\(\frac{x+1}{x+2}=\frac{40}{41}\)
\(x+1=40
\)
\(x=40-1\)
\(x=39\)
Đúng thì ****
MC
1 tháng 7 2018
a) \(|2x|.|3.5|=|-2.8|\)
\(\Rightarrow|2x|.|15|=|-16|\)
\(\Rightarrow|2x|.15=16\)
\(\Rightarrow|2x|=\dfrac{16}{15}\)
Ta có hai trường hợp:
1) \(2x\ge0\)
\(\Rightarrow|2x|=2x=\dfrac{16}{15}\Rightarrow x=\dfrac{8}{15}\)
2) \(2x< 0\)
\(\Rightarrow|2x|=-2x=\dfrac{16}{15}\Rightarrow-x=\dfrac{8}{15}\Rightarrow x=-\dfrac{8}{15}\)
Vậy \(x=\dfrac{8}{15}\) hoặc \(x=-\dfrac{8}{15}\).
b) \(|x-\dfrac{1}{2}|+|x+y|=0\)
\(\Rightarrow\left\{{}\begin{matrix}x-\dfrac{1}{2}=0\\x+y=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=-\dfrac{1}{2}\end{matrix}\right.\)
Vậy \(\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=-\dfrac{1}{2}\end{matrix}\right.\)