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13 tháng 7 2016
a) \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{x\left(x+1\right)}=\frac{2015}{2014}\)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}=\frac{2015}{2014}\)
\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2015}{2014}\)
\(1-\frac{1}{x+1}=\frac{2015}{2014}\)
\(\frac{1}{x+1}=1-\frac{2015}{2014}\)
\(\frac{1}{x+1}=-\frac{1}{2014}\)
\(x+1=-2014\)
\(x=-2015\)
b) \(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{2x\left(x+1\right)}=\frac{2984}{1993}\)
\(\frac{2}{2}+\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{2984}{1993}\)
\(2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2984}{1993}\)
\(2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2984}{1993}\)
\(2\left(1-\frac{1}{x+1}\right)=\frac{2984}{1993}\)
\(1-\frac{1}{x+1}=\frac{1492}{1993}\)
\(\frac{1}{x+1}=\frac{501}{1993}\)
\(501\left(x+1\right)=1993\)không tồn tại số tự nhiên x
18 tháng 5 2020
\(\frac{x+2}{3}=\frac{2x-1}{5}\)
=> \(\left(x+2\right)\cdot5=3\left(2x-1\right)\)
=> \(5x+10=6x-3\)
=> \(6x-5x=10+3\)
=> \(x=13\)
\(\frac{-x}{4}=\frac{-9}{x}\)
=> \(-x^2=4\cdot\left(-9\right)\)
=> \(-x^2=-36\)
=> \(x^2=36\)
=> \(\orbr{\begin{cases}x^2=6^2\\x^2=\left(-6\right)^2\end{cases}}\Rightarrow\orbr{\begin{cases}x=6\\x=-6\end{cases}}\)
18 tháng 5 2020
Quỳnh ơi, chuyển 6x sang sẽ là -6x mà viết như cậu phải là -6x+5x :)
a, \(\frac{x+2}{3}=\frac{2x-1}{5}\)
\(\Leftrightarrow\frac{5x+10}{15}=\frac{6x-3}{15}\Leftrightarrow5x+10=6x-3\Leftrightarrow-x+13=0\Leftrightarrow x=-13\)
b, \(\frac{-x}{4}=\frac{-9}{x}\)\(\Leftrightarrow x^2=36\Leftrightarrow x=\pm6\)
VD
7 tháng 6 2020
1/2(2/3.5+2/5.7+2/7.9+...+2/(2x+1)(2x+3))=15/93
1/2(1/3-1/5+1/5-1/7+1/7-1/9+...+1/2x+1-1/2x+3)=15/93
1/2(1/3-1/2x+3)=15/93
=>1/3-1/2x+3=10/31
=>1/2x+3=1/93
=>2x+3=93
2x=93-3=90
=>x=45
7 tháng 6 2020
Đặt \(A=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{\left(2x+1\right)\left(2x+3\right)}=\frac{15}{93}\)
\(\Rightarrow2A=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{\left(2x+1\right)\left(2x+3\right)}=\frac{10}{31}\)
\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{2x+1}-\frac{1}{2x+3}=\frac{10}{31}\)
\(\frac{1}{3}-\frac{1}{2x+3}=\frac{10}{31}\)
\(\frac{1}{2x+3}=\frac{1}{3}-\frac{10}{31}\)
\(\frac{1}{2x+3}=\frac{1}{93}\)
\(\Rightarrow2x+3=93\)
\(2x=90\)
\(x=45\)
Vậy \(x=45\).
N
1 tháng 8 2019
a) \(-\frac{3}{4}x+\frac{1}{6}x=1-2\frac{5}{9}\)
\(\left(-\frac{3}{4}+\frac{1}{6}\right).x=1-\frac{23}{9}\)
\(-\frac{7}{12}.x=-\frac{14}{9}\)
\(x=-\frac{14}{9}:\left(-\frac{7}{12}\right)\)
\(x=\frac{8}{3}\)
Vậy x = ...
b) \(\left|2x-\frac{3}{8}\right|+2\frac{3}{4}=3\frac{1}{16}\)
\(\left|2x-\frac{3}{8}\right|+\frac{11}{4}=\frac{49}{16}\)
\(\left|2x-\frac{3}{8}\right|=\frac{49}{16}-\frac{11}{4}\)
\(\left|2x-\frac{3}{8}\right|=\frac{5}{16}\)
\(\Rightarrow\left|2x-\frac{3}{8}\right|\in\text{{}\frac{5}{16};-\frac{5}{16}\)}
Nếu, \(2x-\frac{3}{8}=\frac{5}{16}\)
\(2x=\frac{11}{16}\)
\(x=\frac{11}{32}\)
Nếu, \(2x-\frac{3}{8}=-\frac{5}{16}\)
\(2x=\frac{1}{16}\)
\(x=\frac{1}{32}\)
Vậy \(x\in\text{{}\frac{1}{32};\frac{11}{32}\)}
AH
11 tháng 11 2016
Đặt A = \(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{\left(2x+1\right)\left(2x+3\right)}=\frac{5}{31}\)
2A = \(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{\left(2x+1\right)\left(2x+3\right)}=\frac{10}{31}\)
2A = \(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{\left(2x+1\right)}-\frac{1}{2x+3}=\frac{10}{31}\)
2A = \(\frac{1}{3}-\frac{1}{2x+3}=\frac{10}{31}\)
Ta có : \(\frac{1}{3}-\frac{1}{2x+3}=\frac{10}{31}\)
\(\frac{1}{2x+3}=\frac{1}{3}-\frac{10}{31}\)
\(\frac{1}{2x+3}=\frac{1}{93}\)
=> 2x + 3 = 93
2x = 90
x = 45
PH
2
7 tháng 5 2020
a, \(\frac{x-1}{9}=\frac{8}{3}\Leftrightarrow\frac{x-1}{9}=\frac{24}{9}\Leftrightarrow x-1=24\Leftrightarrow x=25\)
b, \(\frac{x+2}{3}=\frac{2x-1}{5}\Leftrightarrow\frac{5x+10}{15}=\frac{6x-3}{15}\Leftrightarrow5x+10=6x-3\)
\(\Leftrightarrow5x+10-6x+3=0\Leftrightarrow-x+13=0\Leftrightarrow x=13\)
7 tháng 5 2020
a) \(\frac{x-1}{9}=\frac{8}{3}\)
\(\Leftrightarrow3\left(x-1\right)=8.9\)
\(\Leftrightarrow3x-3=72\)
\(\Leftrightarrow3x=75\)
\(\Leftrightarrow x=25\)
b) \(\frac{x+2}{3}=\frac{2x-1}{5}\)
\(\Leftrightarrow5\left(x+2\right)=3\left(2x-1\right)\)
\(\Leftrightarrow5x+10=6x-3\)
\(\Leftrightarrow5x-6x=-3-10\)
\(\Leftrightarrow-x=-13\)
\(\Leftrightarrow x=13\)
TL
3
TT
27 tháng 2 2016
7/x+1=5/x-1
=>7(x-1)=5(x+1)
=>7x-7=5x+5
=>7x-5x=7+5
=>2x=12
=>x=6
b/2x/9=-8/-x=>2x*x=9*8
=>2x2=72=>x2=36=>x=6