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

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

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

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

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

\(\Rightarrow x+1=10\)

\(\Leftrightarrow x=9\)

Vậy x = 9 

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Zoro  cảm ơn nhiều

Với \(a\inℕ;a>0\), ta có:

\(\frac{2}{a\left(a+1\right)\left(a+2\right)}=\frac{a+2-a}{a\left(a+1\right)\left(a+2\right)}=\frac{a+2}{a\left(a+1\right)\left(a+2\right)}-\frac{a}{a\left(a+1\right)\left(a+2\right)}\)

\(=\frac{1}{a\left(a+1\right)}-\frac{1}{\left(a+1\right)\left(a+2\right)}\)

Theo đề :

\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right)x=\frac{22}{45}\)

\(\Leftrightarrow\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{8.9.10}\right)x=\frac{44}{45}\)

\(\Leftrightarrow\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right)x=\frac{44}{45}\)

\(\Leftrightarrow\left(\frac{1}{2}-\frac{1}{90}\right)x=\frac{44}{45}\Leftrightarrow\frac{44}{90}x=\frac{44}{45}\Leftrightarrow x=\frac{44}{45}:\frac{44}{90}=\frac{90}{45}=2\)

Vậy \(x=2\)

Chúc bạn học tốt!

P/S: đây là 1 trong những dạng bài toán hay của lớp 6: tính toán trên dãy số. Để làm được dạng này thì mình có 2 quy tắc tổng quát như sau:

1)Khử liên tiếp

2)Đưa về những dãy số cơ bản đã biết cách tính (VD: 1+2+3+...; ...)

( 1/1.2 + 1/2.3 + 1/3.4 + 1/4.5 +1/5.6 ) x 10 - x = 0

= ( 1- 1/2 +1/2 -1/3 +1/3 - 1/4 + 1/4 - 1/5 +1/5 -1/6 ) x 10 - x = 0

= ( 1 - 1/6 ) x 10 - x = 0

= 5/6 x 10 - x =0

=   25/3 - x =0

               x = 25/3 - 0

                x  = 25/3

\(\left(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+\frac{1}{5\times6}\right)\times10-x=0\)

\(\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\right)\times10-x=0\)

\(\left(\frac{1}{1}-\frac{1}{6}\right)\times10-x=0\)

\(\frac{5}{6}\times10-x=0\)

\(\frac{25}{3}-x=0\)

x              =\(\frac{25}{3}-0=\frac{25}{3}\)

\(=\frac{47}{6}\)

Ta thấy :

1/1x2 = 1/1 - 1/2

1/2x3 = 1/2 - 1/3 

....

=>( 1/1x2 + 1/2x3 + 1/3x4 + 1/5x6 ) x 10 - x = ( 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + 1/5 - 1/6 ) x 10 - x

= ( 1/1 - 1/6 ) x 10 - x =0

 5/6 x 10 - x = 0 

 25/3 - x = 0

=> x = 25/3

\(\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{5.6}\right).10-x=0\)

\(\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{30}\right).10-x=0\)

\(\left(1-\frac{1}{4}+\frac{1}{30}\right).10-x=0\)

\(\left(\frac{3}{4}+\frac{1}{30}\right).10-x=0\)

\(\frac{47}{60}.10-x=0\)

\(\frac{47}{6}-x=0\)

\(x=\frac{47}{6}-0\)

\(x=\frac{47}{6}\)

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

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

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

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

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

Ta co:

 \(\left[\left(\frac{29}{25}-x\right).\frac{21}{4}\right]:\left[\left(\frac{95}{9}-\frac{29}{4}\right).\frac{36}{17}\right]=\frac{3}{4}\)

\(\left[\left(\frac{29}{25}-x\right).\frac{21}{4}\right]:\left(\frac{119}{36}.\frac{36}{17}\right)=\frac{3}{4}\)

\(\left[\left(\frac{29}{25}-x\right).\frac{21}{4}\right]:7=\frac{3}{4}\)

\(\left(\frac{29}{25}-x\right).\frac{21}{4}=\frac{3}{4}.7=\frac{21}{4}\)

\(\frac{29}{25}-x=\frac{21}{4}:\frac{21}{4}=1\)

\(x=\frac{29}{25}-1=\frac{4}{25}\)

\(\frac{1}{x-1}-\frac{3x^2}{x^3-1}=\frac{2x}{x^2+x+1}\left(x\ne1\right)\)

\(\Leftrightarrow\frac{1}{x-1}-\frac{3x^2}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{2x}{x^2+x+1}=0\)

\(\Leftrightarrow\frac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{3x^2}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=0\)

\(\Leftrightarrow\frac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{3x^2}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{2x^2-2x}{\left(x-1\right)\left(x^2+x+1\right)}=0\)

\(\Leftrightarrow\frac{1}{\left(x-1\right)\left(x^2+x+1\right)}\left(x^2+x+1-3x^2-2x^2+2x\right)=0\)

\(\Leftrightarrow-4x^2+3x+1=0\left(\frac{1}{\left(x-1\right)\left(x^2+x+1\right)}\ne0\right)\)

\(\Leftrightarrow-4x^2+4x-x+1=0\)

\(\Leftrightarrow-4x\left(x-1\right)-\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(-4x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\-4x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\-4x=1\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=1\left(loại\right)\\x=\frac{-1}{4}\end{cases}}}\)

Vậy \(x=\frac{-1}{4}\)

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

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

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

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

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

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

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

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

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

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

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

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

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