*vn:vô nghiệm.

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

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

\(\Leftrightarrow\left[{}\begin{matrix}\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)=0\\\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}=0\left(vn\right)\end{matrix}\right.\)

\(\Leftrightarrow x=\pm\sqrt{2}\)

-Vậy \(S=\left\{\pm\sqrt{2}\right\}\).

b. \(16x^2-8x+5=0\)

\(\Leftrightarrow16x^2-8x+1+4=0\)

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

-Vậy S=∅.

c. \(2x^3-x^2-8x+4=0\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\pm2\end{matrix}\right.\)

-Vậy \(S=\left\{\dfrac{1}{2};\pm2\right\}\).

d. \(3x^3+6x^2-75x-150=0\)

\(\Leftrightarrow3x^2\left(x+2\right)-75\left(x+2\right)=0\)

\(\Leftrightarrow3\left(x+2\right)\left(x^2-25\right)=0\)

\(\Leftrightarrow3\left(x+2\right)\left(x+5\right)\left(x-5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\pm5\end{matrix}\right.\)

-Vậy \(S=\left\{-2;\pm5\right\}\)

a,\(x\in\left\{5;1,5;\dfrac{-4}{3}\right\}\)

a: Ta có: \(x^2+3x+4=0\)

\(\text{Δ}=3^2-4\cdot1\cdot4=9-16=-7< 0\)

Do đó: Phương trình vô nghiệm

a)  \(x^3+3x^2+3x+2=0\)

<=>  \(x^3+x^2+x+2x^2+2x+2=0\)

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

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

tự làm

b) \(x^4-2x^3+2x-1=0\)

<=>  \(\left(x^4-3x^3+3x^2-x\right)+\left(x^3-3x^2+3x-1\right)=0\)

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

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

<=>  \(\left(x-1\right)^3\left(x+1\right)=0\)

tự làm

c)   \(x^4-3x^3-6x^2+8x=0\)

<=>   \(x\left(x^3-3x^2-6x+8\right)=0\)

<=>  \(x\left[\left(x^3+x^2-2x\right)-\left(4x^2+4x-8\right)\right]=0\)

<=>\(x\left[x\left(x^2+x-2\right)-4\left(x^2+x-2\right)\right]=0\)

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

<=> \(x\left(x-4\right)\left(x-1\right)\left(x+2\right)=0\)

tự làm

mk lưu nhầm ảnh ở bài dưới của câu

a ) \(x^3+3x^2+3x+2=0\)

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

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

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

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

\(\Leftrightarrow x=-2\)

Vậy \(x=-2\)

b ) \(x^4-2x^3+2x-1=0\)

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

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

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

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

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

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

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

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

a, \(x^3+3x^2+3x+2=0\)

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

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

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

b, \(x^4-2x^3+2x-1=0\)

\(\Leftrightarrow\left(x^4-x^3\right)-\left(x^3-x^2\right)-\left(x^2-x\right)+\left(x-1\right)=0\)

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

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

\(\Leftrightarrow x=1\)

a ) x^3+3x^2+3x+2=0x3+3x2+3x+2=0

\Leftrightarrow x^3+3x^2+3x+1+1=0⇔x3+3x2+3x+1+1=0

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

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

\Leftrightarrow x+1=-1⇔x+1=−1

\Leftrightarrow x=-2⇔x=−2

Vậy x=-2x=−2

b ) x^4-2x^3+2x-1=0x4−2x3+2x−1=0

\Leftrightarrow x^4-1-2x\left(x^2-1\right)=0⇔x4−1−2x(x2−1)=0

\Leftrightarrow\left(x^2-1\right)\left(x^2+1\right)-2x\left(x^2-1\right)=0⇔(x2−1)(x2+1)−2x(x2−1)=0

\Leftrightarrow\left(x^2-1\right)\left(x^2+1-2x\right)=0⇔(x2−1)(x2+1−2x)=0

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

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

\Leftrightarrow\left[{}\begin{matrix}\left(x-1\right)^3=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x=-1\end{matrix}\right.⇔[(x−1)3=0x+1=0​⇔[x−1=0x=−1​

\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.⇔[x=1x=−1​

Vậy \left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.[x=1x=−1​  

20) -5-(x + 3) = 2 - 5x ⇔ -5 - x - 3 = 2 -5x ⇔ 4x = 10 ⇔ x = \(\frac{5}{2}\)

Vậy...

1) 16 - 8x = 0 ⇔ 8(2 - x) = 0⇔ 2 - x = 0 ⇔ x = 2

Vậy phương trình có nghiệm là x = 2

1) \(x^4-6x^3-x^2+54x-72=0\)

\(\Leftrightarrow x^3\left(x-2\right)-4x^2\left(x-2\right)-9x\left(x-2\right)+36\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^3-4x^2-9x+36\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left[x^2\left(x-4\right)-9\left(x-4\right)\right]=0\)

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

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

Tự làm nốt...

2) \(x^4-5x^2+4=0\)

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

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

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

Tự làm nốt...

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

\(\Leftrightarrow x^3\left(x-2\right)-6x\left(x-2\right)-4\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^3-6x-4\right)=0\)

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

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

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

\(\Leftrightarrow\left(x-2\right)\left(x+2\right)\left(x-1-\sqrt{3}\right)\left(x-1+\sqrt{3}\right)=0\)

...

\(2x^4-13x^3+20x^2-3x-2=0\)

\(\Leftrightarrow2x^3\left(x-2\right)-9x^2\left(x-2\right)+2x\left(x-2\right)+\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(2x^3-9x^2+2x+1\right)=0\)

Bí

\(2x^3-9x^2+2x+1\)

\(=2x^3-x^2-8x^2+4x-2x+1\)

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

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

\(=\left(2x-1\right)\left(x^2-4x+4-5\right)\)

\(=\left(2x-1\right)\left[\left(x-2\right)^2-5\right]\)

.......