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\(\frac{3}{\left(x+2\right)\left(x+3\right)}\)+\(\frac{3}{\left(x+3\right)\left(x+5\right)}\)+ \(\frac{3}{\left(x+5\right)\left(x+7\right)}\)+ ... + \(\frac{3}{\left(x+99\right)\left(x+101\right)}\)
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14 tháng 12 2017
Ta có: \(A=\frac{1}{\left(x+1\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+5\right)}+.....+\frac{1}{\left(x+9\right)\left(x+11\right)}\)
\(\Rightarrow A=\frac{1}{x+1}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+5}+....+\frac{1}{x+9}-\frac{1}{x+11}\)
\(\Rightarrow A=\frac{1}{x+1}-\frac{1}{x+11}\)
\(\Rightarrow A=\frac{x+11-x+1}{\left(x+1\right)\left(x+11\right)}=\frac{12}{\left(x+1\right)\left(x+11\right)}\)
22 tháng 12 2018
A= \(\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+\frac{2}{x+3}-...+\frac{8}{x+5}-\frac{8}{x+6}\)
A=\(\frac{1}{x+1}+\frac{1}{x+3}+\frac{2}{x+4}+\frac{4}{x+5}-\frac{8}{x+6}\)
Rồi tiếp tục làm nhé bạn.
VM
8 tháng 8 2019
c) \(\frac{x-1}{2009}+\frac{x-2}{2008}=\frac{x-3}{2007}+\frac{x-4}{2006}\)
\(\Leftrightarrow\left(\frac{x-1}{2009}-1\right)+\left(\frac{x-2}{2008}-1\right)=\left(\frac{x-3}{2007}-1\right)+\left(\frac{x-4}{2006}-1\right)\)
\(\Leftrightarrow\frac{x-2010}{2009}+\frac{x-2010}{2008}-\frac{x-2010}{2007}-\frac{x-2010}{2006}=0\)
\(\Leftrightarrow\left(x-2010\right).\left(\frac{1}{2009}+\frac{1}{2008}-\frac{1}{2007}-\frac{1}{2006}\right)=0\)
\(\Leftrightarrow x-2010=0\)
\(\Leftrightarrow x=0+2010\)
\(\Rightarrow x=2010\)
Vậy \(x=2010.\)
Mình chỉ làm câu c) thôi nhé.
Chúc bạn học tốt!
29 tháng 11 2016
\(A=\frac{1}{\left(x+1\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+7\right)}+\frac{1}{\left(x+7\right)\left(x+9\right)}+\frac{1}{\left(x+9\right)\left(x+11\right)}\)
\(=\frac{1+1+1+1+1}{\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)\left(x+9\right)\left(x+11\right)}\)
\(=\frac{5}{\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)\left(x+9\right)\left(x+11\right)}\)
\(=\frac{5}{\left(x+1\right)\left(x+11\right)\left(x+3\right)\left(x+9\right)\left(x+5\right)\left(x+7\right)}\)
\(=\frac{5}{\left(x^2+11x+x+11\right)\left(x^2+9x+3x+27\right)\left(x^2+7x+5x+35\right)}\)
\(=\frac{5}{\left(x^2+12x+11\right)\left(x^2+12x+27\right)\left(x^2+12x+35\right)}\)
29 tháng 11 2016
A=\(\frac{1}{x+1}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+7}+\frac{1}{x+7}-\frac{1}{x+9}+\frac{1}{x+9}-\frac{1}{x+11}\)
Rút gọn hết đi ta có \(\frac{1}{x+1}-\frac{1}{x+11}\)=\(\frac{x+11}{\left(x+1\right).\left(x+11\right)}-\frac{x+1}{\left(x+1\right).\left(x+11\right)}\)
A=\(\frac{x+11-x-1}{\left(x+1\right).\left(x+11\right)}\)
A=\(\frac{10}{x^2+12x+11}\)
9 tháng 6 2017
\(\frac{1}{x+2}-\frac{1}{x+5}+...+\frac{1}{x+10}-\frac{1}{x+17}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\frac{1}{x+2}-\frac{1}{x+7}=\frac{x}{\left(x+2\right)\left(x+7\right)}\)
\(\Rightarrow x=1\)
2 tháng 12 2016
\(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{x+5}\)
\(=\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}\)
\(=\frac{1}{x}\)
2 tháng 12 2016
ta có: \(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{x+5}\)
=\(\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}\)
= \(\frac{1}{x}\)
\(\frac{3}{\left(x+1\right)\left(x+3\right)}=\frac{3}{2}.\frac{\left(x+3\right)-\left(x+1\right)}{\left(x+3\right)\left(x+1\right)}=\frac{3}{2}\left(\frac{1}{x+1}-\frac{1}{x+3}\right)\)
Tương tự:
\(\frac{3}{\left(x+3\right)\left(x+5\right)}=\frac{3}{2}.\left(\frac{1}{x+3}-\frac{1}{x+5}\right)\)
\(\frac{3}{\left(x+5\right)\left(x+7\right)}=\frac{3}{2}\left(\frac{1}{x+5}-\frac{1}{x+7}\right)\)
.....
\(\frac{3}{\left(x+99\right)\left(x+101\right)}=\frac{3}{2}\left(\frac{1}{x+99}-\frac{1}{101}\right)\)
Cộng các vế lại ta có:
\(\frac{3}{\left(x+1\right)\left(x+3\right)}+\)\(\frac{3}{\left(x+3\right)\left(x+5\right)}+\)\(\frac{3}{\left(x+5\right)\left(x+7\right)}+\)...\(+\frac{3}{\left(x+99\right)\left(x+101\right)}\)
=\(\frac{3}{2}\left(\frac{1}{x+1}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+7}+...+\frac{1}{x+99}-\frac{1}{x+101}\right)\)
=\(\frac{3}{2}\left(\frac{1}{x+1}-\frac{1}{x+101}\right)\)