\(x^3-6x^2+5x=0\Leftrightarrow x\left(x^2-6x+5\right)=0\)

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

\(\Leftrightarrow x\left[x\left(x-1\right)-5\left(x-1\right)\right]=0\Leftrightarrow x\left(x-1\right)\left(x-5\right)=0\Leftrightarrow x=0;x=1;x=5\)

\(x^3-6x^2+5x=0\)

\(\Leftrightarrow x.\left(x^2-6x+5\right)=0\)

\(\Leftrightarrow x.\left(x^2-x-5x+5\right)=0\)

\(\Leftrightarrow x.\left[\left(x^2-x\right)-\left(5x-5\right)\right]=0\)

\(\Leftrightarrow x.\left[x\left(x-1\right)-5\left(x-1\right)\right]=0\)

\(\Leftrightarrow x\left(x-1\right)\left(x-5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\\x-5=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=5\end{matrix}\right.\)

Vậy \(x\in\left\{0;1;5\right\}\)

a) Ta có: \(x^3+x^2+x+1=0\)

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

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

mà \(x^2+1>0\forall x\)

nên x+1=0

hay x=-1

Vậy: S={-1}

b) Ta có: \(x^3-6x^2+11x-6=0\) 

\(\Leftrightarrow x^3-x^2-5x^2+5x+6x-6=0\)

\(\Leftrightarrow x^2\left(x-1\right)-5x\left(x-1\right)+6\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^2-5x+6\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-2\right)\left(x-3\right)=0\)

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

Vậy: S={1;2;3}

c) Ta có: \(x^3-x^2-21x+45=0\)

\(\Leftrightarrow x^3-3x^2+2x^2-6x-15x+45=0\)

\(\Leftrightarrow x^2\left(x-3\right)+2x\left(x-3\right)-15\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x^2+2x-15\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x^2+5x-3x-15\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)

Vậy: S={3;-5}

d) Ta có: \(x^4+2x^3-4x^2-5x-6=0\)

\(\Leftrightarrow x^4-2x^3+4x^3-8x^2+4x^2-8x+3x-6=0\)

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

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

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

\(\Leftrightarrow\left(x-2\right)\left[x^2\left(x+3\right)+\left(x+1\right)\left(x+3\right)\right]=0\)

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

mà \(x^2+x+1>0\forall x\)

nên (x-2)(x+3)=0

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

Vậy: S={2;-3}

Bài 1

a/ \(x\left(x^2+1\right)+2\left(x^2+1\right)=0\)

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

b/

\(\Leftrightarrow x^3-6x^2+9x+5x^2-30x+45=0\)

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

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

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

1.

c/ \(\Leftrightarrow x^3+2x^2+2x+x^2+2x+2=0\)

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

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

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

d/

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

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

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

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

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

Bài 1:

e/ \(\Leftrightarrow x^4+2x^2-8x+5=0\)

\(\Leftrightarrow x^4-2x^3+x^2+2x^3-4x^2+2x+5x^2-10x+5=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x^2+2x+5=0\left(vn\right)\\x=1\end{matrix}\right.\)

Bài 2:

a/ Đặt \(x^2-5x=t\)

\(t^2+10t+24=0\Rightarrow\left[{}\begin{matrix}t=-4\\t=-6\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-5x=-4\\x^2-5x=-6\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-5x+4=0\\x^2-5x+6=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=4\\x=2\\x=3\end{matrix}\right.\)

a) \(\Rightarrow x\left(x+3\right)=0\)

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

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

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

c) \(\Rightarrow\left(x-1\right)\left(5x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\)

d) \(\Rightarrow2\left(x+5\right)-x\left(x+5\right)=0\Rightarrow\left(x+5\right)\left(2-x\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)

e) \(\Rightarrow2x^2-10x-3x-2x^2=26\)

\(\Rightarrow-13x=26\Rightarrow x=-2\)

f) \(\Rightarrow\left(x-2012\right)\left(5x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=2012\\x=\dfrac{1}{5}\end{matrix}\right.\)

vậy giỏi zữ vậy

So sánh điều kiện, không có giá trị nào của x thỏa mãn.

Vậy không có giá trị x nào thỏa mãn yêu cầu đề bài.

a) \(\Rightarrow5x\left(x-200\right)-\left(x-200\right)=0\)

\(\Rightarrow\left(x-200\right)\left(5x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=200\\x=\dfrac{1}{5}\end{matrix}\right.\)

b) \(\Rightarrow x\left(x^2-11\right)=0\)

\(\Rightarrow x\left(x-\sqrt{11}\right)\left(x+\sqrt{11}\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=\sqrt{11}\\x=-\sqrt{11}\end{matrix}\right.\)

a) 5x(x-200)-(x-200)=0

(x-200)(5x-1)=0

Th1 : x-200=0

X=200

Th2 : 5x-1=0

5x=1

X=1/5

Vậy S={200;1/5}

\(a,5x.\left(x-200\right)-x+200=0\)

\(\Rightarrow5x.\left(x-200\right)-\left(x-200\right)=0\)

\(\Rightarrow\left(5x-x\right).\left(x-200\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}5x-1=0\\x-200=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}5x=1\\x=200\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=200\end{matrix}\right.\)

\(b.x^3-11x=0\)

\(\Rightarrow x.\left(x^2-11\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x^2-11=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x^2=11\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\\left[{}\begin{matrix}x=\sqrt{11}\\x=-\sqrt{11}\end{matrix}\right.\end{matrix}\right.\)