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VL
1
DP
3 tháng 8 2017
\(\left(\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+...+\frac{2}{3001.3004}\right)\cdot\left(x+1\right)=\frac{9009}{1502}\)
\(\Leftrightarrow\frac{2}{3}\cdot\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{3001}-\frac{1}{3004}\right)\cdot\left(x+1\right)=\frac{9009}{1502}\)
\(\Leftrightarrow\frac{2}{3}\cdot\left(1-\frac{1}{3004}\right)\cdot\left(x+1\right)=\frac{9009}{1502}\)
\(\Leftrightarrow\frac{2}{3}\cdot\frac{3003}{3004}\cdot\left(x+1\right)=\frac{9009}{1502}\)
\(\Leftrightarrow\frac{1001}{1502}\cdot\left(x+1\right)=\frac{9009}{1502}\)
\(\Leftrightarrow x+1=\frac{9009}{1502}\div\frac{1001}{1502}\)
\(\Leftrightarrow x+1=9\Rightarrow x=8\)
2 tháng 6 2018
a) \(A=98+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}\)(có 98 phân số nên ta cộng 1 vào mỗi phân số)
\(A=\left(\frac{1}{2}+1\right)+\left(\frac{1}{3}+1\right)+...+\left(\frac{1}{99}+1\right)\)
\(A=\frac{3}{2}+\frac{4}{3}+...+\frac{100}{99}\)
Và \(B=\frac{3}{2}+\frac{4}{3}+...+\frac{100}{99}\)
\(\Rightarrow\frac{A}{B}=\frac{\frac{3}{2}+\frac{4}{3}+...+\frac{100}{99}}{\frac{3}{2}+\frac{4}{3}+...+\frac{100}{99}}=1\)
b) \(A=2018+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2019}\)(có 2018 phân số nên ta cộng 1 vào mỗi phân số)
\(A=\left(\frac{1}{2}+1\right)+\left(\frac{1}{3}+1\right)+...+\left(\frac{1}{2019}+1\right)\)
\(A=\frac{3}{2}+\frac{4}{3}+...+\frac{2020}{2019}\)
Và \(B=\frac{3}{2}+\frac{4}{3}+...+\frac{2020}{2019}\)
\(\Rightarrow\frac{A}{B}=\frac{\frac{3}{2}+\frac{4}{3}+...+\frac{2020}{2019}}{\frac{3}{2}+\frac{4}{3}+...+\frac{2020}{2019}}=1\)
c) \(A=\frac{99}{1}+\frac{98}{2}+...+\frac{1}{99}\)
\(A=99+\frac{98}{2}+...+\frac{1}{99}\)(có 98 phân số nên ta cộng 1 vào từng phân số)
\(A=\left(\frac{98}{2}+1\right)+\left(\frac{97}{3}+1\right)+...+\left(\frac{1}{99}+1\right)+1\)
\(A=\frac{100}{2}+\frac{100}{3}+...+\frac{100}{99}+1\)
\(A=100\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}\right)\)
Và \(B=\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}\)
\(\Rightarrow\frac{A}{B}=\frac{100\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}}=100\)
DL
2 tháng 6 2018
a)\(B=\frac{3}{2}+\frac{4}{3}+\frac{5}{4}+...+\frac{100}{99}\)
\(B=\left(1+\frac{1}{2}\right)+\left(1+\frac{1}{3}\right)+\left(1+\frac{1}{4}\right)+...+\left(1+\frac{1}{99}\right)\)
\(\Rightarrow B=\left(1+1+1+...+1\right)+\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}\right)\)
\(\Rightarrow B=98+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}\)
\(\Rightarrow A:B=\frac{A}{B}=\frac{98+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}}{98+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}}=1.\)
Vậy \(A:B=1.\)
b)\(B=\left(1+\frac{1}{2}\right)+\left(1+\frac{1}{3}\right)+\left(1+\frac{1}{4}\right)+...+\left(1+\frac{1}{2019}\right)\)
\(\Rightarrow B=\left(1+1+1+...+1\right)+\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}\right)\)
\(\Rightarrow B=2018+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}\)
\(\Rightarrow A:B=\frac{A}{B}=\frac{2018+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}}{2018+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}}=1.\)
Vậy \(A:B=1.\)
c)\(A=\left(1+1+...+1\right)+\frac{98}{2}+\frac{97}{3}+...+\frac{2}{98}+\frac{1}{99}\)
\(A=\left(1+\frac{98}{2}\right)+\left(1+\frac{97}{3}\right)+...+\left(1+\frac{2}{98}\right)+\left(1+\frac{1}{99}\right)\)
\(A=\frac{100}{2}+\frac{100}{3}+...+\frac{100}{98}+\frac{100}{99}\)
\(A=100\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{98}+\frac{1}{99}\right)\)
\(\Rightarrow A:B=\frac{A}{B}=\frac{100\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{98}+\frac{1}{99}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{98}+\frac{1}{99}}=1.\)
Vậy \(A:B=1.\)
TD
8
TA
9 tháng 6 2023
a, \(2\dfrac{1}{3}+4\dfrac{1}{5}+4\dfrac{1}{3}\)
\(=\dfrac{7}{3}+\dfrac{21}{5}+\dfrac{13}{3}\)
\(=\dfrac{1}{3}\left(7+13\right)+\dfrac{21}{5}\)
\(=\dfrac{20}{3}+\dfrac{21}{5}=\dfrac{100+63}{15}=\dfrac{163}{15}\)
b, \(5\dfrac{3}{4}-4\dfrac{1}{2}.3\dfrac{7}{8}\)
\(=\dfrac{23}{4}-\dfrac{9}{2}.\dfrac{31}{8}\)
\(=\dfrac{23}{4}-\dfrac{279}{16}=\dfrac{92-279}{16}=-\dfrac{187}{16}\)
c, \(1\dfrac{1}{2}.3\dfrac{2}{3}.4\dfrac{3}{4}\)
\(=\dfrac{3}{2}.\dfrac{11}{3}.\dfrac{19}{4}=\dfrac{209}{8}\)
d, \(6\dfrac{4}{5}:2\dfrac{3}{4}:1\dfrac{1}{2}\)
\(=\dfrac{34}{5}:\dfrac{11}{4}:\dfrac{3}{2}\)
\(=\left(\dfrac{34}{5}.\dfrac{4}{11}\right).\dfrac{2}{3}=\dfrac{136}{55}.\dfrac{2}{3}=\dfrac{272}{165}\)
NM
1
25 tháng 2 2017
\(\frac{1}{2.2}< \frac{1}{1.2}\)
\(\frac{1}{3.3}< \frac{1}{2.3}\)
......
\(\frac{1}{100.100}< \frac{1}{99.100}\)
\(\Rightarrow\frac{1}{2.2}+\frac{1}{3.3}+...+\frac{1}{100.100}< \frac{1}{1.2}+..+\frac{1}{99.100}\)
\(\Rightarrow\frac{1}{2.2}+..+\frac{1}{100.100}< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}...+\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow\frac{1}{2.2}+..+\frac{1}{100.100}< 1-\frac{1}{100}< 1\).Suy ra điều phải chứng minh. câu b tương tự. bấm đúng cho mình nha
NA
2
A=1+2+3+4+...+1502
=>\(A=\frac{1502.\left(1502+1\right)}{2}\)
=>\(A=\frac{1502.1503}{2}\)
=>\(A=\frac{2257506}{2}\)
=>\(A=1128753\)
l-i-k-e cho mình nha bạn.
Số số hạng có trong dãy số trên là:
(1502 - 1) : 1 + 1 = 1502 (số)
Tổng trên là:
(1502 + 1) x 1502 : 2 = 1128753
Đáp số: 1128573