a: \(\Leftrightarrow\left(x+1\right)^2=3^2=9\)

=>x+1=3 hoặc x+1=-3

=>x=2 hoặc x=-4

b: \(\Leftrightarrow\left(x-1\right)^2=16\)

=>x-1=4 hoặc x-1=-4

=>x=5 hoặc x=-3

a) \(\dfrac{x+1}{3}=\dfrac{3}{x+1}\)

⇔ \(\left(x+1\right)^2=9\)

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

⇒ \(\left[{}\begin{matrix}x=2\\x=-4\end{matrix}\right.\)

Vây ...

b) Tương tự câu a

=>2x=-18

hay x=-9

\(\Rightarrow-1.2x=6.3=18\\ \Rightarrow2x=18:\left(-1\right)\\ \Rightarrow2x=-18\\ \Rightarrow x=-18:2\\ \Rightarrow x=-9\)

\(\dfrac{-1}{6} = \dfrac{3}{2x}\) 

\(=> \) \(-1 . 2x = 3 . 6 \) 

\(=> 2x = \dfrac{ 3 . 6}{ -1 }\) 

\(=> 2x = -18 \) 

\(=> x = \dfrac{ -18 }{2}\) 

\(=> x = 9 \)

Vậy x = 9 

a: \(A=\dfrac{x-1+2x^2+2x+2-x^2-2x}{\left(x-1\right)\left(x^2+x+1\right)}\)

\(=\dfrac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{1}{x-1}\)

`x/2=2/x`

`=>x.x=2.2`

`=>x^2=4`

`=>x=+-2`

a, \(\dfrac{x}{2}=-\dfrac{5}{y}\Rightarrow xy=-10\Rightarrow x;y\inƯ\left(-10\right)=\left\{\pm1;\pm2;\pm5;\pm10\right\}\)

c, \(\dfrac{3}{x-1}=y+1\Rightarrow\left(y+1\right)\left(x-1\right)=3\Rightarrow x-1;y+1\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)

b: =>xy=12

\(\Leftrightarrow\left(x,y\right)\in\left\{\left(12;1\right);\left(6;2\right);\left(4;3\right)\right\}\)

\(A=\left(\dfrac{x}{x^2-4}+\dfrac{2}{2-x}+\dfrac{1}{x+2}\right):\left(x-2+\dfrac{10-x^2}{x+2}\right)\)

\(\Rightarrow A=\left(\dfrac{x-2\left(x+2\right)+1\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\right):\left(\dfrac{\left(x-2\right)\left(x+2\right)+10-x^2}{x+2}\right)\)

\(\Rightarrow A=\left(\dfrac{-6}{x^2-4}\right):\left(\dfrac{6}{x+2}\right)\)

\(\Rightarrow A=-\dfrac{6}{x^2-4}.\dfrac{x+2}{6}=-\dfrac{6\left(x+2\right)}{\left(x-2\right)\left(x+2\right)6}=-\dfrac{1}{x-2}\)

để A<0 thì :

\(\left\{{}\begin{matrix}x-2\ne0\\x-2\notin Z-\end{matrix}\right.\)\(\Leftrightarrow x\in\left\{3;4;5;6;7;8;9;....n\right\}\)

( Z- là tập hợp số nguyên âm )

Để A có giá trị nguyên thì :

\(\left\{{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)

Bài 1: 

b) ĐKXĐ: \(x\ne3\)

Ta có: \(\dfrac{3-x}{20}=\dfrac{-5}{x-3}\)

\(\Leftrightarrow\dfrac{x-3}{-20}=\dfrac{-5}{x-3}\)

\(\Leftrightarrow\left(x-3\right)^2=100\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=10\\x-3=-10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=13\left(nhận\right)\\x=-7\left(nhận\right)\end{matrix}\right.\)

Vậy: \(x\in\left\{13;-7\right\}\)