\(a.\left(x^2+4x+4\right)+\left(x^2-6x+9\right)=2x^2+14x\)

\(x^2+4x+4+x^2-6x+9-2x^2-14x=0\)

\(-18x+13=0\)

\(x=\dfrac{13}{18}\)

Vậy \(S=\left\{\dfrac{13}{18}\right\}\)

\(b.\left(x-1\right)^3-125=0\)

\(\left(x-1\right)^3=125\)

\(x-1=5\)

\(x=6\)

Vậy \(S=\left\{6\right\}\)

\(c.\left(x-1\right)^2+\left(y +2\right)^2=0\)

\(Do\left(x-1\right)^2\ge0\forall x;\left(y+2\right)^2\ge0\forall y\)

\(\Rightarrow\left(x-1\right)^2+\left(y+2\right)^2\ge0\forall x,y\)

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

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

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

Vậy \(S=\left\{1;-2\right\}\)

\(d.x^2-4x+4+x^2-2xy+y^2=0\)

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

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

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

Vậy \(S=\left\{2;2\right\}\)

\(\Leftrightarrow x\cdot\left(x-\sqrt{13}\right)\left(x+\sqrt{13}\right)=0\)

hay \(x\in\left\{0;\sqrt{13};-\sqrt{13}\right\}\)

\(x\left(x^2-13\right)=0\)

\(\left[{}\begin{matrix}x=0\\x^2-13=0\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=0\\\left[{}\begin{matrix}x=\sqrt{13}\\x=-\sqrt{13}\end{matrix}\right.\end{matrix}\right.\)

a: Ta có: \(\left(x^2+2\right)\left(x-4\right)-\left(x+2\right)^3=-16\)

\(\Leftrightarrow x^3-4x^2+2x-8-x^3-6x^2-12x-8=-16\)

\(\Leftrightarrow-10x^2-10x=0\)

\(\Leftrightarrow-10x\left(x+1\right)=0\)

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

c: Ta có: \(x^3+3x^2+3x+28=0\)

\(\Leftrightarrow\left(x+1\right)^3=-27\)

\(\Leftrightarrow x+1=-3\)

hay x=-4

a) \(\Rightarrow x^2\left(x^2-64\right)=0\Rightarrow x^2\left(x-8\right)\left(x+8\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=8\\x=-8\end{matrix}\right.\)

b) \(\Rightarrow x^2\left(x-6\right)+3x\left(x-6\right)+21\left(x-6\right)=0\Rightarrow\left(x-6\right)\left(x^2+3x+21\right)=0\)

\(\Rightarrow x=6\)

Bài 1.

Ta có

x 3   +   3 x 2   +   3 x   +   1   =   0   ⇔   ( x   +   1 ) 3   =   0

ó x + 1 = 0 ó x = -1

Vậy x = -1

Đáp án cần chọn là: A

a: =>3x+10-2x=0

hay x=-10

c: \(\Leftrightarrow3x^2-3x^2+6x=36\)

=>6x=36

hay x=6

a) 2x² - 3x - 2 = 0.

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

<=> 2x + 1 = 0 hoặc x - 2 = 0

<=> x = -1/2 hoặc x = 2

b) 3x² - 7x - 10 = 0.

<=> (x + 1)(3x - 10) = 0

<=> x = -1 hoặc x = 10/3

c) 2x² - 5x + 3 = 0.

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

<=> x = 1 hoặc x = 3/2

viết lại đề đi thiếu đề r

\(x^3-x^2=4x^2-8x+4\\ \Rightarrow x^3-5x^2+8x-4=0\\ \Rightarrow\left(x^3-x^2\right)-\left(4x^2-4x\right)+\left(4x-4\right)=0\\ \Rightarrow x^2\left(x-1\right)-4x\left(x-1\right)+4\left(x-1\right)=0\\ \Rightarrow\left(x^2-4x+4\right)\left(x-1\right)=0\\ \Rightarrow\left(x-2\right)^2\left(x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}\left(x-2\right)^2=0\\x-1=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=1\end{matrix}\right.\)

Ta có: x³ – x²= x²(x -1); 4x² – 8x + 4 = 4(x² – 2x + 1) = 4(x – 1)²

Vậy x² (x -1) = 4(x – 1)² ⇒ x²(x -1) - 4(x – 1)² = 0

⇒ (x – 1)(x² – 4x + 4) = 0 ⇒ (x – 1)(x – 2)² = 0

⇒ x – 1 = 0 hoặc x – 2 = 0 ⇒ x = 1 hoặc x = 2.