K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
29 tháng 4 2016
a/ \(=\frac{21}{23}+\frac{125}{143}-\frac{101.21}{101.23}-\frac{1001.125}{1001.143}=0\)
b/ \(=\frac{4}{20}+\frac{8}{21}+\frac{2}{5}-\frac{3}{5}+\frac{2}{21}-\frac{10}{21}+\frac{3}{20}=\frac{7}{20}-\frac{1}{5}=\frac{4}{20}\)
c/ \(\frac{C}{2}=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{420}=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{20.21}\)
\(\frac{C}{2}=\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{21-20}{20.21}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{20}-\frac{1}{21}\)
\(\frac{C}{2}=\frac{1}{2}-\frac{1}{21}=\frac{19}{42}\Rightarrow C=\frac{19}{21}\)
DP
14 tháng 8 2017
Đặt \(B=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.....+\frac{1}{210}\)
\(\frac{1}{2}B=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{420}\)
\(\frac{1}{2}B=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{20.21}\)
\(\frac{1}{2}B=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{20}-\frac{1}{21}\)
\(\frac{1}{2}B=\frac{1}{2}-\frac{1}{21}\)
\(\Rightarrow B=\frac{\frac{1}{2}-\frac{1}{21}}{\frac{1}{2}}=\frac{19}{21}\)
DP
14 tháng 8 2017
\(A=\frac{1}{1+2}+\frac{1}{1+2+3}+....+\frac{1}{1+2+3+...+50}\)
\(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{\frac{\left(1+50\right).50}{2}}\)
\(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+....+\frac{1}{1275}\)
\(A=\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+....+\frac{2}{2550}\)
\(A=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+..+\frac{2}{50.51}\)
\(A=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{50}-\frac{1}{51}\right)\)
\(A=2\left(\frac{1}{2}-\frac{1}{51}\right)=2\cdot\frac{49}{102}=\frac{49}{51}\)
B
30 tháng 3 2019
\(B=\left(1-\frac{1}{3}\right)\left(1-\frac{1}{6}\right)\left(1-\frac{1}{10}\right)\left(1-\frac{1}{15}\right)...\left(1-\frac{1}{210}\right)\)
\(B=\frac{2}{3}.\frac{5}{6}.\frac{9}{10}.\frac{14}{15}...\frac{209}{210}\)
\(B=\frac{1}{3}.\frac{1}{3}.\frac{9}{2}.\frac{14}{15}...\frac{209}{210}\)
\(B=\frac{1}{6}.\frac{9}{2}.\frac{14}{15}...\frac{209}{210}\)
\(B=\frac{1}{2}.\frac{1}{1}.\frac{7}{5}...\frac{209}{210}\)
\(B=\frac{7}{10}...\frac{209}{210}\)
\(B=\frac{62}{210}\)
NM
1
BA
5 tháng 7 2018
\(A=\left\{X\in N/k.k+1;K\in N;k\le13\right\}\)
A có 14 phần tử.
\(B=\left\{X\in N/x=k.\left(k+i\right):2;K\in N;k\le20\right\}\)
B có 21 phần tử
~HỌC TỐT~
S
4 tháng 3 2017
A = \(\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+...+\frac{1}{210}\)
\(\frac{1}{2}A=\frac{1}{2}\left(\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+...+\frac{1}{210}\right)\)
\(\frac{1}{2}A=\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{420}\)
\(\frac{1}{2}A=\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{20.21}\)
\(\frac{1}{2}A=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{20}-\frac{1}{21}\)
\(\frac{1}{2}A=\frac{1}{3}-\frac{1}{21}\)
\(\frac{1}{2}A=\frac{2}{7}\)
A = \(\frac{2}{7}:\frac{1}{2}\)
A = \(\frac{4}{7}\)
Q
20 tháng 11 2017
a) (x+10)(2y-5) = 143
=> (x+10);(2y-5) thuộc Ư(143)={-1,-143,1,143}
\(\orbr{\begin{cases}x+10=-143\\2y-5=-1\end{cases}}\Rightarrow\orbr{\begin{cases}x=-153\\y=2\end{cases}}\)
\(\orbr{\begin{cases}x+10=-1\\2y-5=-143\end{cases}}\Rightarrow\orbr{\begin{cases}x=-11\\y=-69\end{cases}}\)
\(\orbr{\begin{cases}x+10=1\\2y-5=143\end{cases}}\Rightarrow\orbr{\begin{cases}x=-9\\y=74\end{cases}}\)
\(\orbr{\begin{cases}x+10=143\\2y-5=1\end{cases}}\Rightarrow\orbr{\begin{cases}x=133\\y=3\end{cases}}\)
Vậy ta có các cặp x,y thõa mãn : (-153,2);(-11,-69);(-9,74);(113,3)
b) x+(x+1)+(x+2)+..+(x+30)=1240
=> (x+x+x+...+x)+(1+2+3+...+30)=1240
=> 31x+465=1240
31x = 1240-465
31x = 775
x = 775 : 31
x= 25
c) 1+2+3+...+x=210
\(\frac{\left(x-1\right)}{1}+1=x\)
=> \(\frac{\left(x+1\right).x}{2}=210\)
(x+1)x = 210:2
(x+1)x = 105
chắc ko có x thõa mãn
d) 2+4+6+...+2x=210
=> 2(1+2+3+...+x)=210
1+2+3+..+x= 210:2 = 105
\(\frac{\left(x-1\right)}{1}+1\) = x
\(\frac{\left(x+1\right).x}{2}=105\)
(x+1)x = 105:2
(x+1)x = 52,5
ko có x thõa mãn đề bài
S
20 tháng 11 2017
a, x + 10 và 2y - 5 thuộc Ư(143) = {1;-1;143;-143}
| x + 10 | 1 | -1 | 143 | -143 |
| 2y - 5 | 143 | -143 | 1 | -1 |
| x | -9 | -11 | 133 | -153 |
| y | 74 | -69 | 3 | 2 |
b, x+(x+1)+(x+2)+........+(x+30) = 1240
=> x+x+1+x+2+...+x+30=1240
=> 31x+(1+2+...+30) = 1240
=> 31x + 465 = 1240
=> 31x = 775
=> x = 25
c, 1+2+...+x=210
=> \(\frac{x\left(x+1\right)}{2}=210\)
=> x(x+1) = 420
Mà 420 = 20.21
=> x = 20
d, 2+4+...+2x = 210
=> 2(1+2+...+x) = 210
=> \(\frac{2x\left(x+1\right)}{2}=210\)
=> x(x + 1) = 210
Mà 210 = 14.15
=> x = 14
e, 1+3+5+...+(2x-1) = 225
=> \(\frac{\left[\left(2x-1\right)+1\right].x}{2}=225\)
=> \(\frac{2x^2}{2}=225\)
=> x2 = \(\left(\pm15\right)^2\)
=> x = 15 hoặc x = -15
Lời giải:
$\frac{A}{2}=\frac{1}{2}-\frac{1}{6}-\frac{1}{12}-\frac{1}{20}-...-\frac{1}{420}$
$\frac{A}{2}=\frac{1}{2}-\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{420}\right)$
Xét:
$\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{420}=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{20.21}$
$=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{20}-\frac{1}{21}$
$=\frac{1}{2}-\frac{1}{21}$
Do đó:
$\frac{A}{2}=\frac{1}{2}-(\frac{1}{2}-\frac{1}{21})=\frac{1}{21}$
$\Rightarrow A=\frac{2}{21}$