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\(3+\frac{3}{5}+\frac{3}{25}+\frac{3}{125}+\frac{3}{625}\)
\(=3\times\left(1+\frac{1}{5}+\frac{1}{25}+\frac{1}{125}+\frac{1}{625}\right)\)
\(=3\times\left(\frac{625}{625}+\frac{125}{625}+\frac{25}{625}+\frac{5}{625}+\frac{1}{625}\right)\)
\(=3\times\frac{625+125+25+5+1}{625}\)
\(=3\times\frac{781}{625}=\frac{3\times781}{625}=\frac{2343}{625}\)
Mình không bày bn cách giải, nhưng sẽ gợi ý:
2 bài tương tự nhau, mẫu gấp nhau 3 lần nhé
=[2/3+2/99]+[2/35+2/63]+2/15
=24/99+[4/45+2/15]
=24/99+2/9
=46/99
\(=\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+\frac{2}{7\times9}+\frac{2}{9\times11}\)
\(=\frac{1}{2}\times\left(\frac{2}{1}-\frac{2}{3}+\frac{2}{3}-\frac{2}{5}+\frac{2}{5}-\frac{2}{7}+\frac{2}{7}-\frac{2}{9}+\frac{2}{9}-\frac{2}{11}\right)\)
\(=\frac{1}{2}\times\frac{20}{11}=\frac{10}{11}\)
vì 1/4 bé hơn 1; 3/8 bé hơn 1; 1700/1808 bé hơn 1 nên :
(1/4+3/8+1700/1808) <3
\(\frac{1}{3}+x\times\frac{2}{7}=\frac{11}{12}\)
\(x\times\frac{2}{7}=\frac{11}{12}-\frac{1}{3}\)
\(x\times\frac{2}{7}=\frac{7}{12}\)
\(x=\frac{7}{12}:\frac{2}{7}\)
\(x=\frac{49}{24}\)
~Moon~
\(\frac{1}{3}+x.\frac{2}{7}=\frac{11}{12}\)
\(x.\frac{2}{7}=\frac{11}{12}-\frac{1}{3}\)
\(x.\frac{2}{7}=\frac{7}{12}\)
\(x=\frac{7}{12}\div\frac{2}{7}\)
\(x=\frac{49}{24}\)
a) \(\frac{12}{25}+\frac{3}{5}+\frac{13}{25}\)
= \(\left(\frac{12}{25}+\frac{13}{25}\right)+\frac{3}{5}\)
= \(\frac{25}{25}+\frac{3}{5}\)
= \(1+\frac{3}{5}\)
= \(\frac{1}{1}+\frac{3}{5}\)
= \(\frac{5}{5}+\frac{3}{5}\)
= \(\frac{8}{5}\)
bài này tớ mà làm sai thì tớ ......................ko lm người........-_-
a) \(\frac{12}{25}+\frac{3}{5}+\frac{13}{25}\)
= \(\frac{12}{25}+\frac{15}{25}+\frac{13}{25}\)
= \(\frac{12+15+13}{25}\)
= \(\frac{40}{25}\)
= \(\frac{8}{5}\)
b) \(\frac{3}{2}+\frac{2}{3}+\frac{4}{3}\)
= \(\frac{9}{6}+\frac{4}{6}+\frac{8}{6}\)
= \(\frac{9+4+8}{6}\)
= \(\frac{21}{6}\)
= \(\frac{7}{2}\)
Sao mà mình hỏi bài này từ lâu lắm rồi mà vẫn chưa có bạn nào trả lời nhỉ?
A) \(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}\)
2A= \(1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}\)
2A-A = \(1-\dfrac{1}{32}\)
A= \(\dfrac{31}{32}\)
Đặt tổng trên là: \(A\)
\(A=\dfrac{3}{2}+\dfrac{3}{8}+\dfrac{3}{32}+\dfrac{3}{128}+\dfrac{3}{512}\)
\(\Rightarrow A.4=6+\dfrac{3}{2}+\dfrac{3}{8}+\dfrac{3}{32}+\dfrac{3}{128}\)
\(\Rightarrow A.4-A=\left(6+\dfrac{3}{2}+\dfrac{3}{8}+\dfrac{3}{32}+\dfrac{3}{128}\right)-\left(\dfrac{3}{2}+\dfrac{3}{8}+\dfrac{3}{32}+\dfrac{3}{128}+\dfrac{3}{512}\right)\)
\(\Rightarrow A.3=6-\dfrac{3}{512}=\dfrac{3069}{512}\)
\(\Rightarrow A=\dfrac{3069}{512}:3=\dfrac{1023}{512}\)
\(\frac{3}{2}+\frac{3}{8}+\frac{3}{32}+\frac{3}{128}+\frac{3}{512}\)
\(=3.\left(\frac{1}{2}+\frac{1}{2^3}+\frac{1}{2^5}+\frac{1}{2^7}+\frac{1}{2^9}\right)\)
\(=3.A\)với \(A=\frac{1}{2}+\frac{1}{2^3}+\frac{1}{2^5}+\frac{1}{2^7}+\frac{1}{2^9}\)
\(\Rightarrow2^2A=\left(2+\frac{1}{2}+\frac{1}{2^3}+\frac{1}{2^5}+\frac{1}{2^7}\right)\)
\(\Rightarrow2^2A-A=\left(2+\frac{1}{2}+\frac{1}{2^3}+\frac{1}{2^5}+\frac{1}{2^7}\right)-\left(\frac{1}{2}+\frac{1}{2^3}+\frac{1}{2^5}+\frac{1}{2^7}+\frac{1}{2^9}\right)\)
\(\Rightarrow4A-A=2-\frac{1}{2^9}\)
\(\Rightarrow3A=2-\frac{1}{512}=\frac{1023}{512}\Rightarrow A=\frac{1023}{512}:3\)
\(\Rightarrow\frac{3}{2}+\frac{3}{8}+\frac{3}{32}+\frac{3}{128}+\frac{3}{512}=3.\left(\frac{1023}{512}:3\right)=\frac{1023}{512}\)