So sánh hay chứng minh vậy bạn

a) \(6-2x-8=-4\Leftrightarrow2x=2\Leftrightarrow x=1\)

b) \(6⋮x\Rightarrow x=\left\{-6;-3;-2;-1;1;2;3;6\right\}\)

c) \(2x-3=-7\Leftrightarrow2x=-4\Leftrightarrow x=-2\)

d) \(231-x+6=103\Leftrightarrow x=124\)

e) hình như sai đề

f) \(x^3=27\Leftrightarrow x=3\)

g) \(x+2=3\Leftrightarrow x=1\)

h) \(\left(x+5\right)^2=25\Leftrightarrow x+5=5\Leftrightarrow x=0\)

i) \(x^2=9\Leftrightarrow\orbr{\begin{cases}x=-3\\x=3\end{cases}}\)

j) \(14⋮\left(x+3\right)\Rightarrow x=\left\{-17;-10;-5;-4;-2;4;11\right\}\)

dài thế này bố nó cũng trả lời được

nghĩ sao cho dài vậy

Với tất cả các câu, mk chỉ làm ngắn gọn. Nếu bn muốn đầy đủ, thì bn tự lập bảng rồi xét.

1. \(13⋮\left(x-3\right)\)

\(\Leftrightarrow\left(x-3\right)\inƯ\left(13\right)=\left\{\pm1;\pm13\right\}\)

\(\Rightarrow x\in\left\{2;4;-10;16\right\}\)

Vậy x = ......................

2. \(\left(x+13\right)⋮\left(x-4\right)\)

\(\Leftrightarrow\left(x-4\right)+17⋮\left(x-4\right)\)

\(\Leftrightarrow17⋮x-4\)

\(\Leftrightarrow\left(x-4\right)\inƯ\left(17\right)=\left\{\pm1;\pm17\right\}\)

\(\Rightarrow x\in\left\{3;5;-13;21\right\}\)

Vậy x = ...................

3. \(\left(2x+108\right)⋮\left(2x+3\right)\)

\(\Leftrightarrow\left(2x+3\right)+105⋮\left(2x+3\right)\)

\(\Leftrightarrow105⋮\left(2x+3\right)\)

\(\Leftrightarrow\left(2x+3\right)\inƯ\left(105\right)\)\(=\left\{\pm1;\pm3;\pm5;\pm7;\pm15;\pm21;\pm35;\pm105\right\}\)

\(\Rightarrow x=-2;-1;-3;0;-4;1;-5;2;...............\)

4. \(17x⋮15\)

\(\Leftrightarrow x⋮15\) ( vì \(\left(15,17\right)=1\) )

Do đó : Với mọi x thuộc Z thì \(17x⋮15\)

6. \(\left(x+16\right)⋮\left(x+1\right)\)

\(\Leftrightarrow\left(x+1\right)+15⋮\left(x+1\right)\)

\(\Leftrightarrow15⋮\left(x+1\right)\)

\(\Leftrightarrow\left(x+1\right)\inƯ\left(15\right)=\left\{\pm1;\pm3;\pm5;\pm15\right\}\)

\(\Rightarrow x\in\left\{-2;0;-4;2;-6;4;-16;14\right\}\)

Vậy x = .....................

7. \(x⋮\left(2x-1\right)\)

Mà \(\left(2x-1\right)\) lẻ

Nên : Với mọi x thuộc Z là số lẻ thì \(x⋮\left(2x-1\right)\)

8. \(\left(2x+3\right)⋮\left(x+5\right)\)

\(\Leftrightarrow\left(2x+10\right)-7⋮\left(x+5\right)\)

\(\Leftrightarrow2.\left(x+5\right)-7⋮\left(x+5\right)\)

\(\Leftrightarrow7⋮\left(x+5\right)\)

\(\Leftrightarrow\left(x+5\right)\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)

\(\Rightarrow x\in\left\{-6;-4;-12;2\right\}\)

Vậy x = .........................

9. \(\left(x+11\right)⋮\left(x-1\right)\)

\(\Leftrightarrow\left(x-1\right)+12⋮\left(x-1\right)\)

\(\Leftrightarrow12⋮\left(x-1\right)\)

\(\Leftrightarrow\left(x-1\right)\inƯ\left(12\right)=\left\{\pm1;\pm2;\pm3;\pm4;\pm6;\pm12\right\}\)

\(\Rightarrow x\in\left\{-2;0;-3;1;-4;2;-5;3;-7;5;-13;11\right\}\)

Vậy x = ................................

10. \(15⋮\left(2x+1\right)\)

\(\Leftrightarrow\left(2x+1\right)\inƯ\left(15\right)=\left\{\pm1;\pm3;\pm5;\pm15\right\}\)

\(\Rightarrow x\in\left\{-1;0;-2;1;-3;2;-8;7\right\}\)

Vậy x = .......................

B1

B = 5² . 4 - ( 18 + 6 . 7 ) : 81 : 3³

= 25 . 4 - ( 18 + 42 ) : 3⁴ : 3³

= 100 - 60 : 3

= 100 - 20

= 80

B2

5^x+1 + 52 = 6² + ( 7⁹ : 7⁷ - 2³ )

=> 5^x+1 + 52 = 36 + ( 7² - 8 )

=> 5^x+1 + 52 = 36 + 41

=> 5^x+1 + 52 = 77

=> 5^x+1 = 25

=> 5^x+1 = 5²

=> x + 1 = 2

=> x = 1

\(+)18⋮x-3\)

\(\Rightarrow x-3\inƯ\left(18\right)\)

mà \(Ư\left(18\right)=\left\{1;2;3;6;9;18\right\}\)

\(\Rightarrow\hept{\begin{cases}x-3=1;x-3=6\\x-3=2;x-3=9\\x-3=3;x-3=18\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=4;x=9\\x=5;x=12\\x=6;x=21\end{cases}}\)

\(26⋮\left(x+1\right)\)

\(\Rightarrow\left(x+1\right)\inƯ\left(26\right)\)

mà \(Ư\left(26\right)=\left\{1;2;13;26\right\}\)

\(\Rightarrow\orbr{\begin{cases}x+1=1\\x+1=2\end{cases}}\orbr{\begin{cases}x+1=13\\x+1=26\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\orbr{\begin{cases}x=12\\x=25\end{cases}}\)

cần gấp bài 1 và bài 3, bài 2 k có cx đc 

a) \(6-2\left(x+4\right)=-4\)

\(\Leftrightarrow6-2x-8=-4\)

\(\Leftrightarrow-2-2x=-4\)

\(\Leftrightarrow-2x=-4+2\)

\(\Leftrightarrow-2x=-2\)

\(\Leftrightarrow x=1\)

Vậy \(x=1\)

b) tự làm.

c) \(2x-15:5=-7\)

\(\Leftrightarrow2x-7,5=-7\)

\(\Leftrightarrow2x=-7+7,5\)

\(\Leftrightarrow2x=0,5\)

\(\Leftrightarrow x=0,25\)

\(\Leftrightarrow x=\dfrac{1}{4}\)

Vậy \(x=\dfrac{1}{4}\)

d) \(231-\left(x-6\right)=1339:13\)

\(\Leftrightarrow231-x+6=103\)

\(\Leftrightarrow237-x=103\)

\(\Leftrightarrow237-x+6=103\)

\(\Leftrightarrow-x=103-237\)

\(\Leftrightarrow-x=-134\)

\(\Leftrightarrow x=134\)

e) đề j mà có 2 dấu bằng >> sai :v

f) \(28-x^3=1\)

\(\Leftrightarrow-x^3=1-28\)

\(\Leftrightarrow-x^3=-27\)

\(\Leftrightarrow x^3=27\)

\(\Leftrightarrow x^3=3^3\)

\(\Leftrightarrow x=3\)

Vậy \(x=3\)

g) \(\left(x+2\right)^3=27\)

\(\Leftrightarrow\left(x+2\right)^3=3^3\)

\(\Leftrightarrow x+2=3\)

\(\Leftrightarrow x=3-2\)

\(\Leftrightarrow x=1\)

Vậy \(x=1\)

h) \(81-\left(x-5\right)^2=56\)

\(\Leftrightarrow-\left(x-5\right)^2=56-81\)

\(\Leftrightarrow-\left(x-5\right)^2=-25\)

\(\Leftrightarrow\left(x-5\right)^2=25\)

\(\Leftrightarrow x-5=\pm5\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=5\\x-5=-5\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5+5\\x=-5+5\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=10\\x=0\end{matrix}\right.\)

Vậy \(x_1=0;x_2=10\)

j) \(x^2=9\)

\(\Leftrightarrow x=\pm3\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)

Vậy \(x_1=-3;x_2=3\)

l) tự làm.

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