1/3*5+1/5*7+...+1/(2x+1)*(2x+3)=15/93
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NT
1
DD
Đoàn Đức Hà
Giáo viên
26 tháng 11 2021
\(C=\frac{1}{3\times5}+\frac{1}{5\times7}+\frac{1}{7\times9}+...+\frac{1}{\left(2x+1\right)\times\left(2x+3\right)}\)
\(=\frac{1}{2}\times\left(\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{\left(2x+1\right)\times\left(2x+3\right)}\right)\)
\(=\frac{1}{2}\times\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2x+1}-\frac{1}{2x+3}\right)\)
\(=\frac{1}{2}\times\left(\frac{1}{3}-\frac{1}{2x+3}\right)=\frac{15}{93}\)
\(\Leftrightarrow2x+3=93\)
\(\Leftrightarrow x=45\).
DD
0
9 tháng 8 2016
\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+.....+\frac{1}{\left(2x+1\right).\left(2x+3\right)}=\frac{15}{93}\)
\(2.\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+.....+\frac{1}{\left(2x+1\right).\left(2x+3\right)}\right)=2.\frac{15}{93}\)
\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+.....+\frac{2}{\left(2x+1\right).\left(2x+3\right)}=\frac{10}{31}\)
\(\frac{1}{3}-\frac{1}{2x+3}=\frac{10}{31}\)
\(\frac{1}{2x+3}=\frac{1}{3}-\frac{10}{31}\)
\(\frac{1}{2x+3}=\frac{1}{93}\)
\(\Rightarrow2x+3=93\)
\(\Rightarrow2x=90\)
\(\Rightarrow x=45\)
TA
10 tháng 8 2015
1-1/6+1/6-1/11+....+1/(5x+1)-1/(5x+2)=2010/2011 <=>1-1/(5x+2)=2010/2011 <=>1/2011=1/(5x+2) <=>x=401
3 tháng 2 2021
1/ 3-2x+4+6x=x+7+3x
⇔-2x+6x-x-3x=0
⇔0x=0 (Vô số nghiệm)
2/-6(1,5-2x)=3(-15+2x)
⇔-9+12x=-45+6x
⇔6x+36=0
⇔6(x+6)=0
⇔x+6=0
⇔x=-6
Vậy S ϵ {-6}
3/ 3(2x-5)+5(x-1)=4(x+1)
⇔6x-15+5x-5=4x+4
⇔7x=24
⇔x=\(\dfrac{24}{7}\)
Vậy S ϵ {\(\dfrac{24}{7}\)}
3 tháng 2 2021
1) Ta có: \(3-2x+4+6x=x+7+3x\)
\(\Leftrightarrow4x+7=4x+7\)
\(\Leftrightarrow4x+7-4x-7=0\)
\(\Leftrightarrow0x=0\)(luôn đúng)
Vậy: S={x|\(x\in R\)}
2) Ta có: \(-6\cdot\left(1.5-2x\right)=3\left(-15+2x\right)\)
\(\Leftrightarrow-9+12x=-45+6x\)
\(\Leftrightarrow12x-9+45-6x=0\)
\(\Leftrightarrow6x+36=0\)
\(\Leftrightarrow6x=-36\)
hay x=-6
Vậy: S={-6}
3) Ta có: \(3\left(2x-5\right)+5\left(x-1\right)=4\left(x+1\right)\)
\(\Leftrightarrow6x-15+5x-5=4x+4\)
\(\Leftrightarrow11x-20-4x-4=0\)
\(\Leftrightarrow7x-24=0\)
\(\Leftrightarrow7x=24\)
\(\Leftrightarrow x=\dfrac{24}{7}\)
Vậy: \(S=\left\{\dfrac{24}{7}\right\}\)
11 tháng 8 2018
a) \(\left(2x-1\right)+\frac{3}{15}=\frac{3}{2}\)
\(\Rightarrow2x-1=\frac{3}{2}-\frac{3}{15}=\frac{13}{10}\)
\(\Rightarrow2x=\frac{13}{10}+1=\frac{23}{10}\)
\(\Rightarrow x=\frac{23}{20}\)
b) \(x+\frac{46}{15}=1,5\)
\(\Rightarrow x+\frac{46}{15}=\frac{3}{2}\)
\(\Rightarrow x=\frac{3}{2}-\frac{46}{15}\)
\(\Rightarrow x=\frac{-47}{30}\)
c) \(\left(-2x+1\right)+\frac{3}{15}=\frac{5}{3}\)
\(\Rightarrow-2x+1=\frac{5}{3}-\frac{3}{15}=\frac{22}{15}\)
\(\Rightarrow-2x=\frac{7}{15}\Rightarrow x=\frac{-7}{30}\)
\(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{\left(2x+1\right)\left(2x+3\right)}=\frac{15}{93}\)
\(2\left(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{\left(2x+1\right)\left(2x+3\right)}\right)=2.\frac{15}{93}\)
\(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{\left(2x+1\right)\left(2x+3\right)}=\frac{10}{31}\)
\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2x+1}-\frac{1}{2x+3}=\frac{10}{31}\)
\(\frac{1}{3}-\frac{1}{2x+3}=\frac{10}{31}\)
\(\frac{1}{2x+3}=\frac{1}{93}\)
\(\Rightarrow2x+3=93\)
\(2x=90\)
\(x=\frac{90}{2}=45\)
Vậy \(x=45\)