2: \(\Leftrightarrow x+2\in\left\{1;-1\right\}\)

hay \(x\in\left\{-1;-3\right\}\)

a)Ta có : \(x-5⋮x+2=>x-5-\left(x+2\right)⋮x-2=>-7⋮x-2\)

\(=>x-2\inƯ\left(7\right)\left\{-7;-1;1;7\right\}\)

\(=>x\in\left\{-5;1;3;9\right\}\)

b)Ta có : \(2x+1⋮2x-1=>2x+1-\left(2x-1\right)⋮2x-1=>2⋮2x-1\)

\(=>2x-1\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\)

\(=>2x\in\left\{-1;0;2;3\right\}\)

\(=>x\in\left\{0;1\right\}\)(vì \(x\in Z\))

c)\(\left(x+5\right)-3\left(x+5\right)+2⋮x+5=>2⋮x+5=>x+5\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\)

\(=>x\in\left\{-7;-6;-4;-3\right\}\)

d)\(x+1⋮x+2=>x+2-1⋮x+2\)

\(=>1⋮x+2=>x+2\inƯ\left(1\right)=\left\{1;-1\right\}=>x\in\left\{-1;-3\right\}\)

X+5 

VÌ 5 CHIA HẾT CHO 5

NÊN X+5 CHIA HẾT CHO 5

B, X-18 CHIA HET 6

VÌ 18 CHIA HẾT CHO 6

NÊN X-18 CHIA HEETS CHO 6

C, 21+X CHIA HẾT CHO 7

VÌ 21 CHI HẾT CHO 7\

NÊN 21+X CHIA HÉT CJO 7

K MIK NHA

a) x có dạng: x = 5k ; k ∈ N

b) x có dạng: x = 5k + l; x = 5k+2; x = 5k + 3; x = 5k+4  k ∈ N

dễ mà bạn

là câu b 

( x + 6y ) chia hết cho 5

Sửa đề: \(A=2+2^2+2^3+\cdots+2^{10}+1+5+5^2+\cdots+5^{14}\)

a: Đặt \(B=2+2^2+2^3+\cdots+2^{10}\)

\(=\left(2+2^2+2^3+2^4+2^5\right)+\left(2^6+2^7+2^8+2^9+2^{10}\right)\)

\(=2\left(1+2+2^2+2^3+2^4\right)+2^6\left(1+2+2^2+2^3+2^4\right)\)

\(=31\cdot\left(2+2^6\right)\) ⋮31

Đặt \(C=1+5+5^2+\cdots+5^{14}\)

\(=\left(1+5+5^2\right)+\left(5^3+5^4+5^5\right)+\cdots+\left(5^{12}+5^{13}+5^{14}\right)\)

\(=\left(1+5+5^2\right)+5^3\left(1+5+5^2\right)+\cdots+5^{12}\left(1+5+5^2\right)\)

\(=31\left(1+5^3+\cdots+5^{12}\right)\) ⋮31

Ta có: \(A=2+2^2+2^3+\cdots+2^{10}+1+5+5^2+\cdots+5^{14}\)

=B+C

mà B⋮31 và C⋮31

nên A⋮31

b: Ta có: \(B=2+2^2+2^3+\cdots+2^{10}\)

\(=\left(2+2^2\right)+\left(2^3+2^4\right)+\cdots+\left(2^9+2^{10}\right)\)

\(=\left(2+2^2\right)+2^2\left(2+2^2\right)+\cdots+2^8\left(2+2^2\right)\)

\(=6\left(1+2^2+\cdots+2^8\right)\) ⋮6

Ta có: \(C=1+5+5^2+\cdots+5^{14}\)

\(=1+\left(5+5^2\right)+\left(5^3+5^4\right)+\cdots+\left(5^{13}+5^{14}\right)\)

\(=1+5\left(1+5\right)+5^3\left(1+5\right)+\cdots+5^{13}\left(1+5\right)\)

\(=1+6\left(5+5^3+\cdots+5^{13}\right)\)

=>C chia 6 dư 1

Ta có: A=B+C

\(=6\left(1+2^2+\cdots+2^8\right)+1+6\left(5+5^3+\cdots+5^{13}\right)\)

=>A chia 6 dư 1

\(a,A=\dfrac{9-3x+x^2+10x+25-x^2+1}{\left(x-1\right)\left(x+5\right)}\\ A=\dfrac{7x+35}{\left(x-1\right)\left(x+5\right)}=\dfrac{7\left(x+5\right)}{\left(x-1\right)\left(x+5\right)}=\dfrac{7}{x-1}\\ b,A\in Z\\ \Leftrightarrow x-1\inƯ\left(7\right)=\left\{-7;-1;1;7\right\}\\ \Leftrightarrow x\in\left\{-6;0;2;8\right\}\left(tm\right)\\ b,A< 0\Leftrightarrow x-1< 0\left(7>0\right)\\ \Leftrightarrow x< 1;x\ne-5\\ c,\left|A\right|=3\Leftrightarrow\dfrac{7}{\left|x-1\right|}=3\Leftrightarrow\left|x-1\right|=\dfrac{7}{3}\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}+1=\dfrac{10}{3}\left(tm\right)\\x=-\dfrac{7}{3}+1=-\dfrac{4}{3}\left(tm\right)\end{matrix}\right.\)

1) Ta có x+3=x+1+2

=> 2 chia hết cho x+1

=> x+1 \(\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\)

Ta có bảng

2) Ta có 2x+5=2(x+2)+1

=> 1 chia hết cho x+2

=> x+2 =Ư (1)={-1;1}

Nếu x+2=-1 => x=-3

Nếu x+2=1 => x=-1

3, Ta có 3x+5=3(x-2)+11

=> 11 chia hết cho x-2

=> x-2 thuộc Ư (11)={-11;-1;1;11}

Ta có bảng

4) Ta có x²-x+2=(x-1)²-x

=> x chia hết cho x-1

Ta có x=x-1+1

=> 1 chia hết cho x+1 

=> x+1 thuộc Ư (1)={-1;1}

Nếu x+1=-1 => x=-2

Nếu x+1=1 => x=0

5) Ta có x²+2x+4=(x+2)²-2x

=> 2x chia hết cho x+1

Ta có 2x=2(x+1)-2

=> x+1 thuộc Ư (2)={-2;-1;1;2}

Ta có bảng

a) \(3x+24⋮x-4\)

\(\Rightarrow3x+24-3\left(x-4\right)⋮x-4\)

\(\Rightarrow3x+24-3x+12⋮x-4\)

\(\Rightarrow36⋮x-4\)

\(\Rightarrow x-4\in\left\{-1;1;-2;2;-3;3;-4;4;-9;9;-12;12;-18;18;-36;36\right\}\)

\(\Rightarrow x\in\left\{3;5;2;6;1;7;0;8;-5;13;-8;16;-14;22;-32;40\right\}\left(x\in Z\right)\)

b) \(x^2+5⋮x+1\)

\(\Rightarrow x^2+5-x\left(x+1\right)⋮x+1\)

\(\Rightarrow x^2+5-x^2-x⋮x+1\)

\(\Rightarrow5-x⋮x+1\)

\(\Rightarrow5-x+\left(x+1\right)⋮x+1\)

\(\Rightarrow5-x+x+1⋮x+1\)

\(\Rightarrow6⋮x+1\)

\(\Rightarrow x+1\in\left\{-1;1;-2;2;-3;3;-6;6\right\}\)

\(\Rightarrow x\in\left\{-2;0;-3;1;-4;2;-7;5\right\}\left(x\in Z\right)\)

Bài cuối tương tự bạn tự làm nhé, thanks!