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\(A=\left(\frac{878787}{959595}-\frac{8787}{9595}\right).\frac{1234321}{5678765}\)
\(=\left(\frac{87}{95}-\frac{87}{95}\right).\frac{1234321}{5678765}\)
\(=0.\frac{1234321}{5678765}\)
\(=0\)
\(A=\left(\frac{878787}{959595}+\frac{-8787}{9595}\right).\frac{1234321}{5678765}\)
\(=\left(\frac{87.10101}{95.10101}-\frac{87.101}{95.101}\right).\frac{1234321}{5678765}\)
\(=\left(\frac{87}{95}-\frac{87}{95}\right).\frac{1234321}{5678765}\)
= 0
\(A=\left(\frac{87}{95}+\frac{-87}{95}\right).\frac{1234321}{5678765}\)
=>\(A=0.\frac{1234321}{5678765}\)
=>A=0
\(\left[\frac{878787}{959595}+\left(-\frac{8787}{9595}\right)\right]\times\frac{1234321}{5678765}=\left[\frac{87}{95}+\left(-\frac{87}{95}\right)\right]\times\frac{1234321}{5678765}=0\times\frac{1234321}{5678765}=0\)
\(A=\left(\frac{878787}{959595}+-\frac{8787}{9595}\right).\frac{1234231}{5678765}\)
\(=\left(\frac{87}{95}+-\frac{87}{95}\right).\frac{1234231}{5678765}\)
\(=0.\frac{1234231}{5678765}=0\)
Ta có :
\(A=\left(\frac{878787}{959595}+\frac{-8787}{9595}\right)\)\(.\frac{1234231}{5678765}\)
\(A=\left(\frac{878787\div10101}{959595\div10101}-\frac{8787\div101}{9595\div101}\right)\)\(.\frac{1234231}{5678765}\)
\(A=\left(\frac{87}{95}-\frac{87}{95}\right)\)\(.\frac{1234231}{5678765}\)
\(A=0.\frac{1234231}{5678765}\)
\(A=0\)
Vậy A=0 .
(878787/959595 + -8787/9595) x 1234321/5678765
=(87/95+ -87/95)x 1234321/5678765
=87+(-87)/95x 1234321/5678765
=0x1234321/5678765
=0
Câu 1 :
\(P=\frac{2.3.4-2.3.4.9+2.3.4.11-2.3.4.13}{5.6.7-5.6.7.9+5.6.7.11-5.6.7.13}\)
\(\Rightarrow P=\frac{2.3.4.\left(1-9+11-13\right)}{5.6.7\left(1-9+11-13\right)}=\frac{2.3.4}{5.6.7}=\frac{4}{35}\)
Câu 3 :
\(S=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{20}}< 1\)
\(\Rightarrow2S=2\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{20}}\right)\)
\(S=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{19}\)
Do đó : 2S - S = \(1-\frac{1}{2^{20}}\)
S = \(1-\frac{1}{2^{20}}\) < 1 (đpcm )
Câu 6 :
\(\overline{abcdeg}=\overline{ab}.10000+\overline{cd}.100+\overline{eg}\)
=\(\overline{ab}.9999+\overline{cd}.99+\left(\overline{ab}+\overline{cd}+\overline{eg}\right)\)
ta thấy :
\(\left[{}\begin{matrix}\overline{ab}.9999⋮11\\\overline{cd}.99⋮11\\\left(\overline{ab}+\overline{cd}+\overline{eg}\right)⋮11\end{matrix}\right.\)
( cái phần( \(\overline{ab}+\overline{cd}+\overline{eg}\))⋮11 theo đề bài )
=>\(\overline{abcdeg}⋮11\)
A= \(\left(\frac{878787}{959595}+-\frac{8787}{9595}\right).\frac{1234321}{5678765}\)
A= \(\left(\frac{87}{95}+\frac{-87}{95}\right).\frac{1234321}{5678765}\)
A= \(0.\frac{1234321}{5678765}\)
A= 0