\(\frac{2}{5}x+\frac{3}{10}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.........+\frac{1}{9}-\frac{1}{10}\)

\(\frac{2}{5}x+\frac{3}{10}=1-\frac{1}{10}=\frac{9}{10}\)

\(\frac{2}{5}x=\frac{9}{10}-\frac{3}{10}=\frac{3}{5}\)

\(x=\frac{\frac{3}{5}}{\frac{2}{5}}=\frac{3}{2}\)

Ta có: \(\frac{1}{1x2}\)+ \(\frac{1}{2x3}\)+ \(\frac{1}{3x4}\)+ \(\frac{1}{4x5}\)+ .....+ \(\frac{1}{9x10}\)

         = \(1-\left(\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-.....-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\)

        = 1 - \(\frac{1}{10}\)

        =  \(\frac{9}{10}\)

X = \(\frac{3}{2}\) là đúng đó bạn

1-1/x+1=2015/2016

=>1/x+1=1-2015/2016=1/2016

=>x+1=2016=>x=2015

mình không ghi lại đề nha:

\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2015}{2016}\)

<=>\(1-\frac{1}{x+1}=\frac{2015}{2016}\)

<=>\(\frac{x}{x+1}=\frac{2015}{2016}\)

=>x=

Đến đó bạn tự giải tiếp ha

=>(1/1-1/2)+(1/2-1/3)+(1/3-1/4)+...+(1/x-1/x+1)=2015/2016
=>1/1-1/x+1=2015/2016
=>x/x+1=2015/2016
=>x=2015

\(\Leftrightarrow10\times\left(\frac{1}{1\times2}+\frac{1}{2\times3}+...+\frac{1}{x\times\left(x+1\right)}\right)=9\)

\(\Leftrightarrow\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}\right)=9\div10\)

\(\Leftrightarrow\frac{1}{1}-\frac{1}{x+1}=\frac{9}{10}\)

\(\Leftrightarrow\frac{1}{x+1}=1-\frac{9}{10}\)

\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{10}\)

\(\Rightarrow x+1=10\)

\(\Leftrightarrow x=9\)

Vậy x = 9 

<=> (1-1/10)(x-1)+x/10=x-9/10

<=> 9x/10-9/10+x/10=x-9/10

<=> x=x

Như vậy, phương trình thỏa mãn với mọi x

\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+..+\frac{1}{x\left(x+1\right)}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\)

\(=1-\frac{1}{x+1}\)

\(=\frac{x+1-1}{x+1}=\frac{x}{x+1}\)

\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{x.\left(x+1\right)}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\)

\(=1-\frac{1}{x+1}\)

\(=\frac{x+1}{x+1}-\frac{1}{x+1}\)

\(=\frac{x}{x+1}\)

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{x\left(x+1\right)}\)

\(\Leftrightarrow\frac{1}{1}.\frac{1}{2}+\frac{1}{2}\frac{1}{3}+\frac{1}{3}.\frac{1}{4}+....+\frac{1}{x}\frac{1}{x+1}\)

\(\Leftrightarrow\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{x}-\frac{1}{x+1}\)

\(\Leftrightarrow1-\frac{1}{x+1}\)

\(\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+....+\frac{1}{2009\cdot2010}\right)\cdot x=2009\)

\(\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2009}-\frac{1}{2010}\right)\cdot x=2009\)

\(\left(1-\frac{1}{2010}\right)\cdot x=2009\)

\(\frac{2009}{2010}\cdot x=2009\)

\(x=2009:\frac{2009}{2010}\)

\(x=2010\)

\(\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}+....+\frac{1}{2009}-\frac{1}{2010}\right).x=2009\)

\(\left(\frac{1}{1}-\frac{1}{2010}\right).x=2009\)

\(\frac{2009}{2010}.x=2009\)

\(x=2009:\frac{2009}{2010}\)

\(x=2010\)

x = 2010