a: \(x^3-4x^2-x+4=0\)

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

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

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

=>\(\left[{}\begin{matrix}x-4=0\\x^2-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x^2=1\end{matrix}\right.\Leftrightarrow x\in\left\{2;1;-1\right\}\)

b: Sửa đề: \(x^3+3x^2+3x+1=0\)

=>\(x^3+3\cdot x^2\cdot1+3\cdot x\cdot1^2+1^3=0\)

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

=>x+1=0

=>x=-1

c: \(x^3+3x^2-4x-12=0\)

=>\(\left(x^3+3x^2\right)-\left(4x+12\right)=0\)

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

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

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

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

d: \(\left(x-2\right)^2-4x+8=0\)

=>\(\left(x-2\right)^2-\left(4x-8\right)=0\)

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

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

=>(x-2)(x-6)=0

=>\(\left[{}\begin{matrix}x-2=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=6\end{matrix}\right.\)

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

\(\left[\left(2x\right)^2+2.2x.1+1\right]-4=0\)

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

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

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

\(\Rightarrow\orbr{\begin{cases}2x-1=0\\2x+3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{3}{2}\end{cases}}}\)

Vậy \(\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{3}{2}\end{cases}}\)

\(x^4-3x^3-x+3=0\)

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

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

\(\Rightarrow\orbr{\begin{cases}x-3=0\\x^3-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}}\)

Vậy \(\orbr{\begin{cases}x=3\\x=1\end{cases}}\)

\(x^2.\left(x-1\right)-4x^2+8x-4=0\)

\(x^2.\left(x-1\right)-\left[\left(2x\right)^2-2.2x.2+2^2\right]=0\)

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

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

\(\left(x-1\right).\left[x^2-4.\left(x-1\right)\right]=0\)

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

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

\(\Rightarrow\orbr{\begin{cases}x-1=0\\x-2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=2\end{cases}}}\)

Vậy \(\begin{cases}x=1\\x=2\end{cases}\)

Tham khảo nhé~

`4x(x-5)-(x-1) (4x-3)-5=0`

`=> 4x*x - 4x*5 - ( x*4x-3*x-1*4x+ 1*3) -5=0`

`=> 4x^2 - 20x-(4x^2 -3x-4x+3)-5=0`

`=>  4x^2 - 20x-4x^2+3x+4x-3-5=0`

`=>-13x-8=0`

`=> -13x=8`

`=> x=-8/13`

Vậy `x=-8/13`

`4x(x-5)-(x-1)(4x-3)-5 = 0`

`=> 4x^2 - 20x - (4x^2 -3x-4x+3)= 5`

`=> 4x^2 - 20x - 4x^2 + 3x + 4x -3 = 5`

`=> (4x^2 - 4x^2) - (20x - 3x - 4x) = 8`

`=> -13x = 8`

`=> x    = -8/13`

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

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

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

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

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

\(\Leftrightarrow\)\(\orbr{\begin{cases}-x-3=0\\3x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-3\\x=\frac{1}{3}\end{cases}}}\)

Vậy \(x=-3\) hoặc \(x=\frac{1}{3}\)

Chúc bạn học tốt ~ 

TH1:\(x\ge\frac{1}{4}\) khi đó phương trình tương đương với:

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

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

\(\Leftrightarrow16x^2-12x+2=0\)

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

\(\Rightarrow x=\frac{1}{2};x=\frac{1}{4}\left(TM\right)\)

Tương tự với TH còn lại

a) \(x^3+4x=0\)

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

\(\Rightarrow\left[\begin{array}{nghiempt}x=0\\x^2+4=0\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=0\\x^2=-4\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=0\\x\in\phi\end{array}\right.\)

Vậy: \(x=0\)

b) \(2\left(5-x\right)=4x-3\)

\(\Rightarrow10-2x=4x-3\)

\(\Rightarrow10+3=4x+2x\)

\(\Rightarrow13=6x\)

\(\Rightarrow x=\frac{13}{6}\)

x³+ 4x=0

<=> x(x²+4)=0

=> x=0 hoặc x²+4=0

Mà: x²+4 >4

=>x=0

b) (5-x).2=4x-3

<=>10-2x=4x-3

<=>10+3= 2x+4x

<=>13=6x

=>x=\(\frac{13}{6}\)