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
16 tháng 5 2018
\(-\frac{123456}{123457}>-\frac{123457}{123455}\)
HH
28 tháng 3 2020
\(a,\frac{6}{10}=\frac{3}{5};\frac{6}{16}=\frac{3}{8};-\frac{15}{20}=\frac{3}{4};-\frac{10}{30}=\frac{-1}{3}\)
\(b,\frac{42}{28}=\frac{3}{2};\frac{54}{-21}=-\frac{10}{7};-\frac{27}{33}=-\frac{9}{11};\frac{25}{14}=\frac{25}{14}\)
\(c,\frac{125}{1000}=\frac{1}{8};\frac{198}{126}=\frac{11}{7};\frac{3}{243}=\frac{1}{81};\frac{103}{2090}=\frac{103}{2090}\)
\(d,\frac{2.3}{9.14}=\frac{28}{3}\)
23 tháng 6 2019
\(a)16^{19}=\left(8\times2\right)^{19}=8^{19}\times2^{19}>8^{19}>8^{15}\)
\(\Rightarrow16^{19}>8^{15}\)
\(b)81^8=\left(3^4\right)^8=3^{24}< 3^{33}=\left(3^3\right)^{11}=27^{11}\)
\(\Rightarrow27^{11}>81^8\)
\(c)625^5=\left(5^4\right)^5=5^{20}< 5^{21}=\left(5^3\right)^7=125^7\)
\(\Rightarrow125^7>625^5\)
\(d)244^{11}>243^{11}=\left(3^5\right)^{11}=3^{55}>3^{52}=\left(3^4\right)^{13}=81^{13}>80^{13}\)
\(\Rightarrow244^{11}>80^{13}\)
\(d)31^{17}>17^{17}>17^{14}\)
\(\Rightarrow31^{17}>17^{14}\)
27 tháng 6 2019
\(a,\frac{1}{2}+\frac{2}{3}x=\frac{4}{5}\)
=> \(\frac{2}{3}x=\frac{4}{5}-\frac{1}{2}=\frac{3}{10}\)
=> \(x=\frac{3}{10}:\frac{2}{3}=\frac{9}{20}\)
Vậy \(x\in\left\{\frac{9}{20}\right\}\)
\(b,x+\frac{1}{4}=\frac{4}{3}\)
=> \(x=\frac{4}{3}-\frac{1}{4}=\frac{13}{12}\)
Vậy \(x\in\left\{\frac{13}{12}\right\}\)
\(c,\frac{3}{5}x-\frac{1}{2}=-\frac{1}{7}\)
=> \(\frac{3}{5}x=-\frac{1}{7}+\frac{1}{2}=\frac{5}{14}\)
=> \(x=\frac{5}{14}:\frac{3}{5}=\frac{25}{42}\)
Vậy \(x\in\left\{\frac{25}{42}\right\}\)
\(d,\left|x+5\right|-6=9\)
=> \(\left|x+5\right|=9+6=15\)
=> \(\left[{}\begin{matrix}x+5=15\\x+5=-15\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=15-5=10\\x=-15-5=-20\end{matrix}\right.\)
Vậy \(x\in\left\{10;-20\right\}\)
\(e,\left|x-\frac{4}{5}\right|=\frac{3}{4}\)
=> \(\left[{}\begin{matrix}x-\frac{4}{5}=\frac{3}{4}\\x-\frac{4}{5}=-\frac{3}{4}\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=\frac{3}{4}+\frac{4}{5}=\frac{31}{20}\\x=-\frac{3}{4}+\frac{4}{5}=\frac{1}{20}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{31}{20};\frac{1}{20}\right\}\)
\(f,\frac{1}{2}-\left|x\right|=\frac{1}{3}\)
=> \(\left|x\right|=\frac{1}{2}-\frac{1}{3}\)
=> \(\left|x\right|=\frac{1}{6}\)
=> \(\left[{}\begin{matrix}x=\frac{1}{6}\\x=-\frac{1}{6}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{1}{6};-\frac{1}{6}\right\}\)
\(g,x^2=16\)
=> \(\left|x\right|=\sqrt{16}=4\)
=> \(\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)
vậy \(x\in\left\{4;-4\right\}\)
\(h,\left(x-\frac{1}{2}\right)^3=\frac{1}{27}\)
=> \(x-\frac{1}{2}=\sqrt[3]{\frac{1}{27}}=\frac{1}{3}\)
=> \(x=\frac{1}{3}+\frac{1}{2}=\frac{5}{6}\)
Vậy \(x\in\left\{\frac{5}{6}\right\}\)
\(i,3^3.x=3^6\)
\(x=3^6:3^3=3^3=27\)
Vậy \(x\in\left\{27\right\}\)
\(J,\frac{1,35}{0,2}=\frac{1,25}{x}\)
=> \(x=\frac{1,25.0,2}{1,35}=\frac{5}{27}\)
Vậy \(x\in\left\{\frac{5}{27}\right\}\)
\(k,1\frac{2}{3}:x=6:0,3\)
=> \(\frac{5}{3}:x=20\)
=> \(x=\frac{5}{3}:20=\frac{1}{12}\)
Vậy \(x\in\left\{\frac{1}{12}\right\}\)
T
22 tháng 3 2018
Sửa lại cái dòng này một tí:
\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{99.200}-\frac{1}{200.201}\)
Còn lại đúng hết! Không cần lo
T
22 tháng 3 2018
a) Đặt \(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{199.200.201}\). Ta xét:
\(\frac{1}{1.2}-\frac{1}{2.3}=\frac{1}{1.2.3}\); \(\frac{1}{2.3}-\frac{1}{3.4}=\frac{1}{2.3.4}\); \(\frac{1}{3.4}-\frac{1}{4.5}=\frac{1}{3.4.5}\);.......;\(\frac{1}{99.200}-\frac{1}{200.201}=\frac{1}{99.100.101}\)
Qua công thức trên ta rút ra tổng quát ( nói thêm cho dễ hiểu)
\(\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}=\frac{2}{n\left(n+1\right)\left(n+2\right)}\)
\(\Rightarrow2A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+....+\frac{2}{199.200.201}\)
\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+....+\frac{1}{199.200.201}\)
Ta thấy: \(-\frac{1}{2.3}+\frac{1}{2.3}=0\);\(-\frac{1}{3.4}+\frac{1}{3.4}=0\); . . . . .
\(\Rightarrow2A=\frac{1}{2}-\frac{1}{200.201}=\frac{1}{2}-\frac{1}{40200}\)
\(\Rightarrow A=\frac{\frac{1}{2}-\frac{1}{40200}}{2}\)
XO
26 tháng 7 2020
a) Ta có\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{110}=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{10.11}\)
\(=2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{10.11}\right)=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{10}-\frac{1}{11}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{11}\right)=1-\frac{2}{11}=\frac{9}{11}\)
b) Ta có \(1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{2048}=1-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{2048}\right)\)(1)
Đặt S = \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}+\frac{1}{2048}\)
=> \(2S=1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{1024}\)
Lấy 2S trừ S ta có :
2S - S \(=\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{1024}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}+\frac{1}{2048}\right)\)
\(S=1-\frac{1}{2048}\)
Khi đó (1) <=> \(1-\left(1-\frac{1}{2048}\right)=1-1+\frac{1}{2048}=\frac{1}{2048}\)
T
26 tháng 7 2020
\(a,\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+\frac{2}{30}+....+\frac{2}{90}+\frac{2}{110}\)
\(=2.\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+.....+\frac{1}{90}+\frac{1}{110}\right)\)
\(=2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+....+\frac{1}{9.10}+\frac{1}{10.11}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-....+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{11}\right)\)
\(=1-\frac{2}{11}\)
\(=\frac{9}{11}\)
TH
6 tháng 7 2018
a/ Ta có: n + 10 \(⋮\) n + 3 ( n \(\in\) Z )
\(\Rightarrow n+3+7⋮n+3\)
\(\Rightarrow\) 7 \(⋮\) n + 3
\(\Rightarrow\) n + 3 \(\in\) Ư(7) = { -1 ; 1 ; -7 ; 7 }
\(\Rightarrow\) n \(\in\) { -4 ; -2 ; -10 ; 4 }
Câu b làm t. tự tách n - 15 thành n + 2 - 17
- 17 \(⋮\) n + 2
Câu c tách 2n - 17 thành 2( n - 3 ) - 11
- 11 \(⋮\) n - 3
d/ Ta có: \(n^2+n+10\) \(⋮\) n + 2 ( n \(\in\) Z )
\(\Leftrightarrow\) n( n + 2 ) - n + 10 \(⋮\) n + 2
\(\Leftrightarrow\) n( n + 2 ) - n + 2 + 8 \(⋮\) n + 2
Vì n( n + 2 ) \(⋮\) n + 2 và ( - n + 2) \(⋮\) n + 2
\(\Rightarrow\) 8 \(⋮\) n + 2
\(\Rightarrow\) n + 2 \(\in\) Ư (8) = { -1 ; 1 ; -2 ; 2 ; -4 ; 4 ; -8 ; 8 }
\(\Rightarrow\) n \(\in\) { -3 ; -1 ; -4 ; 0 ; -6 ; 2 ; -10 ; 6 }
Chúc bạn học tốt!!!
\(=0\)
(-123456) + 123456
= 123456 - 123456
= 0