a) 2436 : x = 12

\(\Rightarrow\) x = 2436 : 12

\(\Rightarrow\) x = 203

b) 6x - 5 = 613

\(\Rightarrow\) 6x = 618

\(\Rightarrow\) x = 103

c) 12(x - 1) = 0

\(\Rightarrow\) x - 1 = 0

\(\Rightarrow\) x = 1

d) 0 : x = 0    (đúng \(\forall\)x  \(\in\) N)

\(\Rightarrow\) x \(\in\) N

a) Ta có: 2436:x=12

nên x=2436:12

hay x=203

b) Ta có: 6x-5=613

nên 6x=618

hay x=103

c) Ta có: 12(x-1)=0

mà 12>0

nên x-1=0

hay x=1

d) Ta có: 0:x=0

nên \(x\in R;x\ne0\)

a.2436:x=12

            x=2436:12

            x=203

b.6x-5=613

   6x   =613+5

   6x   =618

      x  =618:6

       x=103

c, 12(x-1)=0

          x-1=0:12

           x-1=0

              x= 0+1

              x=1

d, x:0=0

Vì theo quy luật, 0 chia số nào cx bằng 0. Suy ra x là số bất kỳ.

Nhớ like cho mik nha

\(a,\Leftrightarrow x^2+14x+49-x^2+3x=12\\ \Leftrightarrow17x=-37\Leftrightarrow x=-\dfrac{37}{17}\\ b,\Leftrightarrow x^2-x-2x+2=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

a) \(x^2+2x7+49-x^2+3x=12\Leftrightarrow17x=-37\Leftrightarrow x=\dfrac{-37}{17}\)

b) \(x^2-2x-x+2=0\Leftrightarrow x\left(x-2\right)-\left(x-2\right)=0\Leftrightarrow\left(x-1\right)\left(x-2\right)=\left(0\right)\Leftrightarrow x=1,x=2\)

`a,`

`(x+7)^2-x(x-3)=12`

`<=>x^2+14x+49 - x^2 +3x=12`

`<=> 17x =-37`

`<=>x=(-37)/17`

Vậy `x=(-37)/17`

`b,`

`x^2-3x+2=0`

`<=>x^2-2x-x+2=0`

`<=>x(x-2)-(x-2)=0`

`<=>(x-2)(x-1)=0`

giải ra `x=2,x=1`

Vậy `x=2,x=1`

\(2x^3+x^2-4x-12\)

\(=2x^3+5x^2+6x-4x^2-10x-12\)

\(=\left(2x^3+5x^2+6x\right)-\left(4x^2+10x+12\right)\)

\(=x\left(2x^2+5x+6\right)-2\left(2x^2+5x+6\right)\)

\(=\left(x-2\right)\left(2x^2+5x+6\right)\)

\(a,2x^3+x^2-4x-12=\left(2x^3-4x^2\right)+\left(5x^2-10x\right)+\left(6x-12\right)=2x^2\left(x-2\right)+5x\left(x-2\right)+6\left(x-2\right)=\left(x-2\right)\left(2x^2+5x+6\right)\)

\(b,x^5-xy^4+x^4y-y^5=x\left(x^4-y^4\right)+y\left(x^4-y^4\right)=\left(x+y\right)\left(x^4-y^4\right)=\left(x+y\right)\left(x^2-y^2\right)\left(x^2+y^2\right)=\left(x+y\right)^2\left(x-y\right)\left(x^2+y^2\right)\)

\(c,\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)-9=\left[\left(x+1\right)\left(x+7\right)\right]\left[\left(x+3\right)\left(x+5\right)\right]-9=\left(x^2+8x+7\right)\left(x^2+8x+15\right)-9\)

đặt \(x^2+8x+11=y\)

\(\left(x^2+8x+7\right)\left(x^2+8x+15\right)-9=\left(y-4\right)\left(y+4\right)-9=y^2-16-9=y^2-25=\left(y-5\right)\left(y+5\right)=\left(x^2+8x+11-5\right)\left(x^2+8x+11+5\right)=\left(x^2+8x+6\right)\left(x^2+8x+16\right)=\left(x^2+8x+6\right)\left(x+4\right)^2\)

a) \(\dfrac{5}{24}+x=\dfrac{7}{12}\)

<=> \(x=\dfrac{7}{12}-\dfrac{5}{24}=\dfrac{14}{24}-\dfrac{5}{24}=\dfrac{9}{24}=\dfrac{3}{8}\)

b) \(x-\dfrac{3}{4}=\dfrac{1}{2}\)

<=> \(x=\dfrac{1}{2}+\dfrac{3}{4}=\dfrac{2}{4}+\dfrac{3}{4}=\dfrac{5}{4}\)

c) bn ghi rõ đề chút

ok bạn

(2\7.x+3\7) :2 1\5-3\7=1

hai phần bảy nhân x cộng ba phần bảy chia hai một phần năm trừ 3 phần bảy bằng một

Giải:

a) \(\dfrac{5}{24}+x=\dfrac{7}{12}\) 

              \(x=\dfrac{7}{12}-\dfrac{5}{24}\) 

              \(x=\dfrac{3}{8}\) 

b) \(x-\dfrac{3}{4}=\dfrac{1}{2}\) 

            \(x=\dfrac{1}{2}+\dfrac{3}{4}\) 

            \(x=\dfrac{5}{4}\) 

c) \(\left(\dfrac{2}{7}.x+\dfrac{3}{7}\right):2\dfrac{1}{5}-\dfrac{3}{7}=1\) 

           \(\left(\dfrac{2}{7}.x+\dfrac{3}{7}\right):\dfrac{11}{5}=1+\dfrac{3}{7}\) 

           \(\left(\dfrac{2}{7}.x+\dfrac{3}{7}\right):\dfrac{11}{5}=\dfrac{10}{7}\) 

                        \(\dfrac{2}{7}.x+\dfrac{3}{7}=\dfrac{10}{7}.\dfrac{11}{5}\) 

                        \(\dfrac{2}{7}.x+\dfrac{3}{7}=\dfrac{22}{7}\)           

                                \(\dfrac{2}{7}.x=\dfrac{22}{7}-\dfrac{3}{7}\) 

                                \(\dfrac{2}{7}.x=\dfrac{19}{7}\) 

                                    \(x=\dfrac{19}{7}:\dfrac{2}{7}\) 

                                    \(x=\dfrac{19}{2}\)

a/ x - 7 = 12

=> x = 12 + 7 = 19

b/ 9 + 4(x - 5) = 13

=> 4x - 20 = 4

=> 4x = 24

=> x = 6

c/ (x+2)³ = 64

=> x + 2 = 4

=> x = 2

a) x-7=12

     x=12+7

     x=19

b) 9+4.(x-5)=13

       4.(x-5)=13-9

       4.(x-5)=4

          (x-5)=4:4

          (x-5)=1

          x=5+1

            x=6

cau cuoi mik ko bt lam nha!

chuc ban hoc tot

a/ x - 7 = 12

=> x = 12 + 7 = 19

b/ 9 + 4(x - 5) = 13

=> 4x - 20 = 4

=> 4x = 24

=> x = 6

c/ (x+2)³ = 64

=> x + 2 = 4

=> x = 2

Câu 1: 

a) Ư(-12)={1;-1;2;-2;3;-3;4;-4;6;-6;12;-12}

b) B(-4)={0;-4;-8;4;8;...}

Câu 2: 

a) Ta có: \(x-15=11-\left(-32\right)\)

\(\Leftrightarrow x-15=11+32\)

\(\Leftrightarrow x=43+15=58\)

Vậy: x=58

b) Ta có: \(13-\left(5-x\right)=7\)

\(\Leftrightarrow13-5+x=7\)

\(\Leftrightarrow x+8=7\)

hay x=-1

Vậy: x=-1

c) Ta có: \(x-\dfrac{3}{5}=5.12\)

\(\Leftrightarrow x-0.6=5.12\)

hay x=5,72

Vậy: x=5,72

d) Ta có: \(\dfrac{x+1}{-2}=\dfrac{-8}{x+1}\)

\(\Leftrightarrow\left(x+1\right)^2=\left(-2\right)\cdot\left(-8\right)=16\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=4\\x+1=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)

Vậy: \(x\in\left\{3;-5\right\}\)

a,Ư(-12)={1,-1,2,-2,3,-3,4,-4,6,-6,12,-12}

b,5 bội của 4 là:0,5,10,-10,-5

a, \(\left|x-1\right|=20\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=20\\x-1=-20\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=21\\x=-19\end{matrix}\right.\)

b, đk : x =< 18/3 

\(\Leftrightarrow\left[{}\begin{matrix}x-2=18-3x\\x-2=3x-18\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=8\left(ktm\right)\end{matrix}\right.\)

c, <=> | x - 2 | = 18 - 4x 

đk : x =< 18/4 = 9 /2 

\(\Leftrightarrow\left[{}\begin{matrix}x-2=18-4x\\x-2=4x-18\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{16}{3}\left(ktm\right)\end{matrix}\right.\)

\(\left|x-1\right|-8=12\)

\(\left|x-1\right|=12+8\)

\(\left|x-1\right|=20\)

\(x-1=+-20\)

a)\(\left|x-1\right|-8=12\Rightarrow\left|x-1\right|=20\)

   \(\Rightarrow\left[{}\begin{matrix}x-1=20\\x-1=-20\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=21\\x=-19\end{matrix}\right.\)

b)\(\left|x-2\right|=18-3x\)

   \(\Rightarrow\left[{}\begin{matrix}x-2=18-3x\\x-2=3x-18\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}4x=20\\-2x=-16\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=8\end{matrix}\right.\)

c)\(\left|x-2\right|-18+4x=0\)

   \(\Leftrightarrow\left|x-2\right|=18-4x\)

   Làm tương tự câu b

a) \(\left(x^2+x+1\right)\left(x^2+x+2\right)=12\)

\(\Leftrightarrow\left(x^2+x+1\right)^2+\left(x^2+x+1\right)-12=0\)

\(\Leftrightarrow\left(x^2+x+1\right)^2-3\left(x^2+x+1\right)+4\left(x^2+x+1\right)-12=0\)

\(\Leftrightarrow\left(x^2+x+1\right)\left(x^2+x+1-3\right)+ 4\left(x^2+x+1-3\right)=0\)

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

\(\Leftrightarrow x^2+x+4=0\) hay \(x^2+x-2=0\)

\(\Leftrightarrow x^2+2.\dfrac{1}{2}x+\dfrac{1}{4}+\dfrac{15}{4}=0\) hay \(x^2-x+2x-2=0\)

\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2+\dfrac{15}{4}=0\) (pt vô nghiệm) hay\(x\left(x-1\right)+2\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+2\right)=0\)

\(\Leftrightarrow x=1\) hay \(x=-2\)

-Vậy \(S=\left\{1;-2\right\}\)

b) \(x^3+5x^2-10x-8=0\)

\(\Leftrightarrow x^3-2x^2+7x^2-14x+4x-8=0\)

\(\Leftrightarrow x^2\left(x-2\right)+7x\left(x-2\right)+4\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^2+7x+4\right)=0\)

\(\Leftrightarrow x=2\) hay \(x^2+2.\dfrac{7}{2}+\dfrac{49}{4}-\dfrac{33}{4}=0\)

\(\Leftrightarrow x=2\) hay \(\left(x+\dfrac{7}{2}\right)^2-\dfrac{33}{4}=0\)

\(\Leftrightarrow x=2\) hay \(\left(x+\dfrac{7}{2}+\dfrac{\sqrt{33}}{2}\right)\left(x+\dfrac{7}{2}-\dfrac{\sqrt{33}}{2}\right)=0\)

\(\Leftrightarrow x=2\) hay \(x=\dfrac{-7-\sqrt{33}}{2}\) hay \(x=\dfrac{-7+\sqrt{33}}{2}\)

-Vậy \(S=\left\{2;\dfrac{-7-\sqrt{33}}{2};\dfrac{-7+\sqrt{33}}{2}\right\}\)