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Ta có:
\(A=1+\frac{3}{2^3}+\frac{4}{2^4}+\frac{5}{2^5}+...+\frac{100}{2^{100}}\)
\(2A=2+\frac{3}{2^2}+\frac{4}{2^3}+\frac{5}{2^4}+...+\frac{100}{2^{99}}\)
\(2A-A=\left(2+\frac{3}{2^2}+\frac{4}{2^3}+\frac{5}{2^4}+...+\frac{100}{2^{99}}\right)-\left(1+\frac{3}{2^3}+\frac{4}{2^4}+...+\frac{99}{2^{99}}+\frac{100}{2^{100}}\right)\)
\(A=2+\frac{3}{2^2}+\frac{4}{2^3}+\frac{5}{2^4}+...+\frac{100}{2^{99}}-1-\frac{3}{2^3}-\frac{4}{2^4}-...-\frac{99}{2^{99}}-\frac{100}{2^{100}}\)
\(A=\left(2-1\right)+\frac{3}{2^2}+\left(\frac{4}{2^3}-\frac{3}{2^3}\right)+\left(\frac{5}{2^4}-\frac{4}{2^4}\right)+...+\left(\frac{100}{2^{99}}-\frac{99}{2^{99}}\right)-\frac{100}{2^{100}}\)
\(A=1+\frac{3}{4}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{99}}-\frac{100}{2^{100}}\)
Đặt \(B=\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{99}}\)
\(\Rightarrow A=1+\frac{3}{4}+B-\frac{100}{2^{99}}\) (1)
Ta có:
\(B=\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}...+\frac{1}{2^{99}}\)
\(\Rightarrow2B=\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}...+\frac{1}{2^{98}}\)
\(2B-B=\left(\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{98}}\right)-\left(\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{98}}+\frac{1}{2^{99}}\right)\)
\(B=\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{98}}-\frac{1}{2^3}-\frac{1}{2^4}-...-\frac{1}{2^{98}}-\frac{1}{2^{99}}\)
\(B=\frac{1}{2^2}+\left(\frac{1}{2^3}-\frac{1}{2^3}\right)+\left(\frac{1}{2^4}-\frac{1}{2^4}\right)+...+\left(\frac{1}{2^{98}}-\frac{1}{2^{98}}\right)-\frac{1}{2^{99}}\)
\(B=\frac{1}{4}+0+0+...+0-\frac{1}{2^{99}}\)
\(B=\frac{1}{4}-\frac{1}{2^{99}}\)
Từ (1)
\(\Rightarrow A=1+\frac{3}{4}+\left(\frac{1}{4}-\frac{1}{2^{99}}\right)-\frac{100}{2^{100}}\)
\(A=\frac{7}{4}+\frac{1}{4}-\frac{1}{2^{99}}-\frac{100}{2^{100}}\)
\(A=2-\frac{2}{2^{100}}-\frac{100}{2^{100}}\)
\(A=2-\frac{102}{2^{100}}\)
Vậy \(A=2-\frac{102}{2^{100}}\)
Tính:
b) \(-2.\frac{-38}{21}.\frac{-7}{4}.-\frac{3}{8}\)
= \(\frac{\left(-2\right).\left(-38\right).\left(-7\right).\left(-3\right)}{21.4.8}\)
= \(\frac{-1.-19.1}{1.2.4}=\frac{19}{8}\)
// Học tốt!
Cách 1 :
\(M=\left(7-\frac{1}{3}+\frac{3}{4}\right)-\left(6+\frac{2}{3}-\frac{1}{4}\right)\)
\(M=\left(\frac{84}{12}-\frac{4}{12}+\frac{9}{12}\right)-\left(\frac{72}{12}+\frac{8}{12}-\frac{3}{12}\right)\)
\(M=\frac{89}{12}-\frac{77}{12}\)
\(M=\frac{12}{12}\)
\(M=1\)
Cách 2 :
\(M=\left(7-\frac{1}{3}+\frac{3}{4}\right)-\left(6+\frac{2}{3}-\frac{1}{4}\right)\)
\(M=7-\frac{1}{3}+\frac{3}{4}-6-\frac{2}{3}+\frac{1}{4}\)
\(M=\left(7-6\right)-\left(\frac{2}{3}+\frac{1}{3}\right)+\left(\frac{3}{4}+\frac{1}{4}\right)\)
\(M=1-1+1\)
\(M=1\)
Cách 1:
\(M=\left(7-\frac{1}{3}+\frac{3}{4}\right)-\left(6+\frac{2}{3}-\frac{1}{4}\right)\)
\(M=\frac{89}{12}-\frac{77}{12}\)
\(M=1\)
Cách 2:
\(M=\left(7-\frac{1}{3}+\frac{3}{4}\right)-\left(6+\frac{2}{3}-\frac{1}{4}\right)\)
\(M=7-\frac{1}{3}+\frac{3}{4}-6-\frac{2}{3}+\frac{1}{4}\)
\(M=\left(7-6\right)-\left(\frac{1}{3}+\frac{2}{3}\right)+\left(\frac{3}{4}+\frac{1}{4}\right)\)
\(M=1-1+1\)
\(M=1\)
b, \(\left(-\frac{4}{3}\right)-\frac{2}{5}-\frac{3}{2}=-\frac{40}{30}-\frac{12}{30}-\frac{45}{30}=-\frac{97}{30}\)
c, \(\frac{4}{5}+\frac{2}{7}-\frac{7}{10}=\frac{56}{70}+\frac{20}{70}-\frac{49}{70}=\frac{27}{70}\)
d, \(\frac{2}{3}-\left[\left(-\frac{7}{4}\right)-\left(\frac{4}{8}+\frac{3}{8}\right)\right]=\frac{2}{3}-\left[\left(-\frac{7}{4}\right)-\frac{7}{8}\right]=\frac{2}{3}--\frac{21}{8}=\frac{2}{3}+\frac{21}{8}=3\frac{7}{24}\)
tính lại mt cko chắc ăn nha