a: \(8x\left(x-2017\right)-2x+4034=0\)

\(\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x^2-1=0\\x^2+4=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2=1\\x^2=-4\left(vn\right)\end{matrix}\right.\)

\(\Leftrightarrow x=\pm1\)

a/

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-1=0\\x+1=0\end{matrix}\right.\)

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

b/

\(x^5-9x=0\)

\(\Leftrightarrow x\left(x^4-9\right)=x\left(x^2-3\right)\left(x^2+3\right)=0\)

\(\Leftrightarrow x\left(x-\sqrt{3}\right)\left(x+\sqrt{3}\right)=0\)

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

c/

\(\left(x^3-x^2\right)^2-4x^2+8x-4=0\)

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\pm\sqrt{2}\end{matrix}\right.\)

Sos

a: \(x\in\left\{0;25\right\}\)

c: \(x\in\left\{0;5\right\}\)

`@` `\text {Ans}`

`\downarrow`

`1,`

`x^2 - 9 = 0`

`<=> x^2 = 0 + 9`

`<=> x^2 = 9`

`<=> x^2 = (+-3)^2`

`<=> x = +-3`

Vậy, `S = {3; -3}`

`2,`

`25 - x^2 = 0`

`<=> x^2 = 25 - 0`

`<=> x^2 = 25`

`<=> x^2 = (+-5)^2`

`<=> x = +-5`

Vậy,` S= {5; -5}`

`3,`

`-x^2 + 36 = 0`

`<=> -x^2 = 0 - 36`

`<=> -x^2 = -36`

`<=> x^2 = 36`

`<=> x^2 = (+-6)^2`

`<=> x = +-6`

Vậy, `S= {6; -6}`

`4,`

`4x^2 - 4 = 0`

`<=> 4x^2 = 0+4`

`<=> 4x^2 = 4`

`<=> x^2 = 4 \div 4`

`<=> x^2 = 1`

`<=> x^2 = (+-1)^2`

`<=> x = +-1`

Vậy, `S= {1; -1}`

`@` `\text {Kaizuu lv uuu}`

Lớp \(8\) thì nên Vậy \(S=\left\{...\right\}\) nha em ☕

Lập bảng xét dấu là ra          

bài 2: (x-3).(y+2) = -5

    Vì x, y \(\in\)Z   => x-3 \(\in\)Ư(-5) = {5;-5;1;-1}

Ta có bảng: 

bài 3: a(a+2)<0

TH1 : \(\orbr{\begin{cases}a< 0\\a+2>0\end{cases}}\)=>\(\orbr{\begin{cases}a< 0\\a>-2\end{cases}}\)=> -2<a<0 ( TM)

TH2: \(\orbr{\begin{cases}a>0\\a+2< 0\end{cases}}\Rightarrow\orbr{\begin{cases}a>0\\a< -2\end{cases}}\Rightarrow loại\)
 

           Vậy -2<a<0

Bài 5: \(\left(x^2-1\right)\left(x^2-4\right)< 0\)

TH 1 : \(\hept{\begin{cases}x^2-1>0\\x^2-4< 0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x^2>1\\x^2< 4\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x>1\\x< 2\end{cases}}\)\(\Rightarrow\)1 < a < 2

TH 2: \(\hept{\begin{cases}x^2-1< 0\\x^2-4>0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x^2< 1\\x^2>4\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x< 1\\x>2\end{cases}}\)\(\Rightarrow\)loại

                         Vậy 1<a<2

$(x^2-2)^2+4(x-1)^2-4(x^2-2)(x-1)=0$

$\Leftrightarrow(x^2-2)^2-4(x^2-2)(x-1)+4(x-1)^2=0$

$\Leftrightarrow(x^2-2)^2-2\cdot(x^2-2)\cdot2(x-1)+[2(x-1)]^2=0$

$\Leftrightarrow[(x^2-2)-2(x-1)]^2=0$

$\Leftrightarrow(x^2-2-2x+2)^2=0$

$\Leftrightarrow(x^2-2x)^2=0$

$\Leftrightarrow x^2-2x=0$

$\Leftrightarrow x(x-2)=0$

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

Vậy: $x\in\{0;2\}$.

3) \(x\left(x-4\right)+\left(x-4\right)^2=0\Leftrightarrow\left(x-4\right)\left(x+x-4\right)=0\Leftrightarrow2\left(x-4\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)

4x.(x+1)-8(x+1)=0

(4x-8)(x+1)=0

suy ra x=2 hoặc x=-1