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

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

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

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

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

\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\frac{1}{2}.\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+...+\frac{1}{2}.\left(\frac{1}{8.9}-\frac{1}{9.10}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right)\)

\(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{9.10}\right)=\frac{1}{2}.\frac{22}{45}=\frac{11}{45}\)

a)1.2.3.4...9-1.2.3.4...8-1.2.3.4...8.8

=1.2.3.4...8(9-1-8)

=1.2.3.4...8.0

=0

b)(3.4.2¹⁶)²/11.12³.4¹¹-16⁹=(3.2².2¹⁶)²/11.2¹³.2²²-2³⁶=3².2⁴.2³²/11.2³⁵-2³⁶=3².2²⁶/2³⁵.(11-2)

=3².2³⁶/2³⁵.9=3².2³⁶/2³⁵.3²=2

c)70.(131313/565656+131313/727272+131313/909090

=70.(13/56+13/72+13/90)

=70.39/70=39

d)1/4.9+1/9.14+1/14.19+...+1/64.69

=4/4.9.4+4/9.4.14+4/14.19.4+...+4/64.69.4.

=1/4.(4/4.9+4/9.14+4/14.19+...+4/64.69)

=1/4.(1/4-1/9+1/9-1/14+1/14-1/19+...+1/64-1/69)

=1/4.(1/4-1/69)

=1/4.65/276=65/1104

~~~~~~~~Chúc bạn học giỏi nhé !~~~~~~~~

nhìu thế này sao mà làm nổi

d)\(\frac{1}{3}\times\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+...+\frac{1}{64}-\frac{1}{69}\right)\)

\(\frac{1}{3}\times\left(\frac{1}{4}-\frac{1}{69}\right)\)

\(\frac{1}{3}\times\frac{65}{276}\)

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

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

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

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

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

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

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

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

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

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

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

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

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

\(\frac{1}{8}=12,5\%\)  ;  \(\frac{1}{16}=6,25\%\) ; \(\frac{1}{2}=50\%\) ; \(\frac{1}{4}=25\%\) 

Thay vào trên mà tính.

= \(1+\left(\frac{3\left(1x2+2x4x2\right)}{3\left(5+5x3x25\right)}+1\right)-\left(1+\frac{18}{54}\right)-1\) = \(\frac{18}{380}-\frac{18}{54}\)  

\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{2014.2015.2016}\)

\(=\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{2014.2015.2016}\right)\)

\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{2014.2015}-\frac{1}{2015.2016}\right)\)

\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2015.2016}\right)\)