\(\frac{1}{\left(x+1\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+7\right)}+\frac{1}{\left(x+7\right)\left(x+9\right)}+\frac{1}{\left(x+9\right)\left(x+11\right)}\)
K
Khách
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Những câu hỏi liên quan
14 tháng 12 2017
Ta có: \(A=\frac{1}{\left(x+1\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+5\right)}+.....+\frac{1}{\left(x+9\right)\left(x+11\right)}\)
\(\Rightarrow A=\frac{1}{x+1}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+5}+....+\frac{1}{x+9}-\frac{1}{x+11}\)
\(\Rightarrow A=\frac{1}{x+1}-\frac{1}{x+11}\)
\(\Rightarrow A=\frac{x+11-x+1}{\left(x+1\right)\left(x+11\right)}=\frac{12}{\left(x+1\right)\left(x+11\right)}\)
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=\(\frac{10}{x^2+12x+11}\)
HQ
Hà Quang Minh
Giáo viên
19 tháng 9 2023
a)
\(\begin{array}{l}x:{\left( {\frac{{ - 1}}{2}} \right)^3} = - \frac{1}{2}\\x = - \frac{1}{2}.{\left( {\frac{{ - 1}}{2}} \right)^3}\\x = {\left( {\frac{{ - 1}}{2}} \right)^4}\\x = \frac{1}{{16}}\end{array}\)
Vậy \(x = \frac{1}{{16}}\).
b)
\(\begin{array}{l}x.{\left( {\frac{3}{5}} \right)^7} = {\left( {\frac{3}{5}} \right)^9}\\x = {\left( {\frac{3}{5}} \right)^9}:{\left( {\frac{3}{5}} \right)^7}\\x = {\left( {\frac{3}{5}} \right)^2}\\x = \frac{9}{{25}}\end{array}\)
Vậy \(x = \frac{9}{{25}}\).
c)
\(\begin{array}{l}{\left( {\frac{{ - 2}}{3}} \right)^{11}}:x = {\left( {\frac{{ - 2}}{3}} \right)^9}\\x = {\left( {\frac{{ - 2}}{3}} \right)^{11}}:{\left( {\frac{{ - 2}}{3}} \right)^9}\\x = {\left( {\frac{{ - 2}}{3}} \right)^2}\\x = \frac{4}{9}.\end{array}\)
Vậy \(x = \frac{4}{9}\).
d)
\(\begin{array}{l}x.{\left( {0,25} \right)^6} = {\left( {\frac{1}{4}} \right)^8}\\x.{\left( {\frac{1}{4}} \right)^6} = {\left( {\frac{1}{4}} \right)^8}\\x = {\left( {\frac{1}{4}} \right)^8}:{\left( {\frac{1}{4}} \right)^6}\\x = {\left( {\frac{1}{4}} \right)^2}\\x = \frac{1}{{16}}\end{array}\)
Vậy \(x = \frac{1}{{16}}\).
2 tháng 3 2016
\(\frac{2}{t\left(t+2\right)}=\frac{t+2-t}{t\left(t+2\right)}=\frac{1}{t}-\frac{1}{t+2}\)
\(\Rightarrow\frac{1}{x+1}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+5}+...+\frac{1}{x+7}-\frac{1}{x+9}=\frac{2}{5}\)
\(\Leftrightarrow\frac{1}{x+1}-\frac{1}{x+9}=\frac{2}{5}\Leftrightarrow\frac{8}{\left(x+1\right)\left(x+9\right)}=\frac{2}{5}\)
\(\Leftrightarrow\left(x+1\right)\left(x+9\right)=20\Leftrightarrow x^2+10x+9-20=0\)
\(\Leftrightarrow\left(x+11\right)\left(x-1\right)=0\)
\(\Leftrightarrow\int^{x=-11}_{x=1}\)
=> tổng bp các nghiệm là 122
TN
2 tháng 3 2016
Phương trình có tổng bằng 1/5. tìm tổng bình phương các nghiệm x
4 tháng 6 2019
bạn ơi trả lời được câu này kông
( x + 1 ) + ( x - 3 ) + ( x + 5 ) + ............ + ( x +9) = 35