|x+19|+|x+5|+|x+2020|=5x(*)

+)Ta có:|x+19|\(\ge\)0;|x+5|\(\ge\)0;|x+2020|\(\ge\)0

=>VT(*)=|x+19|+|x+5|+|x+2020|\(\ge\)0

Mà |x+19|+|x+5|+|x+2020|=5x

=>5x\(\ge\)0

=>x\(\ge\)0

+)Ta lại có:x\(\ge\)0=>x+19\(\ge\)19=>|x+19|=x+19

                    x\(\ge\)0=>x+5\(\ge\)5=>|x+5|=x+5

                   x\(\ge\)0=>x+2020\(\ge\)2020=>|x+2020|=x+2020

=>VT(*)=x+19+x+5+x+2020=5x

              x+x+x+19+5+2020=5x

             3x+2044        =5x

                  2044        =5x-3x

                  2044        =2x

               => 2x            =2044

                   x             =\(\frac{2044}{2}=1022\)\(\in\)Z

Vậy x=1022

Chúc bn học tốt

cảm ơn bạn nha!!!

Bài 2:

a)|x| < 3

x\(\in\){-2;-1;0;1;2}

b)|x - 4 | < 3

x\(\in\){ 6 ; 5 ; 4 ; 3 ; 2 }

c) | x + 10 | < 2

x\(\in\){ -2 ; -10 }

Bài 1:

A = 1 + 2 - 3 + 4 + 5 - 6 +...+98 - 99

A = (1 + 4 + 7 +...+97) + [(2-3)+(5-6)+...+(98-99)]

A = 1617 + [(-1)+(-1)+...+(-1)]

A = 1617 + (-49)

A = +(1617-49) = A = 1568

B = - 2 - 4 + 6 - 8 + 10 + 12 - .... + 60

B =  

2) 

a) \(x\in\left\{2;1;0;-1;-2\right\}\)

b) \(x\in\left\{6;-6;5;-5;4\right\}\)

c) \(x\in\left\{-9;-11;-10\right\}\)

3)

\(\left(a;b\right)\in\left\{\left(0;1\right);\left(0;-1\right);\left(1;0\right);\left(-1;0\right)\right\}\)

Emma Granger: Cảm ơn bạn nhìu ạ ! :))

Lời giải:
$5x+12\vdots x-2$

$\Rightarrow (5x-10)+22\vdots x-2$

$\Rightarrow 5(x-2)+22\vdots x-2$

$\Rightarrow 22\vdots x-2$

$\Rightarrow x-2\in\left\{1; -1; 2;-2;11;-11;22;-22\right\}$

$\Rightarrow x\in\left\{3; 1; 4; 0; 13; -9; 24; -20\right\}$

bn ... ơi...mik ...bỏ...cuộc ...hu...hu

. Huhu T^T mong sẽ có ai đó giúp mình "((

\(M=5-|x+1|\)

Vì x \(\in\)Z

\(\Rightarrow|x+1|\ge0\)Mà x thuộc z

\(\Rightarrow M=5-|x+1|\ge5\)Mà x thuộc z

\(\Rightarrow M_{max}=5\)

Dấu bằng xảy ra khi : \(|x+1|=0\)

                                  \(\Leftrightarrow x+1=0\)

                                  \(\Leftrightarrow x=1\)

a) Ta có: | x+1|  > 0; dấu “=“ xảy ra khi x=—1

Nên 5–|x+1| > 5

Do đó M > 5

Vậy GTNN của M là 5 khi x=—1

b) Ta có |x+3| > 0; dấu “=“ xảy ra khi x=—3

Nên |x+3| +5 >5

Do đó A > 5

Vậy GTNN của A là 5 khi x=—3

\(x=\dfrac{5}{a-1}< 0\)

\(\Leftrightarrow a-1< 0\Leftrightarrow a< 1\left(1\right)\)

Và \(x=\dfrac{5}{a-1}\in Z\)

\(\Rightarrow a-1\inƯ\left(5\right)=\left\{1;-1;5;-5\right\}\left(2\right)\)

\(\Rightarrow a\left\{2;0;6;-4\right\}\)

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

6   \(n^5+5n=n^5-n+6n=n\left(n^4-1\right)+6n=n\left(n^2-1\right)\left(n^2+1\right)+6n\)

\(=n\left(n-1\right)\left(n+1\right)\left(n^2+1\right)+6n\)

vì n,n-1 là 2 số nguyên lien tiếp  \(\Rightarrow n\left(n-1\right)⋮2\Rightarrow n\left(n-1\right)\left(n+1\right)\left(n^2+1\right)⋮2\)

  n,n-1,n+1 là 3 sô nguyên liên tiếp \(\Rightarrow n\left(n-1\right)\left(n+1\right)⋮3\Rightarrow n\left(n-1\right)\left(n+1\right)\left(n^2+1\right)⋮3\)

\(\Rightarrow n\left(n-1\right)\left(n+1\right)\left(n^2+1\right)⋮2\cdot3=6\)

\(6⋮6\Rightarrow6n⋮6\Rightarrow n\left(n-1\right)\left(n+1\right)\left(n^2+1\right)-6n⋮6\Rightarrow n^5+5n⋮6\)(đpcm)

7   \(n\left(2n+7\right)\left(7n+1\right)=n\left(2n+7\right)\left(7n+7-6\right)=7n\left(n+1\right)\left(2n+7\right)-6n\left(2n+7\right)\)

\(=7n\left(n+1\right)\left(2n+4+3\right)-6n\left(2n+7\right)\)

\(=7n\left(n+1\right)\left(2n+4\right)+21n\left(n+1\right)-6n\left(2n+7\right)\)

\(=14n\left(n+1\right)\left(n+2\right)+21n\left(n+1\right)-6n\left(2n+7\right)\)

n,n+1,n+2 là 3 sô nguyên liên tiếp dựa vào bài 6 \(\Rightarrow n\left(n+1\right)\left(n+2\right)⋮6\Rightarrow14n\left(n+1\right)\left(n+2\right)⋮6\)

\(21⋮3;n\left(n+1\right)⋮2\Rightarrow21n\left(n+1\right)⋮3\cdot2=6\)

\(6⋮6\Rightarrow6n\left(2n+7\right)⋮6\)

\(\Rightarrow14n\left(n+1\right)\left(n+2\right)+21n\left(n+1\right)-6n\left(2n+7\right)⋮6\)

\(\Rightarrow n\left(2n+7\right)\left(7n+1\right)⋮6\)(đpcm)

......................?

mik ko biết

mong bn thông cảm 

nha ................