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\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\)
\(\rightarrow A=\frac{3}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\)
\(\rightarrow A=\frac{7}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\)
\(\rightarrow A=\frac{15}{16}+\frac{1}{32}+\frac{1}{64}\)
\(\rightarrow A=\frac{31}{32}+\frac{1}{64}\)
\(\rightarrow A=\frac{63}{64}\)
\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\Rightarrow64A=32+16+8+4+2+1\Rightarrow64A=63\Rightarrow A=\frac{63}{64}\)
a)\(A=\frac{1}{5}+\frac{1}{5^2}+....+\frac{1}{5^2^5}\) <=>\(5A=1+\frac{1}{5}+\frac{1}{5^2}+....+\frac{1}{5^{24}}\)
<=>\(5A-A=(1+\frac{1}{5}+...+\frac{1}{5^{24}})-(\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{25}})\)
<=>\(4A=1-\frac{1}{5^{25}}\) <=>\(A=\frac{(5^{25^{ }}-1)}{5^{25}}\div4\)
a) \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+.......+\frac{1}{2^{99}}\)
\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{2^2}+...........+\frac{1}{2^{98}}\)
\(\Rightarrow2A-A=A=1-\frac{1}{2^{99}}\)
b) \(B=\frac{1}{2^2}+\frac{1}{2^4}+...........+\frac{1}{2^{100}}\)
\(\Rightarrow2^2.B=4B=1+\frac{1}{2^2}+......+\frac{1}{2^{98}}\)
\(\Rightarrow4B-B=3B=1-\frac{1}{2^{100}}\)
\(\Rightarrow b=\frac{1-\frac{1}{2^{100}}}{3}\)
a) Ta có A = \(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}\)
=> 2A = \(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}\)
=> 2A - A = \(\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}\right)\)
=> A = \(1-\frac{1}{2^{99}}\)
b) Ta có B = \(\frac{1}{2^2}+\frac{1}{2^4}+\frac{1}{2^6}+...+\frac{1}{2^{100}}\)
=> 22B = \(1+\frac{1}{2^2}+\frac{1}{2^4}+...+\frac{1}{2^{98}}=4B\)
=> 4B - B = \(\left(1+\frac{1}{2^2}+\frac{1}{2^4}+...+\frac{1}{2^{98}}\right)-\left(\frac{1}{2^2}+\frac{1}{2^4}+\frac{1}{2^6}+...+\frac{1}{2^{100}}\right)\)
=> 3B = \(1-\frac{1}{2^{100}}\)
=> B = \(\frac{1}{3}-\frac{1}{2^{100}.3}\)
1.
\(B=20182018.2017-20172017.2018\)
\(B=2018.10001.2017-2017.10001.2018\)
\(B=0\)
1 B=20182018.2017-20172017.2018
B=2018.10001.2017-2017.10001.2018
B=0
2 C=12+22+32+...+1002
C=1(1+0)+2(1+1)+3(1+2)+...+100(1+99)
C=1+2+1.2+3+2.3+...+100+99.100
C=(1+2+3+...+100)+(1.2+2.3+...+99.100)
C=[(1+100).100:2]+[(99.100.101):3]
C=5050+333300
C=338350
\(\left(x+1\right)\left(y-2\right)=3\)
\(\Rightarrow\left(x-1\right),\left(y-2\right)\inƯ\left(3\right)=\left\{\pm1;\pm2\right\}\)
\(TH1:\hept{\begin{cases}x+1=1\\y-2=3\end{cases}\Rightarrow\hept{\begin{cases}x=0\\y=5\end{cases}}}\) \(TH2:\hept{\begin{cases}x+1=-1\\y-2=-3\end{cases}\Rightarrow\hept{\begin{cases}x=-2\\y=-1\end{cases}}}\)
\(TH3:\hept{\begin{cases}x+1=3\\y-2=1\end{cases}\Rightarrow\hept{\begin{cases}x=2\\y=3\end{cases}}}\) \(TH4:\hept{\begin{cases}x+1=-3\\y-2=-1\end{cases}\Rightarrow\hept{\begin{cases}x=-4\\y=1\end{cases}}}\)
Vậy ............................
b, Làm tương tự
\(A=\frac{1}{2}-\frac{1}{2^4}+\frac{1}{2^7}-\frac{1}{2^{10}}+...+\frac{1}{2^{55}}-\frac{1}{2^{58}}\)
\(2^3.A=2^3-\frac{1}{2}+\frac{1}{2^4}-\frac{1}{2^7}+...+\frac{1}{2^{53}}-\frac{1}{2^{55}}\)
\(2^3.A+A=\left(2^3-\frac{1}{2}+\frac{1}{2^4}-\frac{1}{2^7}+...+\frac{1}{2^{53}}-\frac{1}{2^{55}}\right)+\left(\frac{1}{2}-\frac{1}{2^4}+\frac{1}{2^7}-\frac{1}{2^{10}}+...+\frac{1}{2^{55}}-\frac{1}{2^{58}}\right)\)
\(8A+A=2^3-\frac{1}{2^{58}}\)
\(9A=8-\frac{1}{2^{58}}\)
\(A=\frac{8-\frac{1}{2^{58}}}{9}\)
Ủng hộ mk nha ^-^
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