1-x^3-x^5-x^7-......-x^2015. Mà x = -1
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1.
\(\left(\frac{3}{1\times3}+\frac{3}{3\times5}+\frac{3}{5\times7}+...+\frac{3}{97\times99}\right)-x:\frac{3}{2}=\frac{7}{3}\\
\left(\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{97\times99}\right):\frac{3}{2}-x:\frac{3}{2}=\frac{7}{3}\\\left[\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)-x\right]:\frac{3}{2}=\frac{7}{3}\\
\left(1-\frac{1}{99}\right)-x=\frac{7}{3}\times\frac{3}{2}\\
\frac{98}{99}-x=\frac{7}{2}\\
x=\frac{98}{99}-\frac{7}{2}=\frac{-497}{198}\)
2.\(\frac{x}{y}=\frac{4}{3}\Rightarrow\hept{\begin{cases}x=4a\\y=3a\\x-y=4a-3a=a\end{cases}}\\ \left(x-y\right)^{2015}=5^{2015}\Rightarrow x-y=5\\ \Rightarrow a=5\Rightarrow\hept{\begin{cases}x=4\times5=20\\y=3\times5=15\end{cases}}\)
Số dấu ngoặc = (2015 - 1):2 +1 = 1008
=> 1008x +(2015+ 1).1008:2 = 1023120
=> 1008x + 1008.1008 = 1008.1015
=> x +1008 = 1015
x = 7
\(\frac{x-1}{2015}+\frac{x-3}{2013}=\frac{x-5}{2011}+\frac{x-7}{2009}\)
=> \(\frac{x-1}{2015}-1+\frac{x-3}{2013}-1=\frac{x-5}{2011}-1+\frac{x-7}{2009}-1\)
=> \(\frac{x-2016}{2015}+\frac{x-2016}{2013}=\frac{x-2016}{2011}+\frac{x-2016}{2009}\)
=> \(\frac{x-2016}{2009}+\frac{x-2016}{2011}-\frac{x-2016}{2013}-\frac{x-2016}{2015}=0\)
=> \(\left(x-2016\right).\left(\frac{1}{2009}+\frac{1}{2011}-\frac{1}{2013}-\frac{1}{2015}\right)\)
Vì \(\frac{1}{2009}>\frac{1}{2013};\frac{1}{2011}>\frac{1}{2015}\)
=> \(\frac{1}{2009}+\frac{1}{2011}-\frac{1}{2013}-\frac{1}{2015}\ne0\)
=> \(x-2016=0\)
=> \(x=2016\)
A x 2 = 2/1 x3 + 2/ 3 x 5 + 2/ 5 x 7 + ................. + 2/ 2013 x 2015
= 1/1 – 1/3 + 1/3 – 1/5 + 1/5 – 1/7 + .................. + 1/2013 – 1/2015
= 1 – 1/2015 = 2014/2015
Vậy A = 2014/2015 : 2 = 2014/4030.