Vì \(4\le\left|x\right|< 5\)

\(\Rightarrow\left|x\right|=4\)

\(\Rightarrow x=\left\{4;-4\right\}\)

Tổng các số nguyên \(4\le\left|x\right|< 5\) là:

(-4)+4=0

Vậy tổng các số nguyên \(4\le\left|x\right|< 5\) là 0

4

a) 18

b) 0

c) 3916

tick nha

a) x = -4;-3;-2;-1;0;1;2;3;4;5;6;7)

tổng : -4+-3+-2+-1+0+1+2+3+4+5+6+7

          = (-4+ 4)+(-3+3)+(-2+2)+(-1+1)+0+5+6+7

          = 0+0+0+0+0+5+6+7

          = 11+7=18

b)x= -11;-10;-9;-8;....;8;9;10;11

  tổng :  -11+-10+-9+-8+...+8+9+10+11

            = (-11+11)+(-10+10)+(-9+9)+(-8+8)+(-7+7)+(-6+6)+(-5+5)+...+0

             = 0+0+0+.....+0

             = 0

c) làm tương tự như trên

Bài 1b:

\(\frac31\) + \(\frac33\) + \(\frac36\) + \(\frac{3}{10}\) + ...+\(\frac{3}{x\left(x+1\right):2}\) = \(\frac{2015}{336}\)

3.(\(\frac11+\frac13+\frac16+\frac{1}{10}+\cdots+\frac{1}{x\left(x+1):2\right.})\) = \(\frac{2015}{336}\)

3.2(\(\frac12+\frac16+\frac{1}{12}+\frac{1}{20}+\cdots+\frac{1}{x\left(x+1\right)})=\) \(\frac{2015}{336}\)

6.(\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\cdots+\frac{1}{x.\left(x+1\right)})\) = \(\frac{2015}{336}\)

6.(\(\frac11-\frac12\) + \(\frac12\)-\(\frac14\) +...+ \(\frac{1}{x}\) - \(\frac{1}{x+1}\)) = \(\frac{2015}{336}\)

6.(\(\frac11\) - \(\frac{1}{x+1}\)) = \(\frac{2015}{336}\)

1 - \(\frac{1}{x+1}\) = \(\frac{2015}{336}\) : 6

1 - \(\frac{1}{x+1}\) = \(\frac{2015}{2016}\)

\(\frac{1}{x+1}\) = 1 - \(\frac{2015}{2016}\)

\(\frac{1}{x+1}\) = \(\frac{1}{2016}\)

\(x+1\) = 2016

\(x\) = 2016 - 1

\(x\) = 2015

Bài 2:

A = \(\frac{6n+1}{4n+3}\) (n ∈ Z\(^{-}\))

A ∈ Z khi và chỉ khi:

(6n + 1) ⋮ (4n + 3)

(12n + 2) ⋮ (4n + 3)

[3(4n + 3) - 7] ⋮ (4n + 3)

7 ⋮ (4n + 3)

(4n + 3) ∈ Ư(7) = {-7; -1; 1; 7}

n ∈ {- 5/2; -1; - 1/2; 1}

Nếu n = - 1 thì A = (-6 + 1)/(-4 + 3) = 5 (loại)

Nếu n = 1 thì: A = (6 + 1).(4+3) = 1 (loại)

Không có giá trị nào thỏa mãn đề bài hay n ∈ ∅

a. 32 = 2⁵ => n thuộc tập 1; 2; 3; 4

b. \(\left(\frac{1}{x}-\frac{2}{3}\right)^2=\frac{1}{16}\)

\(\Rightarrow\frac{1}{x}-\frac{2}{3}=\frac{1}{4}\)

\(\Rightarrow\frac{1}{x}=\frac{1}{4}+\frac{2}{3}=\frac{11}{12}\)

\(\Rightarrow x=\frac{12}{11}\)

c. p nguyên tố => \(p\ge2\) => 5^2p luôn có dạng A25

=> 5^2p+2015 chẵn

=> 2014^2p + q³ chẵn

Mà 2014^2p chẵn => q³ chẵn => q chẵn => q = 2

=> 5^2p + 2015 = 2014^2p+8

=> 5^2p+2007 = 2014^2p

2014 có mũ dạng 2p => 2014^2p có dạng B6

=> 5^2p = B6 - 2007 = ...9 (vl)

(hihi câu này hơi sợ sai)

d. \(17A=\frac{17^{19}+17}{17^{19}+1}=1+\frac{16}{17^{19}+1}\), \(17B=\frac{17^{18}+17}{17^{18}+1}=1+\frac{16}{17^{18}+1}\)

\(17^{19}+1>17^{18}+1\Rightarrow\frac{16}{17^{19}+1}< \frac{16}{17^{18}+1}\)

\(\Rightarrow17A< 17B\)

\(\Rightarrow A< B\)

de thi chon hoc sinh gioi nay

a) \(x\in\left\{-1;0;1;2;3;4;\right\}\)

b) \(x\in\left\{-6;-5;-4;-3;-2;-1\right\}\)

c) \(x\in\left\{1;2;3;4;5;6;7\right\}\)

d) \(x\in\left\{-1;0;1;2;3;4;5\right\}\)

a, \(x\in\left\{-1;0;1;2;3;4\right\}\)

b, \(x\in\left\{-6;-5;-4;-3;-2;-1\right\}\)

c, \(x\in\left\{1;2;3;4;5;6;7\right\}\)

d, \(x\in\left\{\pm1;0;2;3;4;5\right\}\)