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Cách chứng minh đề sai : Số số phân số là
(10101-3):5+1=\(\frac{10103}{5}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
bạn ơi câu a mk ko pik vì nó có số 5 ở cuối
b) 121212 / 424242 : 10101 / 10101 = 2/7
c) 187187187 / 221221221 : 1001001/1001001 = 11/13
d) mk ko pik nhé bạn thông cảm
ai thấy đúg thì tk nha
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a: \(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{100\cdot101}\)
=1-1/2+1/2-1/3+...+1/100-1/101
=1-1/101=100/101
b: \(A=1+\dfrac{1}{2}+1+\dfrac{1}{6}+1+\dfrac{1}{12}+...+1+\dfrac{1}{10100}\)
\(=100+\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{100}-\dfrac{1}{101}\right)\)
\(=101-\dfrac{1}{101}< 101\)
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a)
\(\overline{ab}\times101=\overline{ab}\times\left(100+1\right)=\overline{ab00}+\overline{ab}=\overline{abab}\)
b)
\(\overline{ab}\times10101=\overline{ab}\times\left(10000+101\right)=\overline{ab0000}+\overline{abab}=\overline{ababab}\)
c)
\(\overline{abc}\times1001=\overline{abc}\times\left(1000+1\right)=\overline{abc000}+\overline{abc}=\overline{abcabc}\)
d)
\(\overline{ab}\times1001=\overline{ab}\times\left(1000+1\right)=\overline{ab000}+\overline{ab}=\overline{ab0ab}\)
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\(10101.\left(\frac{5}{10101}-\frac{5}{20202}+\frac{5}{30303}+\frac{5}{40404}\right)\)
\(=10101.\frac{5}{10101}-10101.\frac{5}{20202}+10101.\frac{5}{30303}+10101.\frac{5}{40404}\)
\(=5-\frac{5}{2}+\frac{5}{3}+\frac{5}{4}\)
\(=\frac{60}{12}-\frac{30}{12}+\frac{20}{12}+\frac{15}{12}\)
\(=\frac{60-30+20+15}{12}\)
\(=\frac{65}{12}\)
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Câu hỏi của Lê Tiến Cường - Toán lớp 6 - Học toán với OnlineMath
\(A=\frac{3}{2}+\frac{7}{6}+\frac{13}{12}+...+\frac{10101}{10100}=\frac{2+1}{2}+\frac{6+1}{6}+\frac{12+1}{12}+...+\frac{10100+1}{10100}\)
\(A=\left(1+\frac{1}{2}\right)+\left(1+\frac{1}{6}\right)+\left(1+\frac{1}{12}\right)+....+\left(1+\frac{1}{10100}\right)\)
\(A=\left(1+\frac{1}{1\times2}\right)+\left(1+\frac{1}{2\times3}\right)+\left(1+\frac{1}{3\times4}\right)+...+\left(1+\frac{1}{100\times101}\right)\)
\(A=\left(1+1+1+....+1\right)+\left(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{100\times101}\right)\)
\(A=100+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{100}-\frac{1}{101}\right)\)
\(A=100+1-\frac{1}{101}=101-\frac{1}{101}< 101=B\)
\(\Rightarrow A< B\)
So easy
121212
= 121212
nha!