Tham khảo thử đúng không nha mn

     \(x^2+x-y^2=0\)

⇔ \(\left(x^2-y^2\right)+x=0\)

⇔ \(\left(x-y\right)\left(x+y\right)+x=0\)

⇒ \(x-y=0\) hoặc \(x+y=0\) hoặc \(x=0\)

⇒ \(x=y=0\)

\(\Leftrightarrow x^2+x=y^2\)

\(\Leftrightarrow4x^2+4x=4y^2\)

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

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

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

Vậy \(\left(x;y\right)=\left(0;0\right);\left(-1;0\right)\)

`x(x - 4) - 3x + 12 = 0`

`<=> x(x - 4) + 3(x - 4) = 0`

`<=> (x + 3)(x - 4) = 0`

`<=>` $\left[\begin{matrix} x + 3 = 0\\ x - 4 = 0\end{matrix}\right.$

`<=>` $\left[\begin{matrix} x = -3\\ x = 4\end{matrix}\right.$

Vậy `S = {-3; 4}`

\(x\left(x-4\right)-3x+12=0\)

\(\Leftrightarrow x^2-4x-3x+12=0\)

\(\Leftrightarrow x\left(x-4\right)-3\left(x-4\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x-4=0\end{matrix}\right.\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)

ta có x³-6x²+11x-6=0

hay x³-x²-5x²-+5x+6x-6=0

=>x(x-1) - 5x(x-1)+6(x-1)=0

(x-1).(x-5x+6)=0 <=> (x-1)(x²-2x-3x+6)=0

(x-1)(x(x-2)-3(x-2)=0

(x-1)(x-2)(x-3)=0 <=> x-1=0 hoặc x-2=0 hoặc x-3=0

<=> x=1 hoặc x=2 hoặc x=3

vậy S ={1;2;3}

x(x²+6x+9) - 3x= x³+6x²+12x+8+1

\(\Leftrightarrow\)x³+6x²+9x-3x=x³+6x²+12x+9

\(\Leftrightarrow\)6x=12x+9

\(\Leftrightarrow\)6x=-9

\(\Leftrightarrow\)x=-3/2

Vậy phương trình có 1 nghiệm duy nhất x=-3/2

x(x + 3)^2  - 3x = (x + 2)^3 + 1

<=> x(x^2 + 6x + 9) = x^3 + 6x^2 + 12x + 8 + 1

<=> x^3 + 6x^2 + 9x = x^3 + 6x^2 + 12x + 9

<=> 3x + 9 = 0

<=> 3x = -9

<=> x = -3

\(x\left(x+3\right)^2-3x=\left(x+2\right)^3+1\)

\(\Leftrightarrow x\left(x^2+6x+9\right)=x^3+6x^2+12x+8+1\)

\(\Leftrightarrow x^3+6x^2+9x=x^3+6x^2+12x+9\)

\(\Leftrightarrow3x+9=0\)

\(\Leftrightarrow3x=-9\)

\(\Leftrightarrow x=-3\)

Thay x=3 vào pt,ta được:

3^2+(m^2-2m)*3-9+12m=0

=>3m^2-6m+12m=0

=>3m^2+6m=0

=>m=0 hoặc m=-2

\(\dfrac{x+5}{x-5}=\dfrac{5}{x^2-5x}+\dfrac{1}{x}\)

\(\Leftrightarrow\dfrac{x+5}{x-5}=\dfrac{5}{x\left(x-5\right)}+\dfrac{1}{x}\)

ĐKXĐ : \(\left\{{}\begin{matrix}x\ne0\\x-5\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\x\ne5\end{matrix}\right.\)

Ta có : \(\dfrac{x+5}{x-5}=\dfrac{5}{x\left(x-5\right)}+\dfrac{1}{x}\)

\(\Leftrightarrow\dfrac{x\left(x+5\right)}{x\left(x-5\right)}=\dfrac{5}{x\left(x-5\right)}+\dfrac{x-5}{x\left(x-5\right)}\)

`=> x (x+5) = 5 +x-5`

`<=> x^2 +5x - 5-x+5=0`

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

`<=> x(x+4)=0`

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

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

Vậy phương trình có nghiệm `x=-4`

CHO TEN ROI NOI

ngọc anh ạ

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

\(\Rightarrow\left(x^2+2x+1\right)-\left(y^2+4y+4\right)-7=0\)

\(\Rightarrow\left(x+1\right)^2-\left(y+2\right)^2=7\)

\(\Rightarrow\left(x+1+y+2\right)\left(x+1-y-2\right)=4\)

\(\Rightarrow\left(x-y-1\right)\left(x+y+3\right)=7\)

Vì \(x,y\) nguyên dương nên \(x+y+3>x-y-1>0\)

\(\Rightarrow\hept{\begin{cases}x+y+3=7\\x-y-1=1\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=3\\y=1\end{cases}}\)

sửa đề : 

\(x^3-3x^2+3x-1=0\)

\(\Leftrightarrow\left(x-1\right)^3=0\Leftrightarrow x=1\)