\(a,\Rightarrow4x\left(x^2-9\right)=0\\ \Rightarrow4x\left(x-3\right)\left(x+3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\\ b,\Rightarrow\left(3x-5-x-1\right)\left(3x-5+x+1\right)=0\\ \Rightarrow\left(2x-6\right)\left(4x-4\right)=0\\ \Rightarrow2\left(x-3\right)4\left(x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)

a) \(\Rightarrow4x\left(x^2-9\right)=0\)

\(\Rightarrow4x\left(x-3\right)\left(x+3\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\)

b) \(\Rightarrow\left(3x-5-x-1\right)\left(3x-5+x+1\right)=0\)

\(\Rightarrow\left(2x-6\right)\left(4x-4\right)=0\)

\(\Rightarrow8\left(x-3\right)\left(x-1\right)=0\)

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

\(a,\Rightarrow\left(x-2000\right)\left(5x-1\right)=0\Rightarrow\left[{}\begin{matrix}x=2000\\x=\dfrac{1}{5}\end{matrix}\right.\\ b,\Rightarrow x\left(x^2-13\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=\sqrt{13}\\x=-\sqrt{13}\end{matrix}\right.\\ c,\Rightarrow3x\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\\ d,\Rightarrow\left(x-5\right)\left(x+3\right)=0\Rightarrow\left[{}\begin{matrix}x=5\\x=-3\end{matrix}\right.\\ e,\Rightarrow\left(3x-2\right)\left(3x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{2}{3}\end{matrix}\right.\)

a,\(x\in\left\{5;1,5;\dfrac{-4}{3}\right\}\)

a, \(2x\left(x-3\right)-15+5x=0\\ \Rightarrow2x\left(x-3\right)-\left(15-5x\right)=0\\ \Rightarrow2x\left(x-3\right)-5\left(3-x\right)=0\\ \Rightarrow\left(2x+5\right)\left(x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{5}{2}\\x=3\end{matrix}\right.\)

b, \(x^3-7x=0\\ \Rightarrow x\left(x^2-7\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=\pm7\end{matrix}\right.\)

c, \(\left(2x-3\right)^2-\left(x+5\right)^2=0\\ \Rightarrow\left(2x-3-x-5\right)\left(2x-3+x+5\right)=0\\ \Rightarrow\left(x-8\right)\left(3x+2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=8\\x=-\dfrac{2}{3}\end{matrix}\right.\)

Xem lại đề câu d 

2:

a: =>x-1=0 hoặc 3x+1=0

=>x=1 hoặc x=-1/3

b: =>x-5=0 hoặc 7-x=0

=>x=5 hoặc x=7

c: =>\(\left[{}\begin{matrix}x-1=0\\x+5=0\\3x-8=0\end{matrix}\right.\Leftrightarrow x\in\left\{1;-5;\dfrac{8}{3}\right\}\)

d: =>x=0 hoặc x^2-1=0

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

Bạn tách ra từng câu thoi nhe .

a. (2x + 1)² - 4x² + 2x² - 2 = 0

<=> (2x + 1 - 2x)(2x + 1 + 2x) + 2(x² - 1) = 0

<=> (4x + 1) + 2x² - 2 = 0

<=> 4x + 1 + 2x² - 2 = 0

<=> 2x² + 4x - 2 + 1 = 0

<=> 2x² + 4x - 1 = 0

<=> 2x² + 4x = 1

<=> 2x(x + 2) = 1

Vì 1 chỉ có tích là 1 . 1 nên:

<=> \(\left[{}\begin{matrix}2x=1\\x+2=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-1\end{matrix}\right.\)

\(a,\Leftrightarrow4x^2+4x+1-4x^2+2x^2-2=0\\ \Leftrightarrow2x^2+4x-1=0\\ \Leftrightarrow2\left(x^2+2x+1\right)-3=0\\ \Leftrightarrow2\left(x+1\right)^2-3=0\\ \Leftrightarrow\left(x+1\right)^2=\dfrac{3}{2}\\ \Leftrightarrow\left[{}\begin{matrix}x+1=\sqrt{\dfrac{3}{2}}\\x+1=-\sqrt{\dfrac{3}{2}}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-2-\sqrt{6}}{2}\\x=\dfrac{-2+\sqrt{6}}{2}\end{matrix}\right.\)

\(b,\left(x-2\right)\left(x+2\right)-\left(x+3\right)^2-2x-5=0\\ \Leftrightarrow x^2-4-x^2-6x-9-2x-5=0\\ \Leftrightarrow-8x=18\\ \Leftrightarrow x=-\dfrac{9}{4}\)

\(\left|x-3\right|+\left|x-\dfrac{1}{2}\right|=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-3=0\\x-\dfrac{1}{2}=0\end{matrix}\right.\)

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

Vậy \(S=\varnothing\)

b: \(\left|x-3\right|+\left|x-\dfrac{1}{2}\right|\ge0\forall x\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x=3\\x=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow x\in\varnothing\)

\(a,\left(8-x\right)\left(x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}8-x=0\\x+5=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=8\\x=-5\end{matrix}\right.\\ b,2x\left(x+81\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x=0\\x+81=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-81\end{matrix}\right.\)

a)\(\left(8-x\right)\left(x+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}8-x=0\\x+5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=8\\x=-5\end{matrix}\right.\)
b)\(2x\left(x+81\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x=0\\x+81=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-81\end{matrix}\right.\)

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 = .........................