1. Tính:
D = 3/5.7 + 3/7.9 + 3/9.11 + ... + 3/53.55
2. Tìm x sao cho:
A = x+2/x-5 là số nguyên dương
3. Chứng tỏ rằng:
S = 1/2 + 1/2^2 + 1/2^3 + ... + 1/2^20 nhỏ hơn 1
A = 1/10 + 1/11 + 1/12 + ... + 1/99 + 1/100 lớn hơn 1
Giúp mình với. Hu hu
1:
Ta có: \(D=\dfrac{3}{5\cdot7}+\dfrac{3}{7\cdot9}+\dfrac{3}{9\cdot11}+...+\dfrac{3}{53\cdot55}\)
\(=\dfrac{3}{2}\left(\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+\dfrac{2}{9\cdot11}+...+\dfrac{2}{53\cdot55}\right)\)
\(=\dfrac{3}{2}\left(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+...+\dfrac{1}{53}-\dfrac{1}{55}\right)\)
\(=\dfrac{3}{2}\left(\dfrac{1}{5}-\dfrac{1}{55}\right)\)
\(=\dfrac{3}{2}\left(\dfrac{11}{55}-\dfrac{1}{55}\right)\)
\(=\dfrac{3}{2}\cdot\dfrac{2}{11}=\dfrac{3}{11}\)
2) Để A là số nguyên dương thì
\(\left\{{}\begin{matrix}x+2⋮x-5\\x-5>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-5+7⋮x-5\\x>5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}7⋮x-5\\x>5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-5\inƯ\left(7\right)\\x>5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-5\in\left\{1;-1;7;-7\right\}\\x>5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\in\left\{6;4;12;-2\right\}\\x>5\end{matrix}\right.\)
\(\Leftrightarrow x\in\left\{6;12\right\}\)
Giải:
1)D=3/5.7+3/7.9+3/9.11+...+3/53.55
D=3/2.(2/5.7+2/7.9+2/9.11+...+2/53.55)
D=3/2.(1/5-1/7+1/7-1/9+1/9-1/11+...+1/53-1/55)
D=3/2.(1/5-1/55)
D=3/2.2/11
D=3/11
( 1/1.3+1/3.5+1/5.7+1/7.9+1/9.11). X = 2/3
\(\text{Ta có:}\) \(\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\right).x=\frac{2}{3}\)
\(\Leftrightarrow2.\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\right).x=\frac{2}{3}.2\)
\(\Leftrightarrow\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right).x=\frac{4}{3}\)
\(\Leftrightarrow\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+.....+\frac{1}{9}-\frac{1}{11}\right).x=\frac{4}{3}\)
\(\Leftrightarrow\left(1-\frac{1}{11}\right)x=\frac{4}{3}\)
\(\Leftrightarrow\frac{10}{11}x=\frac{4}{3}\)
\(\Leftrightarrow x=\frac{4}{3}:\frac{10}{11}=\frac{22}{15}\)
Tính bằng cách thuận tiên nhất:
1):(3/4 x 5/97 + 1/9 x 13/47) x (1/5 - 7/25 x 5/7)
2): 8/17 x 4/15 + 8/17 x 22/15 - 8/15 x 9/17
3): 2021/2 x 1/3 + 4042/4 x 1/5 + 6063/3 x 22/15
4); 4/7 x 2/13 + 8/13 :7/4 + 4/7 : 13/2 + 4/7 x 1/13
5): 2022 x 2021 - 1/ 2021 + 2022 x 2020
6): 18 x 123 + 9 x 4567 x 2 + 3 x 5310 x 6 / (2 + 4 + 6 + 8 + ...+20 + 22) + 48
7): A= 2021 x 2021 x 202020 - 2020 x 2020 x 20212021 / 2020 x 20192019
1) Ta có: \(\left(\dfrac{3}{4}\cdot\dfrac{5}{97}+\dfrac{1}{9}\cdot\dfrac{13}{47}\right)\cdot\left(\dfrac{1}{5}-\dfrac{7}{25}\cdot\dfrac{5}{7}\right)\)
\(=\left(\dfrac{3}{4}\cdot\dfrac{5}{97}+\dfrac{1}{9}\cdot\dfrac{13}{47}\right)\cdot\left(\dfrac{1}{5}-\dfrac{1}{5}\right)\)
=0
2) Ta có: \(\dfrac{8}{17}\cdot\dfrac{4}{15}+\dfrac{8}{17}\cdot\dfrac{22}{15}-\dfrac{8}{15}\cdot\dfrac{9}{17}\)
\(=\dfrac{8}{17}\left(\dfrac{4}{15}+\dfrac{22}{15}-\dfrac{9}{15}\right)\)
\(=\dfrac{8}{17}\cdot\dfrac{15}{15}=\dfrac{8}{17}\)
3) Ta có: \(\dfrac{2021}{2}\cdot\dfrac{1}{3}+\dfrac{4042}{4}\cdot\dfrac{1}{5}+\dfrac{6063}{3}\cdot\dfrac{22}{15}\)
\(=\dfrac{2021}{2}\left(\dfrac{1}{3}+\dfrac{1}{5}\right)+2021\cdot\dfrac{22}{15}\)
\(=\dfrac{2021}{2}\cdot\dfrac{8}{15}+\dfrac{2021}{2}\cdot\dfrac{44}{15}\)
\(=\dfrac{2021}{2}\cdot\dfrac{52}{15}\)
\(=\dfrac{52546}{15}\)
4) Ta có: \(\dfrac{4}{7}\cdot\dfrac{2}{13}+\dfrac{8}{13}:\dfrac{7}{4}+\dfrac{4}{7}:\dfrac{13}{2}+\dfrac{4}{7}\cdot\dfrac{1}{13}\)
\(=\dfrac{4}{7}\left(\dfrac{2}{13}+\dfrac{8}{13}+\dfrac{2}{13}+\dfrac{1}{13}\right)\)
\(=\dfrac{4}{7}\)
a,2/5.7 + 2/7.9 + 2/9.11 + ... + 2/53.55
b,1/2 .1/3 +1/4 . 1/5 + ... + 1/8 . 1/9
c,(1+1/2)(1+1/3)(1=1/4).....(1+1/99)
\(\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\left(1+\frac{1}{4}\right).....\left(1+\frac{1}{99}\right)\)
\(=\left(\frac{2}{2}+\frac{1}{2}\right)\left(\frac{3}{3}+\frac{1}{3}\right)\left(\frac{4}{4}+\frac{1}{4}\right).....\left(\frac{99}{99}+\frac{1}{99}\right)\)
\(=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}.....\frac{100}{99}\)
\(=\frac{3.4.5....100}{2.3.4....99}=\frac{100}{2}=50\)
Tính nhanh tổng sau:
A=1/3+1/6+1/10+1/15+1/21+1/28
B=1/5.7+1/7.9+1/9.11+....+1/23.25
[ [ 250/7 + 11/4 ] - 40/7 + 1/4 ] : [ 11 + x ] = 3
1/5.7 + 1/7.9 + 1/9.11 + ... + 1/2009.2011 + 1/x = 1/5 . 0,5
Các bạn tính giúp mình với, giấu chấm đó là thay dấu nhân nhé.
Tìm x: [12/11-(1/2+1/44].( x-0,2)=1/1.3+1/5.7+1/7.9+1/9.11
Bạn gõ thiều phân số \(\frac{1}{3.5}\)à?
Tìm x: [12/11-(1/2+1/44].(x-0,2)=1/1.3+1/3.5+1/5.7+1/7.9+1/9.11
Mik giải phía dưới rồi đó. Câu lúc nãy bạn đăng ý
Tìm x: [12/11-(1/2+1/44].(x-0,2)=1/1.3+1/3.5+1/5.7+1/7.9+1/9.11
\(\left[\frac{12}{11}-\left(\frac{1}{2}+\frac{1}{44}\right)\right].\left(x-0,2\right)=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\)
\(\frac{25}{44}.\left(x-0,2\right)=\frac{1}{2}.\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{9.11}\right)\)
\(x-0,2=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{11}\right):\frac{25}{44}\)
\(x-\frac{1}{5}=\frac{22}{25}.\left(1-\frac{1}{11}\right)=\frac{22}{25}.\frac{10}{11}=\frac{4}{5}\)
\(x=\frac{4}{5}+\frac{1}{5}\)
\(x=1\)
