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\(A=\frac{1}{5}+\frac{2}{5^2}+\frac{3}{5^3}+....+\frac{10}{5^{10}}+\frac{11}{5^{11}}\)
\(\Rightarrow5A=1+\frac{2}{5}+\frac{3}{5^2}+....+\frac{10}{5^9}+\frac{11}{5^{10}}\)
\(\Rightarrow5A-A=\left(1+\frac{2}{5}+...+\frac{11}{5^{10}}\right)-\left(\frac{1}{5}+\frac{2}{5^2}+...+\frac{11}{5^{11}}\right)\)
\(\Rightarrow4A=1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{10}}-\frac{11}{5^{11}}\)(1)
Đặt \(B=1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{10}}\)
\(\Rightarrow5B=5+1+\frac{1}{5}+...+\frac{1}{5^9}\)
\(\Rightarrow5B-B=\left(5+1+...+\frac{1}{5^9}\right)-\left(1+\frac{1}{5}+...+\frac{1}{5^{10}}\right)\)
\(\Rightarrow4B=5-\frac{1}{5^{10}}< 5\)
\(\Rightarrow B< \frac{5}{4}\)(2)
Thay (2) vào (1) \(\Rightarrow4A< \frac{5}{4}-\frac{11}{5^{11}}< \frac{5}{4}\)
\(\Rightarrow A< \frac{5}{16}\left(đpcm\right)\)
A=1+(2-3-3+5)+(6-7-8+9)+....+(98-99-100+101)+102
=1+0+0+....+102=103
b) |1-2x|>7
=> 1-2x>7 hoặc 1-2x<-7
=> 2x<-6 hoặc 2x>8
=> x<-3 hoặc x>4
3,
a) (−23+37):45+(−13+47):45
= \(-\frac{5}{21}:\frac{4}{5}+\frac{5}{21}:\frac{4}{5}\)
= \(\left(-\frac{5}{21}+\frac{5}{21}\right):\frac{4}{5}\)
= \(0:\frac{4}{5}=0\)
2,
a) \(\frac{-3}{4}\).\(\frac{12}{-5}\).(\(\frac{-25}{6}\))
= \(\frac{-3.4.3.\left(-5\right).5}{4.\left(-5\right).3.3}\)
= \(-5\)
b) (−2).\(\frac{-38}{21}\).\(\frac{-7}{4}\).(\(\frac{-3}{8}\))
= \(\frac{-2.\left(-38\right)\left(-7\right)\left(-3\right)}{\left(-7\right)\left(-3\right)\left(-2\right)\left(-2\right).8}\)
= \(\frac{19}{8}\)
c) (\(\frac{11}{12}:\frac{33}{16}\)).\(\frac{3}{5}\)
= \(\left(\frac{11}{12}.\frac{16}{33}\right).\frac{3}{5}\)
= \(\frac{4}{9}.\frac{3}{5}\)
= \(\frac{4}{15}\)
d) \(\frac{7}{23}\left[\left(\frac{-8}{6}\right)-\frac{45}{18}\right]\)
= \(\frac{7}{23}.\left(\frac{-41}{10}\right)\)
= \(\frac{-287}{203}\)
3. Tính:
a) (\(\frac{-2}{3}+\frac{3}{7}\)):\(\frac{4}{5}\)+(\(\frac{-1}{3}+\frac{4}{7}\)):\(\frac{4}{5}\)
= (\(\frac{-2}{3}+\frac{3}{7}\)\(+\)\(\frac{-1}{3}+\frac{4}{7}\)) : \(\frac{4}{5}\)
= 0 : \(\frac{4}{5}\)
= 0
b) \(\frac{5}{9}\):(\(\frac{1}{11}-\frac{5}{22}\))+\(\frac{5}{9}\):(\(\frac{1}{15}-\frac{2}{3}\))
= \(\frac{5}{9}\): \(\frac{-3}{22}\)+ \(\frac{5}{9}\): \(\frac{-3}{5}\)
= \(\frac{5}{9}\): \(\frac{-81}{110}\)
= \(\frac{-550}{729}\)
A = \(\left(\frac{1}{15}-\frac{1}{15}\right)\)\(+\left(\frac{3}{7}-\frac{3}{7}\right)\)\(+\left(\frac{5}{9}-\frac{5}{9}\right)\)\(+\left(\frac{2}{11}-\frac{2}{11}\right)\)\(+\left(\frac{7}{13}-\frac{7}{13}\right)\)\(-\frac{9}{16}\)
A = 0 + 0 + 0 + 0 + 0 - \(\frac{9}{16}\)
A = \(-\frac{9}{16}\)
\(A=\frac{1}{5}-\frac{3}{7}+\frac{5}{9}-\frac{2}{11}+\frac{7}{13}-\frac{9}{16}-\frac{7}{13}+\frac{2}{11}-\frac{5}{9}+\frac{3}{7}-\frac{1}{5}\)
\(=\left(\frac{1}{5}-\frac{1}{5}\right)-\left(\frac{3}{7}-\frac{3}{7}\right)+\left(\frac{5}{9}-\frac{5}{9}\right)-\left(\frac{2}{11}-\frac{2}{11}\right)+\left(\frac{7}{13}-\frac{7}{13}\right)-\frac{9}{16}\)
\(=0-0+0-0+0-\frac{9}{16}\)
\(=-\frac{9}{16}\)
bài này bạn lấy ở đâu z
\(A=\frac{1}{5}+\frac{2}{5^2}+\frac{3}{5^3}+...+\frac{11}{5^{11}}\Rightarrow5A=1+\frac{2}{5}+\frac{3}{5^2}+...+\frac{11}{5^{10}}\)
\(\Rightarrow4A=5A-A=1+\left(\frac{1}{5}+\frac{1}{5^2}+\frac{...1}{5^{10}}\right)-\frac{11}{5^{11}}\)
\(< 1+\left(\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{10}}\right)< 1+\frac{1}{4}=\frac{5}{4}\)
\(\Rightarrow A< \frac{5}{4}:4=\frac{5}{16}\)
Lưu ý : \(M=\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{10}}\Rightarrow5M=1+\frac{1}{5}+...+\frac{1}{5^9}\Rightarrow4M=5M-M=1-\frac{1}{5^{10}}\)
\(\Rightarrow M=\frac{1}{4}-\frac{1}{5^{10}}:4< \frac{1}{4}\)