dài quá bạn! tách ra nha

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

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

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

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

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

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

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

.......

\(\left(3x-4\right)\left(2x+1\right)\left(5x-2\right)=0\)

\(\Rightarrow\hept{\begin{cases}3x-4=0\\2x+1=0\\5x-2=0\end{cases}\Rightarrow}\hept{\begin{cases}3x=4\\2x=-1\\5x=2\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{4}{3}\\x=-\frac{1}{2}\\x=\frac{2}{5}\end{cases}}}\)

Vậy ...

Ối ối nhầm rồi :(

\(\left(3x-4\right)\left(2x+1\right)\left(5x-2\right)=0\)

\(\Rightarrow\hept{\begin{cases}3x-4=0\\2x+1=0\\5x-2=0\end{cases}\Rightarrow\hept{\begin{cases}3x=4\Leftrightarrow x=\frac{4}{3}\\2x=-1\Leftrightarrow x=-\frac{1}{2}\\5x=2\Leftrightarrow x=\frac{2}{5}\end{cases}}}\)

Vậy ... là nghiệm của pt

(3x - 4)(2x + 1)(5x - 2) = 0

<=> 3x - 4 = 0 hoặc 2x + 1 = 0 hoặc 5x - 2 = 0

<=> x = 4/3 hoặc x = -1/2 hoặc x = 2/5

a: TH1: x>=3 hoặc x<=0

=>x^2-3x=5x

=>x^2-8x=0

=>x=0(nhận) hoặc x=8(nhận)

TH2: 0<x<3

=>3x-x^2=5x

=>-x^2-2x=0

=>x=0(loại) hoặc x=-2(loại)

b: TH1: x>=0 hoặc x<=-5

=>x^2+5x=6x

=>x^2-x=0

=>x=0(nhận) hoặc x=1(nhận)

TH2: -5<x<0

=>-x^2-5x=6x

=>-x^2-11x=0

=>x=0(loại) hoặc x=-11(loại)

c: TH1: x>=0 hoặc x<=-2

=>x^2+2x=-x

=>x^2+3x=0

=>x=0(nhận) hoặc x=-3(nhận)

TH2: -2<x<0

=>-x^2-2x=x

=>x^2+2x=x

=>x^2+x=0

=>x=0(loại) hoặc x=-1(nhận)

c: TH1: x>=1 hoặc x<=0

=>x^2-x=x-1

=>x^2-2x+1=0

=>x=1(nhận)

TH2: 0<x<1

=>-x^2+x=x-1

=>-x^2=-1

=>x=1(loại) hoặc x=-1(loại)