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A = \(\frac{5}{4}.\left(5-\frac{4}{3}\right).\frac{1}{11}\)
A = \(\frac{5}{44}\left(5-\frac{4}{3}\right)\)
A = \(\frac{25}{44}-\frac{5}{33}\)
A = \(\frac{25.3}{4.11.3}-\frac{5.4}{3.11.4}\)
A = \(\frac{75}{132}-\frac{20}{132}\)
A = \(\frac{55}{132}\)
B = \(\frac{3}{4}:\left(-12\right).\left(-\frac{2}{3}\right)\)
B = \(\frac{-1}{16}.\left(-\frac{2}{3}\right)\)
B = \(\frac{1}{24}\)
C = \(\frac{5}{4}:\left(-15\right).\left(-\frac{2}{5}\right)\)
C = \(\frac{-1}{12}.\left(-\frac{2}{5}\right)\)
C = \(\frac{1}{30}\)
D = \(\left(-3\right).\left(\frac{2}{3}-\frac{5}{4}\right):\left(-7\right)\)
D = \(\frac{3}{7}.\left(\frac{8}{12}-\frac{15}{12}\right)\)
D = \(\frac{3}{7}.\left(-\frac{7}{12}\right)\)
D = \(\frac{-3}{12}\)
D = \(\frac{-1}{4}\)
\(C=\frac{5}{4}+\frac{5}{4^2}+\frac{5}{4^3}+...+\frac{5}{4^{99}}\)
\(C=5.\left(\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+...+\frac{1}{4^{99}}\right)\)
đặt \(A=\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+...+\frac{1}{4^{99}}\)
\(4A=1+\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{98}}\)
\(4A-A=\left(1+\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{98}}\right)-\left(\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+...+\frac{1}{4^{99}}\right)\)
\(3A=1-\frac{1}{4^{99}}\)< 1
\(\Rightarrow A=\frac{1-\frac{1}{4^{99}}}{3}< \frac{1}{3}\)
suy ra \(C=5.\left(\frac{1-\frac{1}{4^{99}}}{3}\right)< 5.\frac{1}{3}=\frac{5}{3}\)
Ta có :
\(C=\frac{5}{4}+\frac{5}{4^2}+\frac{5}{4^3}+...+\frac{5}{4^{99}}\)
\(\Rightarrow4.C=4\left(\frac{5}{4}+\frac{5}{4^2}+\frac{5}{4^3}+...+\frac{5}{4^{99}}\right)\)
\(\Rightarrow4C=5+\frac{5}{4}+\frac{5}{4^2}+...+\frac{5}{4^{98}}\)
\(\Rightarrow4C-C=5+\frac{5}{4}+\frac{5}{4^2}+...+\frac{5}{4^{98}}-\frac{5}{4}-\frac{5}{4^2}-...-\frac{5}{4^{99}}\)
\(\Rightarrow3C=\)\(5-\frac{5}{4^{99}}=5\left(1-\frac{1}{4^{99}}\right)\)
\(\Rightarrow C=\frac{5}{3}.\left(1-\frac{1}{4^{99}}\right)< \frac{5}{3}\left(đpcm\right)\)
Kb vs k cho mình nhé!