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1) tự làm (thực hiện từ dưới lên)
2) B = \(\frac{\left(\frac{1}{2}\right)^{10}.5-\left(\frac{1}{4}\right)^5.3}{\frac{\frac{1}{1024}.1}{3}-\left(\frac{1}{2}\right)^{11}}\)
= \(\frac{\left(\frac{1}{2}\right)^{10}.5-\left(\frac{1}{2}\right)^{10}.3}{\left(\frac{1}{2}\right)^{10}.\frac{1}{3}-\left(\frac{1}{2}\right)^{10}.\frac{1}{2}}\)
= \(\frac{\left(\frac{1}{2}\right)^{10}.\left(5-3\right)}{\left(\frac{1}{2}\right)^{10}.\left(\frac{1}{3}-\frac{1}{2}\right)}\)
= \(\frac{2}{-\frac{1}{6}}\)= 2 . (-6) = -12
1) \(5+\frac{1}{1+\frac{1}{1+\frac{2}{1+\frac{3}{4}}}}=5+\frac{15}{7}=\frac{5}{1}+\frac{15}{7}=\frac{50}{7}\)
A= -(1+1/2+1/4+1/8+...+1/1024)
A=-(1+1/2+1/2^2+1.2^3+...1/2^10)
2A= -(2+1+1/2+1/^2+...1/2^9)
A=2A-A = -(2+1+1/2+1/^2+...1/2^9)-(1+1/2+1/2^2+1.2^3+...1/2^10) = -(2+1/2^10) = -2-1/2^10= -(2049/1024)
\(A=-1-\frac{1}{2}-\frac{1}{4}-...-\frac{1}{1024}\)
\(\Rightarrow-A=1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{1024}\)
\(\Rightarrow-2A=\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2056}\)
\(\Rightarrow-2A-A=\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2056}\right)-\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{1024}\right)\)
\(\Rightarrow-3A=\frac{1}{2056}-1\)
\(\Rightarrow-3A=\frac{-2055}{2056}\)
\(\Rightarrow A=\frac{685}{2056}\)
Vậy...
đặt A = \(-1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{1024}\)
ta có:
A = \(-1-\frac{1}{2}-\frac{1}{4}-...-\frac{1}{1024}\)
A = \(-\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{1024}\right)\)
Đặt B = \(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{1024}\)
ta có:
B = \(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{1024}\)
=> 2B = \(2+1+\frac{1}{2}+...+\frac{1}{512}\)
=> 2B - B = \(\left(2+1+\frac{1}{2}+...+\frac{1}{512}\right)-\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{1024}\right)\)
=> B = \(2-\frac{1}{1024}\)
=> B = \(\frac{2048}{1024}-\frac{1}{1024}=\frac{2047}{1024}\)
Thay B vào A ta có:
A = \(\frac{-2047}{1024}\)
vậy A = \(\frac{-2047}{1024}\)
\(A=\left(\frac{1}{3^2}+\frac{1}{3^4}+...+\frac{1}{3^{100}}\right)-\left(\frac{1}{3}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\right)\)
\(3^2A=3^2\left(\frac{1}{3^2}+\frac{1}{3^4}+...+\frac{1}{3^{100}}\right)-3^2\left(\frac{1}{3}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\right)\)
\(9A=\left(1+\frac{1}{3^2}+...+\frac{1}{3^{98}}\right)-\left(3+\frac{1}{3}+...+\frac{1}{3^{97}}\right)\)
\(9A-A=\left(1-\frac{1}{3^{100}}\right)-\left(3-\frac{1}{3^{99}}\right)\)
\(8A=1-3=-2\)
A=\(\frac{-2}{8}=\frac{-1}{4}\)
\(B=4\left|\frac{-1}{4}\right|+\frac{1}{3^{100}}=1+\frac{1}{3^{100}}=1\)
Vậy B=1
-1( 1+1/2+1/4+1/8+...+1/1024)
= -1.( 1+ 1-1/2+1/2-1/4+1/4-1/8+...+1/512-1/1024)
= -1.( 1+ 1-1/1024)
=-( 2- 1/1024)
= - 2047/ 1024
p/s : mk chỉ nghĩ ra cách lm thui, chớ về phần tính toán mk sợ sai, nếu sai mong bạn thông cảm nha! ( mk nghĩ kq sai !)