Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(S=\dfrac{2}{10\cdot12}+\dfrac{2}{12\cdot14}+...+\dfrac{2}{98\cdot100}\)
\(S=\dfrac{2}{10}-\dfrac{2}{12}+\dfrac{2}{12}-\dfrac{2}{14}+...+\dfrac{2}{98}-\dfrac{2}{100}\)
\(S=\dfrac{2}{10}-\dfrac{2}{100}=\dfrac{9}{50}=0,18\)
Vậy \(S>\dfrac{1}{10}\)
\(S=\dfrac{2}{10\cdot12}+\dfrac{2}{12\cdot14}+\dfrac{2}{14\cdot16}+...+\dfrac{2}{98\cdot100}\)
\(S=\dfrac{2}{10}-\dfrac{2}{12}+\dfrac{2}{12}-\dfrac{2}{14}+...+\dfrac{2}{98}-\dfrac{2}{100}\)
\(S=\dfrac{2}{10}-\dfrac{2}{100}\)
\(S=\dfrac{20}{100}-\dfrac{2}{100}\)
\(S=\dfrac{18}{100}=\dfrac{9}{50}=0,18\)
\(\dfrac{1}{10}=0,1\), mà \(0,1< 0,18\)
\(\Rightarrow S>\dfrac{1}{10}\left(đpcm\right)\)
Đặt tổng trên là A ta có
\(2A=\frac{2}{10.12}+\frac{2}{12.14}+\frac{2}{14.16}+...+\frac{2}{48.52}\)
\(2A=\frac{12-10}{10.12}+\frac{14-12}{12.14}+\frac{16-14}{14.16}+...+\frac{50-48}{48.50}\)
\(2A=\frac{1}{10}-\frac{1}{12}+\frac{1}{12}-\frac{1}{14}+\frac{1}{14}-\frac{1}{16}+...+\frac{1}{48}-\frac{1}{50}=\frac{1}{10}-\frac{1}{50}=\frac{2}{25}\)
\(\Rightarrow A=\frac{2A}{2}=\frac{1}{25}\)
Ta có : \(\frac{n+3}{n-3}=\frac{n-3+6}{n-3}=\frac{n-3}{n-3}+\frac{6}{n-3}=1+\frac{6}{n-3}\)( có giá trị là số tự nhiên )
Mà \(1\in N\Leftrightarrow\frac{6}{n-3}\in N\)
\(\Leftrightarrow\left(n-3\right)\inƯ_6=\left\{1;2;3;6\right\}\)
\(\Rightarrow n=\left\{4;5;6;9\right\}\)
Làm luôn nha không ghi đề xin lỗi
a)A=\(\dfrac{2.n+1+3.n+5-4.n+5}{n-3}\)
A=\(\dfrac{5.n+6-4.n+5}{n-3}\)
A=\(\dfrac{n+1}{n-3}\)
A=\(\dfrac{n-3+4}{n-3}\)
A=\(\dfrac{n-3}{n-3}\)+\(\dfrac{4}{n-3}\)
A=1+\(\dfrac{4}{n-3}\)
Để A nguyên thì 4\(⋮\)n-3 hay n-3\(\in\)Ư(4).Ta có bảng sau:
| n-3 | 1 | 2 | 4 | -1 | -2 | -4 |
| n | 4 | 5 | 7 | 2 | 1 |
-1 |
Vậy n\(\in\){ 4;5;7;2;1;-1)
b)Ta có A=\(\dfrac{n+1}{n-3}\)
Gọi ước nguyên tố của n+1 và n-3 là d
Ta có n+1\(⋮\)d
n+3\(⋮\)d
\(\Rightarrow\)n-3-n-1\(⋮\)d
\(\Rightarrow\)4\(⋮\)d
Vì d là ước nguyên tố nên d=2
Ta có n+1\(⋮\)d
n-3\(⋮\)d
\(\Rightarrow\)n+1-2\(⋮\)d
n-1\(⋮\)2
\(\Rightarrow\)n=2.k+1
Vậy n\(\ne\)2.k+1 hay n là số chẵn thì A là phân số tối giản
Đặt \(A=\frac{2}{10\cdot12}+\frac{2}{12\cdot14}+\frac{2}{14\cdot16}+...+\frac{2}{48\cdot50}\)
\(A=\frac{1}{10}-\frac{1}{12}+\frac{1}{12}-\frac{1}{14}+...+\frac{1}{48}-\frac{1}{50}\)
\(A=\frac{1}{10}-\frac{1}{50}=\frac{5}{50}-\frac{1}{50}=\frac{4}{50}=\frac{2}{25}\)
Vậy \(A=\frac{2}{25}\)
= \(\frac{1}{10}-\frac{1}{12}+\frac{1}{12}-\frac{1}{14}+...+\frac{1}{48}-\frac{1}{50}\)
= \(\frac{1}{10}-\frac{1}{50}\)= \(\frac{2}{25}\)
3. Gọi d là ƯCLN(2n + 3, 4n + 8), d ∈ N*
\(\Rightarrow\hept{\begin{cases}2n+3⋮d\\4n+8⋮d\end{cases}\Rightarrow\hept{\begin{cases}2\left(2n+3\right)⋮d\\4n+8⋮d\end{cases}\Rightarrow}\hept{\begin{cases}4n+6⋮d\\4n+8⋮d\end{cases}}}\)
\(\Rightarrow\left(4n+8\right)-\left(4n+6\right)⋮d\)
\(\Rightarrow2⋮d\)
\(\Rightarrow d\in\left\{1;2\right\}\)
Mà 2n + 3 không chia hết cho 2
\(\Rightarrow d=1\)
\(\RightarrowƯCLN\left(2n+3,4n+8\right)=1\)
\(\Rightarrow\frac{2n+3}{4n+8}\) là phân số tối giản.
\(1)\dfrac{1}{5}+\dfrac{2}{30}+\dfrac{121}{156}\le x\le\dfrac{1}{2}+\dfrac{156}{72}+\dfrac{1}{3}\)
\(\dfrac{156}{780}+\dfrac{26}{780}+\dfrac{605}{780}\le x\le\dfrac{3}{6}+\dfrac{13}{6}+\dfrac{2}{6}\)
\(\dfrac{787}{780}\le x\le2\)
\(\Rightarrow x\in\left\{2\right\}\)
Câu 2:
\(N=\dfrac{2a+9+5a+17-3a-4a-23}{a+3}=\dfrac{3}{a+3}\)
Để N là số tự nhiên thì \(\left\{{}\begin{matrix}a>-3\\a+3\in\left\{1;-1;3;-3\right\}\end{matrix}\right.\Leftrightarrow a\in\left\{-2;0\right\}\)
\(A=11\dfrac{3}{13}-\left(2\dfrac{4}{7}+5\dfrac{3}{13}\right)\)
\(A=11\dfrac{3}{13}-5\dfrac{3}{13}-2\dfrac{4}{7}\)
\(A=6-2\dfrac{4}{7}\)
\(A=5\dfrac{7}{7}-2\dfrac{4}{7}\)
\(A=3\dfrac{3}{7}\)
\(B=\left(6\dfrac{4}{9}+3\dfrac{7}{11}\right)-4\dfrac{4}{9}\)
\(B=\left(6\dfrac{4}{9}-4\dfrac{4}{9}\right)+3\dfrac{7}{11}\)
\(B=2+3\dfrac{7}{11}\)
\(B=5\dfrac{7}{11}\)
\(C=\dfrac{-5}{7}.\dfrac{2}{11}+\dfrac{-5}{7}-\dfrac{9}{11}+1\dfrac{5}{7}\)
\(C=\dfrac{-5}{7}.\left(\dfrac{2}{11}+1\right)-\dfrac{9}{11}+1\dfrac{5}{7}\)
\(C=\dfrac{-5}{7}.\dfrac{13}{11}-\dfrac{9}{11}+1\dfrac{5}{7}\)
\(C=\dfrac{-65}{77}-\dfrac{9}{11}+1\dfrac{5}{7}\)
\(C=\dfrac{4}{11}+1\dfrac{5}{7}\)
\(C=\dfrac{160}{11}\)
\(D=0,7.2\dfrac{2}{3}.20.0,375.\dfrac{5}{28}\)
\(D=\dfrac{7}{10}.\dfrac{8}{3}.20.\dfrac{375}{1000}.\dfrac{5}{28}\)
\(D=\dfrac{7}{28}=\dfrac{5}{2}\)
\(E=\left(-6,17+3\dfrac{5}{9}-2\dfrac{36}{97}\right)\left(\dfrac{1}{3}-0,25-\dfrac{1}{12}\right)\)
\(E=\left(-6,17+3\dfrac{5}{9}-2\dfrac{36}{97}\right)\left(\dfrac{1}{3}-\dfrac{1}{4}-\dfrac{1}{12}\right)\)
\(E=\left(-6,17+3\dfrac{5}{9}-2\dfrac{36}{97}\right)\left(\dfrac{1}{12}-\dfrac{1}{12}\right)\)
\(E=\left(-6,17+3\dfrac{5}{9}-2\dfrac{36}{97}\right).0\)
\(\Rightarrow E=0\)
Bài 1:
a) \(\dfrac{2}{5}\cdot x-\dfrac{1}{4}=\dfrac{1}{10}\)
\(\dfrac{2}{5}\cdot x=\dfrac{1}{10}+\dfrac{1}{4}\)
\(\dfrac{2}{5}\cdot x=\dfrac{7}{20}\)
\(x=\dfrac{7}{20}:\dfrac{2}{5}\)
\(x=\dfrac{7}{8}\)
Vậy \(x=\dfrac{7}{8}\).
b) \(\dfrac{3}{5}=\dfrac{24}{x}\)
\(x=\dfrac{5\cdot24}{3}\)
\(x=40\)
Vậy \(x=40\).
c) \(\left(2x-3\right)^2=16\)
\(\left(2x-3\right)^2=4^2\)
\(\circledast\)TH1: \(2x-3=4\\ 2x=4+3\\ 2x=7\\ x=\dfrac{7}{2}\)
\(\circledast\)TH2: \(2x-3=-4\\ 2x=-4+3\\ 2x=-1\\ x=\dfrac{-1}{2}\)
Vậy \(x\in\left\{\dfrac{7}{2};\dfrac{-1}{2}\right\}\).
Bài 2:
a) \(25\%-4\dfrac{2}{5}+0.3:\dfrac{6}{5}\)
\(=\dfrac{1}{4}-\dfrac{22}{5}+\dfrac{3}{10}:\dfrac{6}{5}\)
\(=\dfrac{1}{4}-\dfrac{22}{5}+\dfrac{3}{10}\cdot\dfrac{5}{6}\)
\(=\dfrac{1}{4}-\dfrac{22}{5}+\dfrac{1}{4}\)
\(=\dfrac{5}{20}-\dfrac{88}{20}+\dfrac{5}{20}\)
\(=\dfrac{5-88+5}{20}\)
\(=\dfrac{78}{20}=\dfrac{39}{10}\)
b) \(\left(\dfrac{1}{6}-\dfrac{1}{5^2}\cdot5+\dfrac{1}{30}\right)\left(\dfrac{2011}{2010}+\dfrac{2010}{1009}+\dfrac{2009}{2008}\right)\)
\(=\left(\dfrac{1}{6}-\dfrac{1}{25}\cdot5+\dfrac{1}{30}\right)\left(\dfrac{2011}{2010}+\dfrac{2010}{1009}+\dfrac{2009}{2008}\right)\)
\(=\left(\dfrac{1}{6}-\dfrac{1}{5}+\dfrac{1}{30}\right)\left(\dfrac{2011}{2010}+\dfrac{2010}{1009}+\dfrac{2009}{2008}\right)\)
\(=\left(\dfrac{5}{30}-\dfrac{6}{30}+\dfrac{1}{30}\right)\left(\dfrac{2011}{2010}+\dfrac{2010}{1009}+\dfrac{2009}{2008}\right)\)
\(=\left(\dfrac{5-6+1}{30}\right)\left(\dfrac{2011}{2010}+\dfrac{2010}{1009}+\dfrac{2009}{2008}\right)\)
\(=0\cdot\left(\dfrac{2011}{2010}+\dfrac{2010}{1009}+\dfrac{2009}{2008}\right)\)
\(=0\)
Bài 3:
a) \(\dfrac{4}{19}\cdot\dfrac{-3}{7}+\dfrac{-3}{7}\cdot\dfrac{15}{19}\)
\(=\dfrac{-3}{7}\left(\dfrac{4}{19}+\dfrac{15}{19}\right)\)
\(=\dfrac{-3}{7}\cdot1\)
\(=\dfrac{-3}{7}\)
b) \(7\dfrac{5}{9}-\left(2\dfrac{3}{4}+3\dfrac{5}{9}\right)\)
\(=\dfrac{68}{9}-\dfrac{11}{4}-\dfrac{32}{9}\)
\(=\dfrac{68}{9}-\dfrac{32}{9}-\dfrac{11}{4}\)
\(=4-\dfrac{11}{4}\)
\(=\dfrac{16}{4}-\dfrac{11}{4}\)
\(\dfrac{5}{4}\)
Bài 4:
\(\dfrac{4}{12\cdot14}+\dfrac{4}{14\cdot16}+\dfrac{4}{16\cdot18}+...+\dfrac{4}{58\cdot60}\)
\(=2\left(\dfrac{1}{12\cdot14}+\dfrac{1}{14\cdot16}+\dfrac{1}{16\cdot18}+...+\dfrac{1}{58\cdot60}\right)\)
\(=2\left(\dfrac{1}{12}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{18}+...+\dfrac{1}{58}-\dfrac{1}{60}\right)\)
\(=2\left(\dfrac{1}{12}-\dfrac{1}{60}\right)\)
\(=2\left(\dfrac{5}{60}-\dfrac{1}{60}\right)\)
\(=2\cdot\dfrac{1}{15}\)
\(=\dfrac{2}{15}\)
(1) Để \(\dfrac{2n}{n-2}\) là số nguyên thì 2n⋮n-2
2n-4+4⋮n-2
2n-4⋮n-2⇒4⋮n-2
n-2∈Ư(4)⇒Ư(4)={1;-1;2;-2;4;-4}
n∈{3;1;4;0;6;-2}
(2) \(\dfrac{3}{10.12}+\dfrac{3}{12.14}+...+\dfrac{3}{48.50}\)
=\(\dfrac{3}{2}.\left(\dfrac{2}{10.12}+\dfrac{2}{12.14}+...+\dfrac{2}{48.50}\right)\)
=\(\dfrac{3}{2}.\left(\dfrac{1}{10}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{14}+...+\dfrac{1}{48}-\dfrac{1}{50}\right)\)
=\(\dfrac{3}{2}.\left(\dfrac{1}{10}-\dfrac{1}{50}\right)\)
=\(\dfrac{3}{2}.\dfrac{2}{25}\)
=\(\dfrac{3}{25}\)
Giải:
(1) Để \(\dfrac{2n}{n-2}\) là số nguyên thì \(2n⋮n-2\)
\(2n⋮n-2\)
\(\Rightarrow2n-4+4⋮n-2\)
\(\Rightarrow4⋮n-2\)
\(\Rightarrow n-2\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)
Vậy \(n\in\left\{0;1;3;4;6\right\}\)
(2) \(\dfrac{3}{10.12}+\dfrac{3}{12.14}+\dfrac{3}{14.16}+...+\dfrac{3}{48.50}\)
\(=\dfrac{3}{2}.\left(\dfrac{2}{10.12}+\dfrac{2}{12.14}+\dfrac{2}{14.16}+...+\dfrac{2}{48.50}\right)\)
\(=\dfrac{3}{2}.\left(\dfrac{1}{10}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{16}+...+\dfrac{1}{48}-\dfrac{1}{50}\right)\)
\(=\dfrac{3}{2}.\left(\dfrac{1}{10}-\dfrac{1}{50}\right)\)
\(=\dfrac{3}{2}.\dfrac{2}{25}\)
\(=\dfrac{3}{25}\)
Chúc bạn học tốt!