\(\dfrac{x+3}{12}=\dfrac{3}{x+3}\) (đk: \(x\neq-3\))

\(\Rightarrow\left(x+3\right)\cdot\left(x+3\right)=3\cdot12\)

\(\Rightarrow\left(x+3\right)^2=36\)

\(\Rightarrow\left(x+3\right)^2=\left(\pm6\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}x+3=6\\x+3=-6\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=-9\left(tm\right)\end{matrix}\right.\)

Vậy \(x\in\left\{3;-9\right\}\).

\(\dfrac{x+3}{12}=\dfrac{3}{x+3}\\ \Rightarrow\left(x+3\right)\left(x+3\right)=12.3\\ \Rightarrow x^2+6x+9=36\\ \Rightarrow x^2+6x+9-36=0\\ \Rightarrow x^2+6x-27=0\\ \Rightarrow x^2-3x+9x-27=0\\ \Rightarrow x\left(x-3\right)+9\left(x-3\right)=0\\ \Rightarrow\left(x+9\right)\left(x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-3=0\\x+9=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=3\\x=-9\end{matrix}\right.\)

=>x-1=0

hay x=1

đề bài cho như này 9^x-1 = 1 hay như này vậy 9^x - 1 = 1

Vì tổng số hạt của X là 10 nên ta có:

(1) P+N+E=10 

Mặt khác P=E(2)

Mà, số hạt mang điện nhiều hơn số hạt không mang điện là 2 nên ta có:

(3) (P+E)-N=2

Từ (1), (2), (3) ta lập hpt:

\(\left\{{}\begin{matrix}P+N+E=10\\P=E\\\left(P+E\right)-N=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2P+N=10\\2P-N=2\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}P=E=Z=3\\N=4\end{matrix}\right.\)

bạn kiểm tra lại đề nhé! mình nghĩ A=(x+1)(x+2)(x+3)(x+6) thì đúng hơn

\(A=x^3-2x+n\)

\(B=n-2\)

\(A\text{⋮}B\) ⇒ \(\left(x^3-2x+n\right)\text{⋮}\left(n-2\right)\)

⇒ \(\left[\left(x^3-2x^2\right)+\left(2x^2-4x\right)+\left(2x-4\right)+\left(n+4\right)\right]\text{⋮}\left(n-2\right)\)

⇒ \(\left[x^2\left(x-2\right)+2x\left(x-2\right)+2\left(x-2\right)+\left(n+4\right)\right]\text{⋮}\left(n-2\right)\)

⇒ \(\left[\left(x-2\right)\left(x^2+2x+2\right)+\left(n+4\right)\right]\text{⋮}\left(x-2\right)\)

Vì \(\left(x-2\right)\left(x^2+2x+2\right)\text{⋮}\left(n-2\right)\)

Để \(A\text{⋮}B\)

⇒ \(n+4=0\)

⇒ \(n=-4\)

Bài 1: 

a: x/-2=-18/x

=>x²=36

=>x=6 hoặc x=-6

b: x/2+x/5=17/10

=>7/10x=17/10

hay x=17/7

19.9-63x+18=0

-63x= -(171+18)

-63x= -189

x=189:63

x=3

\(\left(\dfrac{6:\dfrac{3}{5}-\dfrac{17}{16}.\dfrac{6}{7}}{\dfrac{21}{5}.\dfrac{10}{11}+\dfrac{57}{11}}-\dfrac{\left(\dfrac{3}{20}+\dfrac{1}{2}-\dfrac{1}{15}\right).\dfrac{12}{49}}{\dfrac{10}{3}+\dfrac{2}{9}}\right).x=\dfrac{215}{96}\)

\(\Rightarrow\left(\dfrac{\dfrac{509}{56}}{9}-\dfrac{\dfrac{7}{12}.\dfrac{12}{49}}{\dfrac{32}{9}}\right).x=\dfrac{215}{96}\)

\(\Rightarrow\left(\dfrac{509}{504}-\dfrac{\dfrac{1}{7}}{\dfrac{32}{9}}\right).x=\dfrac{215}{96}\)

\(\Rightarrow\left(\dfrac{509}{504}-\dfrac{9}{224}\right).x=\dfrac{215}{96}\)

\(\Rightarrow\dfrac{1955}{2016}.x=\dfrac{215}{96}\)

\(\Rightarrow x=\dfrac{215}{96}:\dfrac{1955}{2016}\)

\(\Rightarrow x=\dfrac{903}{391}\)

`[ 6 : 3/5 - 17/16 . 6/7 : 21/5 . 10/11 + 57/11 -  (3/20 + 1/2 - 1/15) . 12/49 :  10/3 + 2/9 ] . x = 215/96`

`=>[ 6  . 5/3 - 17/16 . 6/7 . 5/21 . 10/11 + 57/11 -  (3/20 + 1/2 - 1/15) . 12/49 . 3/10 + 2/9 ] . x = 215/96`

`=>[10- 51/56 . 6/7 . 5/21 . 10/11 + 57/11 -  (3/20 + 1/2 - 1/15) . 12/49 . 3/10 + 2/9 ] . x = 215/96`

`=> [10- 153/196  . 5/21 . 10/11 + 57/11 -  (3/20 + 1/2 - 1/15) . 12/49 . 3/10 + 2/9 ] . x = 215/96`

`=> [10- 255/1372 . 10/11 + 57/11 -  (3/20 + 1/2 - 1/15) . 12/49 . 3/10 + 2/9 ] . x = 215/96`

`=> [10- 1275/7546 + 57/11 -  (3/20 + 1/2 - 1/15) . 12/49 . 3/10 + 2/9 ] . x = 215/96`

`=> (10- 1275/7546 + 57/11 -  7/12. 12/49 . 3/10 + 2/9 ) . x = 215/96`

`=> ( 10- 1275/7546 + 57/11 -343/600 . 3/10 + 2/9 ) . x = 215/96`

`=> ( 10- 1275/7546 + 57/11 -343/2000 + 2/9 ) . x = 215/96`

`=>15,06357671 . x= 215/96`

`=> x= 215/96: 15,06357671`

`=>x= 0,1486754027`

Đề có phải như vậy không nhỉ ?

Viết lại đề bài:

Tìm số nguyên x sao cho \(\frac{6}{x+1}.\frac{x-1}{3}\)là số nguyên

Giải:

\(\frac{6}{x+1}.\frac{x-1}{3}\text{​​}\)

\(=\frac{3.2}{x+1}.\frac{x-1}{3}\text{​​}\)

\(=\frac{3.2.\left(x-1\right)}{\left(x+1\right).3}\text{​​}\)

\(=\frac{3.2.\left(x-1\right)}{3.\left(x+1\right)}​​\)

\(=\frac{3.2.\left(x-1\right)}{3.\left(x+1\right)}​​\)

\(=\frac{2.\left(x-1\right)}{\left(x+1\right)}​​\)

\(=2.\frac{\left(x-1\right)}{\left(x+1\right)}​​\)

Bí....

Sorr nhak

Ta có:\(\frac{6x}{x+1}=\frac{6x+6-6}{x+1}=\frac{6\left(x+1\right)-6}{x+1}=6-\frac{6}{x+1}\)

Để\(\frac{6x}{x+1}\)là số nguyên \(\Leftrightarrow6⋮x+1\)

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

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

Để\(\frac{x-1}{3}\)là số nguyên\(\Leftrightarrow\left(x-1\right)⋮3\)

\(\Rightarrow x-1=3k\Rightarrow x=3k+1\left(k\in Z\right)\left(2\right)\)

Từ (1) và (2)\(\Rightarrow x\in\left\{-2;1\right\}\)

Vậy \(x\in\left\{-2;1\right\}\)