Mk thấy bài 1 và 2 dễ nên bạn tự làm nha

3

+)Ta có n-2 \(⋮\)n-2

=>2.(n-2)\(⋮\)n-2

=>2n-4\(⋮\)n-2(1)

+)Theo bài ta có:2n+1\(⋮\)n-2(2)

+)Từ (1) và (2)

=>(2n+1)-(2n-4)\(⋮\)n-2

=>2n+1-2n+4\(⋮\)n-2

=>5\(⋮\)n-2

=>n-2\(\in\)Ư(5)={\(\pm\)1;\(\pm\)5}

+)Ta có bảng:

Vậy n\(\in\){1;3;-3;7}

Chúc bn học tốt

a. 5.(–8).( –2).(–3)                                                       b. 4.(–5)2 + 2.(–5) – 20

=(-5).8.(-2).(-3)                                                               ={(-5).2} {4+1}-20

=(-5)(-2)(-3).8                                                                 =(-10).5-20=-50-20=-70

=10.(-24)=-240

a, (2n +1) + 40 = 200 

=> 2n + 1 = 160 

=> 2n = 159 

=> n = 159/2

a) (2n + 1) + 40 = 200

    (2n + 1) = 200 - 40

    (2n + 1) = 160

            2n = 160 - 1

             2n = 159

               n = 159/2 = 79,5

A)(2n+1)+40=200

   2n+1=200-40=160

   2n=160-1=159

   n=159/2=79.5

b)200/x+200=100*6

  200/x=600-200=400

  x=400/200=2

a) \(a\left(b+1\right)=3\left(a;b\inℤ\right)\)

\(\Rightarrow a;\left(b+1\right)\in U\left(3\right)=\left\{-1;1;-3;3\right\}\)

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

b) \(2n+7⋮n+1\left(n\inℤ\right)\)

\(\Rightarrow2n+7-2\left(n+1\right)⋮n+1\)

\(\Rightarrow2n+7-2n-2⋮n+1\)

\(\Rightarrow5⋮n+1\)

\(\Rightarrow n+1\in U\left(5\right)=\left\{-1;1;-5;5\right\}\)

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

c) \(xy+x-y=6\left(x;y\inℤ\right)\)

\(\Rightarrow x\left(y+1\right)-y-1+1=6\)

\(\Rightarrow x\left(y+1\right)-\left(y+1\right)=5\)

\(\Rightarrow\left(x-1\right)\left(y+1\right)=5\)

\(\Rightarrow\left(x-1\right);\left(y+1\right)\in U\left(5\right)=\left\{-1;1;-5;5\right\}\)

\(\Rightarrow\left(x;y\right)\in\left\{\left(-0;-6\right);\left(2;4\right);\left(-4;-2\right);\left(6;0\right)\right\}\)

a) ko có a, b thỏa mãn

b) Giá trị lớn nhất của A = \(\frac{7}{6}\)

c) 16

d)  x = \(\frac{14}{3}\)

e) x=-1

g) n= 7

h) 

j) x=1

k) n=11

\(a,x< 50\Leftrightarrow\sqrt{x}-1< 5\sqrt{2}-1\\ M=\dfrac{\sqrt{x}-1}{2}\in Z\\ \Leftrightarrow\sqrt{x}-1\in B\left(2\right)=\left\{0;2;4;6\right\}\\ \Leftrightarrow\sqrt{x}\in\left\{1;3;5;7\right\}\\ \Leftrightarrow x\in\left\{1;9;25;49\right\}\\ b,\Leftrightarrow\sqrt{x}-5\inƯ\left(9\right)=\left\{-3;-1;1;3;9\right\}\left(\sqrt{x}-5>-5\right)\\ \Leftrightarrow\sqrt{x}\in\left\{2;4;6;8;14\right\}\\ \Leftrightarrow x\in\left\{4;16;36;64;196\right\}\)

a) đề  x³+x²-x +a chia hét cho (x-1)² ?

x³+x²-x +a=x(x²-2x+1)+3(x²-2x+1)+4x-3+a đề sai nhé

b)A(2)=0=> 8-12+10+m=0  => m=6

c)2n²-n+2=2n(n+1)-3(n+1) +5 chia het cho n+1 khi n+1 là ước của 5

n+1=-1;1;-5;5

n=-2;0;-6;4