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\(B=512-\dfrac{512}{2}-\dfrac{512}{2^2}-\dfrac{512}{2^3}-...-\dfrac{512}{2^{10}}\)
\(B=512-512\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+..+\dfrac{1}{2^{10}}\right)\)
Đặt: \(L=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{10}}\)
\(2L=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\)
\(2L-L=1-\dfrac{1}{2^{10}}\Leftrightarrow L=1-\dfrac{1}{2^{10}}\)
Thay Vào B
\(B=512-512\left(1-\dfrac{1}{2^{10}}\right)=512-512+\dfrac{512}{2^{10}}=\dfrac{1}{2}\)
\(B=512-\dfrac{512}{2}-\dfrac{512}{2^2}-....-\dfrac{512}{2^{10}}\)
\(=512-\left(\dfrac{512}{2}+\dfrac{512}{2^2}+....+\dfrac{512}{2^{10}}\right)\)
\(=512-\left[512\left(\dfrac{1}{2}+\dfrac{1}{2^2}+....+\dfrac{1}{2^{10}}\right)\right]\)
Đặt \(A=\dfrac{1}{2}+\dfrac{1}{2^2}+....+\dfrac{1}{2^{10}}\)
\(2A=1+\dfrac{1}{2}+...+\dfrac{1}{2^9}\)
\(\Rightarrow2A-A=1-\dfrac{1}{2^{10}}\)
\(\Rightarrow A=1-\dfrac{1}{2^{10}}\)
\(\Rightarrow B=512-\left(512.A\right)=512-\left[512.\left(1-\dfrac{1}{2^{10}}\right)\right]\)
\(=512-512.\dfrac{1023}{1024}=512-\dfrac{1023}{2}=\dfrac{1}{2}\)
Sửa đề: (2/7)^7*7^7
\(A=\dfrac{\left(2\right)^7+\left(\dfrac{9}{3}:\dfrac{3}{16}\right)^3}{2^7\left(5^2+2^2\right)}\)
\(=\dfrac{\left(2\right)^7+\left(16\right)^3}{2^7\cdot29}\)
\(=\dfrac{2^7+2^7\cdot2^5}{2^7\cdot29}=\dfrac{1+2^5}{29}=\dfrac{33}{29}\)
\(a,=\dfrac{3^6\cdot5^4\cdot9^4-5^{13}\cdot3^{13}\cdot5^{-9}}{3^{12}\cdot5^6+9^6\cdot5^6}=\dfrac{3^{14}\cdot5^4-5^4\cdot3^{13}}{3^{12}\cdot5^6+3^{12}\cdot5^6}\\ =\dfrac{3^{13}\cdot5^4\cdot2}{2\cdot3^{12}\cdot5^6}=\dfrac{3}{5^2}=\dfrac{3}{25}\)
\(b,=\dfrac{\left(\dfrac{2}{5}\cdot5\right)^7+\left(\dfrac{9}{4}\cdot\dfrac{16}{3}\right)^3}{2^7\cdot5^2+2^9}=\dfrac{2^7+12^3}{2^7\left(5^2+2^2\right)}=\dfrac{2^7+4^3\cdot3^3}{2^7\cdot29}=\dfrac{2^6\left(2+3^3\right)}{2^7\cdot29}=\dfrac{1}{2}\)
B=512(1-1/2-1/2^2-1/2^3-...-1/2^10
B=512*1/1024
B=1/2
B=0.5