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\(A=1+\dfrac{3}{2^3}+\dfrac{4}{2^4}+\dfrac{5}{2^5}+...+\dfrac{100}{2^{100}}\)
\(2A=2\left(1+\dfrac{3}{2^3}+\dfrac{4}{2^4}+\dfrac{5}{2^5}+...+\dfrac{100}{2^{100}}\right)\)
\(2A=2+\dfrac{3}{2^2}+\dfrac{4}{2^3}+\dfrac{5}{2^4}+....+\dfrac{100}{2^{99}}\)
\(2A-A=\left(2+\dfrac{3}{2^2}+\dfrac{4}{2^3}+\dfrac{5}{2^4}+...+\dfrac{100}{2^{99}}\right)-\left(1+\dfrac{3}{2^3}+\dfrac{4}{2^4}+\dfrac{5}{2^5}+...+\dfrac{100}{2^{100}}\right)\)\(A=2+\dfrac{3}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{99}}-\dfrac{100}{2^{100}}\)
\(A=\dfrac{11}{4}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{99}}-\dfrac{100}{2^{100}}\)
\(A=\dfrac{11}{4}+\dfrac{1}{2^3}-\dfrac{1}{2^{99}}-\dfrac{100}{2^{100}}\)
Đặt \(D=1+\dfrac{3}{2^3}+\dfrac{4}{2^4}+...+\dfrac{100}{2^{100}}\)
\(2D=2+\dfrac{3}{2^2}+\dfrac{4}{2^3}+...+\dfrac{100}{2^{99}}\)
\(2D-D=\left(2+\dfrac{3}{2^2}+\dfrac{4}{2^3}+...+\dfrac{100}{2^{99}}\right)-\left(1+\dfrac{3}{2^3}+\dfrac{4}{2^4}+...+\dfrac{100}{2^{100}}\right)\)
\(D=2+\dfrac{3}{2^2}+\dfrac{1}{2^3}+....+\dfrac{1}{2^{99}}-\dfrac{100}{2^{100}}\)
\(D=\dfrac{11}{4}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{99}}-\dfrac{100}{2^{100}}\)
\(D=\dfrac{11}{4}+\dfrac{1}{2^3}-\dfrac{1}{2^{99}}-\dfrac{100}{2^{100}}\)
\(a)3\dfrac{1}{2}.\dfrac{4}{49}-\left[2,\left(4\right):2\dfrac{5}{11}\right]:\left(\dfrac{-42}{5}\right)\)
\(=\dfrac{7}{2}.\dfrac{4}{49}-\dfrac{88}{27}:\left(\dfrac{-42}{7}\right)\)
\(=\dfrac{2}{7}-\dfrac{-220}{567}\)
\(=\dfrac{382}{567}\)
các phần con lại dễ nên bn tự lm đi nhé mk bn lắm
Chúc bạn học tốt!
8)\(\frac{4}{9}:\left(-\frac{1}{7}\right)+6\frac{5}{9}:\left(-\frac{1}{7}\right)\)
=\(\frac{4}{9}:\left(-\frac{1}{7}\right)+\frac{59}{9}:\left(-\frac{1}{7}\right)\)
=\(\left(\frac{4}{9}+\frac{59}{9}\right).\left(-7\right)\)
=7.(-7)
=-49
Giải:
a) \(\dfrac{1}{3}x+\dfrac{1}{5}-\dfrac{1}{2}x=1\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{1}{5}-\dfrac{1}{6}x=\dfrac{5}{4}\)
\(\Leftrightarrow\dfrac{1}{6}x=\dfrac{-21}{20}\)
\(\Leftrightarrow x=\dfrac{-63}{10}\)
Vậy ...
b) \(\dfrac{3}{2}\left(x+\dfrac{1}{2}\right)-\dfrac{1}{8}x=\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{3}{2}x+\dfrac{3}{4}-\dfrac{1}{8}x=\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{11}{8}x=\dfrac{-1}{2}\)
\(\Leftrightarrow x=\dfrac{-4}{11}\)
Vậy ...
Các câu sau làm tương tự câu b)
\(A=1+\dfrac{3}{2^3}+\dfrac{4}{2^4}+..+\dfrac{100}{2^{100}}\\ \Rightarrow2A=2+\dfrac{3}{2^2}+\dfrac{4}{2^3}+...+\dfrac{100}{2^{99}}\\ \Rightarrow A=\dfrac{7}{4}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{99}}-\dfrac{100}{2^{100}}\\ B=\dfrac{1}{2^3}+...+\dfrac{1}{2^{99}}\\ \Rightarrow2B=\dfrac{1}{2^2}+...+\dfrac{1}{2^{98}}\\ \Rightarrow B=\dfrac{1}{4}-\dfrac{1}{2^{99}}\\ \Rightarrow A=\dfrac{7}{4}+\dfrac{1}{4}-\dfrac{1}{2^{99}}-\dfrac{100}{2^{100}}\\ =2-\dfrac{1}{2^{99}}-\dfrac{100}{2^{100}}\)
Ngu