Rut gon bieu thuc: S = \(\frac{1}{^{2^1}}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\)
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20 tháng 4 2018
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+.......+\frac{1}{2^{99}}\)
\(\Rightarrow2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+.......+\frac{1}{2^{99}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+.......+\frac{1}{2^{100}}\right)\)
\(A=1-\frac{1}{2^{100}}\)
\(A=\frac{2^{100}-1}{2^{100}}\)
DN
1
20 tháng 4 2018
\(S=1+\frac{1}{3}+\frac{1}{3^2}+........+\frac{1}{3^n}\)
\(3S=3+1+\frac{1}{3}+.......+\frac{1}{3^{n-1}}\)
\(\Rightarrow3S-S=\left(3+1+\frac{1}{3}+......+\frac{1}{3^{n-1}}\right)-\left(1+\frac{1}{3}+\frac{1}{3^2}+......+\frac{1}{3^n}\right)\)
\(\Rightarrow2S=3-\frac{1}{3^n}\Rightarrow2S=\frac{3^{n+1}-1}{3^n}\Rightarrow S=\frac{3^{n+1}-1}{2.3^n}\)
19 tháng 4 2019
Đặt \(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\)
\(\Rightarrow2A=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2011}}\)
\(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2011}}+\frac{1}{2^{2012}}\)
\(\Rightarrow2A-A=A=2-\frac{1}{2^{2012}}\)
7 tháng 4 2016
Áp dụng a/(a^4+a^2+1)=1/2.(1/(a^2-a+1)-1/(a^2+a+1)) ta được
A=1/2.(1/(1^2-1+1)-1/(1^2+1+1)+1/(2^2-2+1)-1/(2^2+2+10)+...+1/(2014^2-2014+1)-1/(2014^2+2014+1))
A=1/2.(1-1/(2014^2+2014+1))
A=-2029105/4058211
(CHẮC CHẮN ĐÚNG)
4 tháng 7 2015
đk: x>=0; x khác 3
a) \(P=\frac{\sqrt{x}-3}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}-\frac{5}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}+\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}-3}=\frac{\sqrt{x}-3-5+x-4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}=\frac{x+\sqrt{x}-12}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\)
\(P=\frac{\left(\sqrt{x}+4\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}=\frac{\sqrt{x}+4}{\sqrt{x}+2}\)
b) \(P=\frac{\sqrt{x}+2+2}{\sqrt{x}+2}=1+\frac{2}{\sqrt{x}+2}\)
ta có: \(x\ge0\Rightarrow\sqrt{x}\ge0\Leftrightarrow\sqrt{x}+2\ge2\Leftrightarrow\frac{2}{\sqrt{x}+2}\le1\Leftrightarrow1+\frac{2}{\sqrt{x}+2}\le2\Rightarrow MaxP=2\Rightarrow x=0\)
6 tháng 9 2019
\(đkxđ\Leftrightarrow\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)
\(A=\)\(\frac{15\sqrt{x}-11}{x+2\sqrt{x}-3}-\frac{3\sqrt{x}-2}{\sqrt{x}-1}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}.\)
\(=\frac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\)\(\frac{3\sqrt{x}-2}{\sqrt{x}-1}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)
\(=\frac{15\sqrt{x}-11-\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)-\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{15\sqrt{x}-11-3x-7\sqrt{x}+6-2x-\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)\(=\frac{-5x+5\sqrt{x}+2\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{-5\sqrt{x}\left(\sqrt{x}-1\right)+2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)\(=\frac{-5\sqrt{x}+2}{\sqrt{x}+3}\)
\(A=\frac{-5\sqrt{x}+2}{\sqrt{x}+3}\)
\(\Rightarrow\frac{-5\sqrt{x}+2}{\sqrt{x}+3}=-\frac{1}{7}\Rightarrow-7\left(-5\sqrt{x}+2\right)=\sqrt{x}+3\)
\(\Rightarrow35\sqrt{x}-14=\sqrt{x}+3\)
\(\Rightarrow34\sqrt{x}=17\)
\(\Rightarrow\sqrt{x}=\frac{1}{2}\Rightarrow x=\frac{1}{4}\left(tm\right)\)
Vậy với \(x=\frac{1}{4}\)thì \(A=-\frac{1}{7}\)
Tinh 2S, roi lay 2S-S=1-1/2^100
ban co the giai thich cu the hon duoc khong?