Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Đặt \(A=\frac{1}{31}+\frac{1}{32}+...+\frac{1}{90}\)
\(=\left(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{45}\right)+\left(\frac{1}{46}+\frac{1}{47}+...+\frac{1}{90}\right)\)
Đặt \(B=\frac{1}{31}+\frac{1}{32}+...+\frac{1}{45}\)
Ta có: \(\frac{1}{31}>\frac{1}{45}\)
\(\frac{1}{32}>\frac{1}{45}\)
....................
\(\frac{1}{45}=\frac{1}{45}\)
\(\Rightarrow B>\frac{1}{45}.15\)
\(\Rightarrow B>\frac{1}{3}\)
Đặt \(C=\frac{1}{46}+\frac{1}{47}+...+\frac{1}{90}\)
Ta có: \(\frac{1}{46}>\frac{1}{90}\)
\(\frac{1}{47}>\frac{1}{90}\)
.....................
\(\frac{1}{90}=\frac{1}{90}\)
\(\Rightarrow C>\frac{1}{90}.45\)
\(\Rightarrow C>\frac{1}{2}\)
\(\Rightarrow B+C>\frac{1}{3}+\frac{1}{2}\)
Hay \(A>\frac{5}{6}\left(1\right)\)
Lại có: \(A=\left(\frac{1}{31}+...+\frac{1}{59}\right)+\left(\frac{1}{60}+...+\frac{1}{90}\right)\)
Đặt \(D=\frac{1}{31}+...+\frac{1}{59}\)
Ta có: \(\frac{1}{31}< \frac{1}{30}\)
. ...................
\(\frac{1}{59}< \frac{1}{30}\)
\(\Rightarrow D< \frac{1}{30}.60\)
\(\Rightarrow D< \frac{1}{2}\)
Đăt \(E=\frac{1}{60}+...+\frac{1}{90}\)
Ta có: \(\frac{1}{60}=\frac{1}{60}\)
.................
\(\frac{1}{90}< \frac{1}{60}\)
\(\Rightarrow E< \frac{1}{60}.31\)
\(\Rightarrow E< \frac{31}{60}< 1\)
\(\Rightarrow E< 1\)
\(\Rightarrow E+D< 1+\frac{1}{2}\)
Hay \(A< \frac{3}{2}\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\frac{5}{6}< A< \frac{3}{2}\)
\(a)\) \(\frac{-11}{12}< \frac{x}{12}< \frac{-3}{4}\)
\(\Leftrightarrow\)\(\frac{-11}{12}< \frac{x}{12}< \frac{-9}{12}\)
\(\Leftrightarrow\)\(-11< x< -9\)
\(\Rightarrow\)\(x=-10\)
\(A=\frac{1}{31}+\frac{1}{32}+...+\frac{1}{60}=\left(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{40}\right)+\left(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{50}\right)+\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{60}\right)\)
Ta có: \(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{40}< \frac{1}{30}+\frac{1}{30}+...+\frac{1}{30}=\frac{10}{30}=\frac{1}{3}\)
\(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{50}< \frac{1}{40}+\frac{1}{40}+...+\frac{1}{40}=\frac{10}{40}=\frac{1}{4}\)
\(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{60}< \frac{1}{50}+\frac{1}{50}+...+\frac{1}{50}=\frac{10}{50}=\frac{1}{5}\)
Do đó \(A< \frac{1}{3}+\frac{1}{4}+\frac{1}{5}=\frac{47}{60}< \frac{48}{60}=\frac{4}{5}\)
Vậy \(A< \frac{4}{5}\)
<br class="Apple-interchange-newline"><div id="inner-editor"></div>141 +142 +...+150 <140 +140 +...+140 =1040 =14
151 +152 +...+160 <150 +150 +...+150 =1050 =15
Do đó A<13 +14 +15 =4760 <4860 =45
Vậy A<45
\(\frac{2}{7}< \frac{x}{3}< \frac{11}{4};x\inℕ\)
=>\(\frac{12.2}{84}< \frac{28x}{84}< \frac{11.21}{84}\)
=>\(\frac{24}{84}< \frac{28x}{84}< \frac{231}{84}\)
=>24<28x<231
=>28x\(\in\){25;26;27;28;.............................;230}
=>Các số chia hết cho 28 là:28;56;84;112;140;168;196;224
=>x (thỏa mãn)\(\in\){1;2;3;4;5;6;7;8}
Vậy x\(\in\) {1;2;3;4;5;6;7;8}
\(\left(4,5m-\frac{3}{4}.5\frac{1}{3}\right).\frac{1}{12}+\frac{1}{2}x=1\frac{1}{2}\)
\(\left(4,5m-\frac{3}{4}.\frac{16}{3}\right).\frac{1}{2}.\frac{1}{6}+\frac{1}{2}x=\frac{3}{2}\)
\(\left(4,5m-\frac{48}{12}\right).\frac{1}{2}.\left(\frac{1}{6}+x\right)=\frac{3}{2}\)
\(\left(4,5m-4\right).\left(\frac{1}{6}+x\right)=\frac{3}{2}:\frac{1}{2}\)
\(\left(4,5m-4\right).\left(\frac{1}{6}+x\right)=\frac{3}{2}.\frac{2}{1}\)
\(\left(4,5m-4\right).\left(\frac{1}{6}+x\right)=\frac{6}{2}\)
\(\left(4,5m-4\right).\left(\frac{1}{6}+x\right)=3\)
=>3\(⋮\)\(\frac{1}{6}+x\)
=>\(\frac{1}{6}+x\)\(\in\)Ư(3)={\(\pm\)1;\(\pm\)3}
Ta có bảng:
| \(\frac{1}{6}+x\) | -1 | 1 | -3 | 3 |
| x | \(-1\frac{1}{6}\) | \(1\frac{1}{6}\) | \(-3\frac{1}{6}\) | 3\(\frac{1}{6}\) |
Vậy x\(\in\){\(-1\frac{1}{6}\);\(1\frac{1}{6}\);\(-3\frac{1}{6}\);\(\frac{1}{6}\)}
Chúc bn học tốt
Ta có : \(A=\frac{1}{5^2}+\frac{2}{5^3}+...+\frac{n}{5^{n+1}}+...+\frac{11}{5^{12}}\)
=> \(5A=\frac{1}{5}+\frac{2}{5^2}+...+\frac{n}{5^n}+...+\frac{11}{5^{11}}\)
Lấy 5A trừ A theo vế ta có :
5A - A = \(\left(\frac{1}{5}+\frac{2}{5^2}+...+\frac{n}{5^n}+...+\frac{11}{5^{11}}\right)-\left(\frac{1}{5^2}+\frac{2}{5^3}+...+\frac{n}{5^{n+1}}+...+\frac{11}{5^{12}}\right)\)
4A = \(\left(\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{11}}\right)-\frac{11}{5^{12}}\)
Đặt B = \(\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{11}}\)
=> 5B = \(1+\frac{1}{5}+...+\frac{1}{5^{10}}\)
Lấy 5B trừ B ta có :
=> 5B - B = \(\left(1+\frac{1}{5}+...+\frac{1}{5^{10}}\right)-\left(\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{11}}\right)\)
=> 4B =\(1-\frac{1}{5^{11}}\)
=> B = \(\frac{1}{4}-\frac{1}{5^{11}.4}\)
Khi đó 4A = \(\frac{1}{4}-\frac{1}{5^{11}.4}-\frac{1}{5^{12}}\)
=> A = \(\frac{1}{16}-\left(\frac{1}{5^{11}.16}+\frac{1}{5^{12}.4}\right)< \frac{1}{16}\left(\text{ĐPCM}\right)\)
cậu ơi , mình quên không ghi 1 dữ liệu ạ
n thuộc N
V ậy có cần phải chỉnh sửa ở trong bài làm không ạ?????
Bài 1 :
Ta có :
\(A=\frac{10^{17}+1}{10^{18}+1}=\frac{\left(10^{17}+1\right).10}{\left(10^{18}+1\right).10}=\frac{10^{18}+10}{10^{19}+10}\)
Mà : \(\frac{10^{18}+10}{10^{19}+10}>\frac{10^{18}+1}{10^{19}+1}\)
Mà \(A=\frac{10^{18}+10}{10^{19}+10}\)nên \(A>B\)
Vậy \(A>B\)
Bài 2 :
Ta có :
\(S=\frac{2013}{2014}+\frac{2014}{2015}+\frac{2015}{2016}+\frac{2016}{2013}\)
\(\Rightarrow S=\frac{2014-1}{2014}+\frac{2015-1}{2015}+\frac{2016-1}{2016}+\frac{2013+3}{2013}\)
\(\Rightarrow S=1-\frac{1}{2014}+1-\frac{1}{2015}+1-\frac{1}{2016}+1+\frac{3}{2013}\)
\(\Rightarrow S=4+\frac{3}{2013}-\left(\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}\right)\)
Vì \(\frac{1}{2013}>\frac{1}{2014}>\frac{1}{2015}>\frac{1}{2016}\)nên \(\frac{3}{2013}-\left(\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}\right)>0\)
Nên : \(M>4\)
Vậy \(M>4\)
Bài 3 :
Ta có :
\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+.......+\frac{1}{100^2}\)
Suy ra : \(A< \frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+....+\frac{1}{99.101}\)
\(\Rightarrow A< \frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{2.4}+......+\frac{2}{99.101}\right)\)
\(\Rightarrow A< \frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-......-\frac{1}{101}\right)\)
\(\Rightarrow A< \frac{1}{2}.\left[\left(1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{99}\right)-\left(\frac{1}{3}+\frac{1}{4}+......+\frac{1}{101}\right)\right]\)
\(\Rightarrow A< \frac{1}{2}.\left(1+\frac{1}{2}-\frac{1}{100}-\frac{1}{101}\right)\)
\(\Rightarrow A< \frac{1}{2}.\left(1+\frac{1}{2}\right)\)
\(\Rightarrow A< \frac{3}{4}\)
Vậy \(A< \frac{3}{4}\)
Bài 4 :
\(a)A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{2015.2017}\)
\(\Rightarrow A=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{1}{2015.2017}\right)\)
\(\Rightarrow A=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.....+\frac{1}{2015}-\frac{1}{2017}\right)\)
\(\Rightarrow A=\frac{1}{2}.\left(1-\frac{1}{2017}\right)\)
\(\Rightarrow A=\frac{1}{2}.\frac{2016}{2017}\)
\(\Rightarrow A=\frac{1008}{2017}\)
Vậy \(A=\frac{1008}{2017}\)
\(b)\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+......+\frac{1}{x\left(x+2\right)}=\frac{1008}{2017}\)
\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+......+\frac{2}{x.\left(x+2\right)}=\frac{2016}{2017}\)
\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.....+\frac{1}{x}-\frac{1}{x+2}=\frac{2016}{2017}\)
\(1-\frac{1}{x+2}=\frac{2016}{2017}\)
\(\Rightarrow\frac{1}{x+2}=1-\frac{2016}{2017}\)
\(\Rightarrow\frac{1}{x+2}=\frac{1}{2017}\)
\(\Rightarrow x+2=2017\)
\(\Rightarrow x=2017-2=2015\)
Vậy \(x=2015\)
A = 1/2 + 1/3 + 1/4 + ... + 1/31
A = (1/2 + 1/3) + (1/4 + 1/5 + 1/6 + 1/7) + (1/8 + 1/9 + ... + 1/15) + (1/16 + 1/17 + ... + 1/31)
A < 1/2 × 2 + 1/4 × 4 + 1/8 × 8 + 1/16 × 16
A < 1 + 1 + 1 + 1
A < 4 ( đpcm)