1/1x2+1/2x3+1/3x4+...+1/9x10
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\(A=\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}....\frac{1}{9x10}=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}...+\frac{1}{9}-\frac{1}{10}=\frac{9}{10}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}=\frac{1}{1}-\frac{1}{10}=\frac{9}{10}\)
=1/1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+............+1/9+1/10
=1-1/10
=10/10-1/10
=9/10
Bài làm:
\(\frac{1}{1\times2}+\frac{1}{2\times3}\)\(+\frac{1}{3\times4}+\frac{1}{4\times5}\)\(+...\frac{1}{9\times10}\)
\(=\frac{1}{1}-\frac{1}{2}\)\(+\frac{1}{2}-\frac{1}{3}\)\(+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}\)\(-\frac{1}{5}\)\(+...\frac{1}{9}-\frac{1}{10}\)
\(=\)\(\frac{1}{1}-\frac{1}{10}\)
\(=\frac{9}{10}\)
1/1*2+1/2*3+1/3*4+...+1/9*10
=1-1/2+1/2-1/3+1/3-1/4+...+1/9-1/10
=1-1/10
=9/10
nho k cho minh voi nhe
\(\frac{1}{1.2}\)+ \(\frac{1}{2.3}\)+ \(\frac{1}{3.4}\)+ ......... + \(\frac{1}{7.8}\)+ \(\frac{1}{8.9}\)+ \(\frac{1}{9.10}\)
\(=\)\(1\)\(-\)\(\frac{1}{10}\)
\(=\)\(\frac{9}{10}\)
=1- 1/2 +1/2- 1/3+....+1/9- 1/10
=1- 1/10
=9/10
1/1x2 + 1/2x3 + 1/3x4 +...+1/9x10 = 1/1 - 1/2 + 1/2 -1/3 + 1/3 -1/4 +...+1/9-1/10
=1+( 1/2 -1/2)+(1/3-1/3)+...+(1/9-1/9)-1/10
=1 + 0 + 0 +...-1/10 = 9/10
\(\Leftrightarrow2\left(x-\dfrac{1}{3}\right)\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)=\dfrac{3}{4}\)
\(\Leftrightarrow2\left(x-\dfrac{1}{3}\right)\left(1-\dfrac{1}{10}\right)=\dfrac{3}{4}\Leftrightarrow\dfrac{9}{10}\left(x-\dfrac{1}{3}\right)=\dfrac{3}{8}\)
\(\Leftrightarrow x-\dfrac{1}{3}=\dfrac{5}{12}\Leftrightarrow x=\dfrac{5}{12}+\dfrac{1}{3}=\dfrac{9}{12}=\dfrac{3}{4}\)
Ta đặt biểu thức là:
A = 1/1 x 2 + 1/2 x 3 + 1/3 x 4 + .... + 1/9 x 10
A = 1 - 1/2 + 1/2 - 1/3 +1/3 - 1/4 + ... + 1/9 - 1/10
A = 1 - 1/10
A = 9/10
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\)
\(=1-\frac{1}{10}\)
\(=\frac{9}{10}\)