\(2x^3+5x^2-12x=0\\ \Leftrightarrow x\left(2x^2+5x-12\right)=0\\ \Leftrightarrow x\left[\left(2x^2+8x\right)-\left(3x+12\right)\right]=0\\ \Leftrightarrow x\left[2x\left(x+4\right)-3\left(x+4\right)\right]=0\\ x\left(2x-3\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{3}{2}\\x=-4\end{matrix}\right.\)

2x³ + 5x² - 12x = 0
<=> x(2x² + 5x - 12) = 0
<=> x(2x² + 8x - 3x - 12) = 0
<=> x[2x(x + 4) - 3(x + 4)] = 0
<=> x(x + 4)(2x - 3) = 0
   1) x = 0
    2) x = -4 
    3) x =  3/2

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

kbhbchbidc

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

a, sửa đề : \(25x^2+4y^2-10x+12y+10=0\)

\(\Leftrightarrow25x^2-10x+1+4y^2+12y+9=0\)

\(\Leftrightarrow\left(5x-1\right)^2+\left(2y+3\right)^2=0\)

Đẳng thức xảy ra khi x = 1/5 ; y = -3/2 

b, \(3x^2+2y^2-12x+12y+30=0\)

\(\Leftrightarrow3\left(x^2-4x+4\right)+2\left(y^2+6y+9\right)=0\)

\(\Leftrightarrow3\left(x-2\right)^2+2\left(y+3\right)^2=0\)

Đẳng thức xảy ra khi x = 2 ; y = -3 

\(a)\)

\(25x^2+4y^2-10x+12x+10=0\)

\(\Leftrightarrow\left(5x\right)^2-10x+1+\left(2y\right)^2+12y+9=0\)

\(\Leftrightarrow[\left(5x\right)^2-10x+1+\left(2y\right)^2+12y+9=0\)

\(\Leftrightarrow[\left(5x\right)^2-2.5x.1-1^2]+[\left(2y\right)^2+2.2y.3+3^{20}]=0\)

\(\Leftrightarrow\left(5x-1\right)^2+\left(2y+3\right)^2=0\)

\(\Leftrightarrow\left(5x-1\right)^2=0\Leftrightarrow5x-1=0\Leftrightarrow x=\frac{1}{5}\)

\(\Leftrightarrow\left(2y+3\right)^2=0\Leftrightarrow2y+3=0\Leftrightarrow2y=-3\Leftrightarrow y=\frac{-3}{2}\)

\(b)\)

\(3x^2+2y^2-12x+12y+30=0\)

\(\Leftrightarrow3x^2-12x+12+2y^2+12y+18=0\)

\(\Leftrightarrow3\left(x-2\right)^2+2\left(y+3\right)^2=0\)

Mà: \(3\left(x-2\right)^2\ge0\forall x;2\left(y+3\right)^2\ge0\forall y\)

\(\Leftrightarrow3\left(x-2\right)^2+2\left(y+3\right)^2=0\)chỉ khi: \(x-2=y+3=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=2\\y=-3\end{cases}}\)

\(4x^5-12x^3=0\Leftrightarrow4x^3\left(x^2-3\right)=0\Leftrightarrow4x^3\left(x-\sqrt{3}\right)\left(x+\sqrt{3}\right)=0\)

\(\Leftrightarrow x=0\text{ hoặc }x=\sqrt{3}\text{ hoặc }x=-\sqrt{3}\)

Pt bậc 3 này ko giải được trong chương trình phổ thông

x^3-6^2+12x-8=1

(x-2)^3=1

=>x-2=1

=>x=3

Câu b tương tự nha

a: \(x^2+12x+36=0\)

\(\Leftrightarrow\left(x+6\right)^2=0\)

\(\Leftrightarrow x+6=0\)

hay x=-6

b: Ta có: \(x^2-1=0\)

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

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

c: Ta có: \(25x^2-9=0\)

\(\Leftrightarrow\left(5x-3\right)\left(5x+3\right)=0\)

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

Lời giải:
a. $x^2+12x+36=0$

$\Leftrightarrow (x+6)^2=0$

$\Leftrightarrow x+6=0$

$\Leftrightarrow x=-6$

b.

$x^2-1=0$

$\Leftrightarrow (x-1)(x+1)=0$
$\Leftrightarrow x-1=0$ hoặc $x+1=0$

$\Leftrightarrow x=1$ hoặc $x=-1$

c. 

$25x^2-9=0$

$\Leftrightarrow (5x)^2-3^2=0$

$\Leftrightarrow (5x-3)(5x+3)=0$

$\Leftrightarrow 5x-3=0$ hoặc $5x+3=0$

$\Leftrightarrow x=\frac{3}{5}$ hoặc $x=-\frac{3}{5}$

4xy - 12x + 5y - 63 = 0

<=> 4x(y - 3) + 5y - 15 - 48 = 0

<=> 4x(y - 3) + 5(y - 3) = 48

<=> (4x + 5)(y - 3) = 48

Lập bảng xét các trường hợp : 

Vậy các cặp (x;y) nguyên là (-1;51) ; (-2;-13)