cho:M=1+2^4+2^8+...+2^2012+2^2016;N=1+2^2+2^4+2...+2^2016+2^2018.Tính N/M
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gọi biểu thức 2^4+2^8+....+2^2016 là A ta có
A=2^4+2^8+.....+2^2016
8A=2^4+2^8+.....+2^2010
8A-A=2-2^2010
7A=1+2-2^2010
A = 1 + 24 + 28 + ...... + 22012 + 22016
24A = 24 + 28 + 212 + ..... + 22016 + 22020
24A - A = (24 + 28 + 212 + ..... + 22016 + 22020) - (1 + 24 + 28 + ...... + 22012 + 22016)
15A = 22010 - 1
\(\Rightarrow A=\frac{2^{2010}-1}{15}\)
B = 1 + 22 + 24 + ....... + 22016 + 22018
22B = 22 + 24 + 26 + ........ + 22018 + 22020
22B - B = (22 + 24 + 26 + ........ + 22018 + 22020) - (1 + 22 + 24 + ....... + 22016 + 22018)
3B = 22010 - 1
\(\Rightarrow B=\frac{2^{2010}-1}{3}\)
\(\Rightarrow\frac{A}{B}=\frac{\frac{2^{2010}-1}{15}}{\frac{2^{2010}-1}{3}}=\frac{\left(2^{2010}-1\right).\frac{1}{15}}{\left(2^{2010}-1\right).\frac{1}{3}}=\frac{\frac{1}{15}}{\frac{1}{3}}=\frac{\frac{1}{3}.\frac{1}{5}}{\frac{1}{3}}=\frac{1}{5}\)
a) ĐKXĐ: \(x\notin\left\{0;2\right\}\)
Ta có: \(\dfrac{x+2}{x-2}-\dfrac{1}{x}=\dfrac{2}{x\left(x-2\right)}\)
\(\Leftrightarrow\dfrac{x\left(x+2\right)}{x\left(x-2\right)}-\dfrac{x-2}{x\left(x-2\right)}=\dfrac{2}{x\left(x-2\right)}\)
Suy ra: \(x^2+2x-x+2-2=0\)
\(\Leftrightarrow x^2+x=0\)
\(\Leftrightarrow x\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=-1\left(nhận\right)\end{matrix}\right.\)
Vậy: S={-1}
M = 1 + 24 + 28 + ............. + 22012 + 22016
16M = 24 + 28 + ............. + 22012 + 22016 + 22020
16M - M = (24 + 28 + ............. + 22012 + 22016 + 22020) - ( 1 + 24 + 28 + ............. + 22012 + 22016)
15M = 22020 - 1
M = \(\frac{2^{2020}-1}{15}\)
N = 1 + 22 + 24 + ............. + 22016 + 22018
4N = 22 + 24 + ............. + 22016 + 22018 + 22020
4N - N = (22 + 24 + ............. + 22016 + 22018 + 22020) - ( 1 + 22 + 24 + ............. + 22016 + 22018)
3N = 22020 - 1
N = \(\frac{2^{2020}-1}{3}\)
\(\frac{N}{M}=\frac{2^{2020}-1}{3}:\frac{2^{2020}-1}{15}=\frac{2^{2020}-1}{3}.\frac{15}{2^{2020}-1}=\frac{15}{3}=5\)