a: =>(x-1)(3x-4)>0

=>x>4/3 hoặc x<1

b: =>x^3-3x^2-10x^2+30x+12x-36>0

=>(x-3)(x^2-10x+12)>0

Th1: x-3>0và x^2-10x+12>0

=>x>5+căn 13

TH2: x-3<0 và x^2-10x+12<0

=>x<3 và 5-căn 13<x<5+căn 13

=>3<x<5+căn 13

a-b=5 và (a,b)/[a,b]. Tim a:b

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

TH1 : \(x=-\frac{1}{2}\); TH2 : \(x^2=-2\)vô lí vì \(x^2\ge0\forall x;-2< 0\)

b, \(\left(x^2+4\right)\left(7x-3\right)=0\)

TH1 : \(x^2=-4\)vô lí vì \(x^2\ge0\forall x;-4< 0\)

TH2 : \(x=\frac{3}{7}\)

c, \(\left(x^2+x+1\right)\left(6-2x\right)=0\)

TH1 : \(x^2+x+1\ne0\)vì \(x^2+x+\frac{1}{4}+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\)

TH2 : \(2x=6\Leftrightarrow x=3\)

d, \(\left(8x-4\right)\left(x^2+2x+2\right)=0\)

TH1 : \(x=\frac{1}{2}\)

TH2 : \(x^2+2x+2\ne0\)vì \(x^2+2x+1+1=\left(x+1\right)^2+1>0\)

dễ

vãi                       b b b b b b

a) Đề:..............

\(\Rightarrow\hept{\begin{cases}2x+1=0\\x^2+2=0\end{cases}\Rightarrow\hept{\begin{cases}2x=1\\x^2=-2\left(vl\right)\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{1}{2}\\\end{cases}}}}\)

Vậy phương trình có nghiệm S = {1/2}

b) Đề:...............

\(\Rightarrow\hept{\begin{cases}x^2+4=0\\7x-3=0\end{cases}\Rightarrow\hept{\begin{cases}x^2=-4\left(vl\right)\\7x=3\end{cases}\Rightarrow}\hept{\begin{cases}x=\frac{3}{7}\\\end{cases}}}\)

Vậy phương trình có nghiệm S = {3/7}

\(2x^3+7x^2+7x+2=0\)

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

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

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

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

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

.......................................................................................

\(x^3-8x^2-8x+1=0\)

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

......................................................................................

cảm ơn 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

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

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

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

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

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

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

.......

Bài 1: 

b: \(\Leftrightarrow x-2=0\)

hay x=2

a) \(x^4+2x^3-3x^2-8x-4=0\)

\(\Leftrightarrow x^4-2x^3+4x^3-8x^2+5x^2-10x+2x-4=0\)

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

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

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

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

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

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

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

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

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

\(\Rightarrow x\in\left\{2;-1;-2\right\}\)

Vậy....

c, \(2x^3+7x^2+7x+2=0\)

\(\Leftrightarrow2\left(x^3+1\right)+7x\left(x+1\right)=0\Leftrightarrow2\left(x+1\right)\left(x^2-x+1\right)+7x\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left[2\left(x^2-x+1\right)+7x\right]=0\)

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

Tập nghiệm của pt: \(S=\left\{-1;-2;-\frac{1}{2}\right\}\)

b, \(\left(x-2\right)\left(x+2\right)\left(x^2-10\right)=72\Leftrightarrow\left(x^2-4\right)\left(x^2-10\right)=72\) (1)

Đặt: \(x^2-7=t\left(t\ge-7\right)\)

Khi đó (1) trở thành: \(\left(t+3\right)\left(t-3\right)=72\Leftrightarrow t^2-9=72\Leftrightarrow\orbr{\begin{cases}t=9\\t=-9\left(loai\right)\end{cases}}\)

\(t=9\Rightarrow x^2-7=9\Leftrightarrow x=\pm4\)

Tập nghiệm của pt là \(S=\left\{\pm4\right\}\)

a, \(x^4+2x^3-3x^2-8x-4=0\)

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

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

\(\Leftrightarrow\left(x+1\right)^2\left(x^2-4\right)=0\Leftrightarrow\orbr{\begin{cases}x=-1\\x=\pm2\end{cases}}\)