Bài 3: 

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

hay \(x\in\left\{0;-1\right\}\)

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

=>x-1=0

hay x=1

d: \(\Leftrightarrow6x^2-3x-4x+2=0\)

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

hay \(x\in\left\{\dfrac{1}{2};\dfrac{2}{3}\right\}\)

\(\dfrac{x-4}{x+2}+\dfrac{x+3}{x+4}=\dfrac{2x+1}{x^2+6x+8}\\ =\dfrac{x-4.2}{x+4}+\dfrac{x+3}{x+4}=\dfrac{2x+1}{x^2+6x+8}\\ =\dfrac{x-8+x+3}{x+4}=\dfrac{2x+1}{x^2+6x+8}\\ =\dfrac{2x-5}{x+4}=\dfrac{2x+1}{x^2+6x+8}\)

2:

a: =>2x^2-4x-2=x^2-x-2

=>x^2-3x=0

=>x=0(loại) hoặc x=3

b: =>(x+1)(x+4)<0

=>-4<x<-1

d: =>x^2-2x-7=-x^2+6x-4

=>2x^2-8x-3=0

=>\(x=\dfrac{4\pm\sqrt{22}}{2}\)

a: \(\Leftrightarrow x^2+6x+9=0\)

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

=>x+3=0

hay x=-3

b: \(\Leftrightarrow x^2+x-12-6x+4=x^2-8x+16\)

=>-7x+8=-8x+16

=>x=8

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

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

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

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

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

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

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

.......

`|5x| = - 3x + 2`

Nếu `5x>=0<=> x>=0` thì phương trình trên trở thành :

`5x =-3x+2`

`<=> 5x +3x=2`

`<=> 8x=2`

`<=> x= 2/8=1/4` ( thỏa mãn )

Nếu `5x<0<=>x<0` thì phương trình trên trở thành :

`-5x = -3x+2`

`<=>-5x+3x=2`

`<=> 2x=2`

`<=>x=1` ( không thỏa mãn ) 

Vậy pt đã cho có nghiệm `x=1/4`

__

`6x-2<5x+3`

`<=> 6x-5x<3+2`

`<=>x<5`

Vậy bpt đã cho có tập nghiệm `x<5`

\(4x^4+4x^3+x^2+3x\ge0\)

\(4x^4+4x^2+1-\left(2x^4+6x^3-2x^2+4x-1\right)=\left(x^2-x+1\right)\sqrt{\left(x^2-x+1\right)\left(2x^2+1\right)+2x^4+6x^3-2x^3+4x-1}\)

\(\Leftrightarrow\left(2x^2+1\right)^2-\left(2x^4+6x^3-2x^2+4x-1\right)=\left(x^2-x+1\right)\sqrt{\left(x^2-x+1\right)\left(2x^2+1\right)+2x^4+6x^3-2x^3+4x-1}\)

\(2x^2+1=u;\sqrt{4x^4+4x^3+x^2+3x}=v\left(u>0;v>0\right)\)

\(\hept{\begin{cases}u^2-\left(2x^4+6x^3-2x^2+4x-1\right)=\left(x^2-x+1\right)v\\v^2-\left(2x^4+6x^3-2x^2+4x-1\right)=\left(x^2-x+1\right)u\end{cases}\Rightarrow u^2-v^2=\left(x^2-x+1\right)\left(v-u\right)\Leftrightarrow\orbr{\begin{cases}u=v\\u+v+x^2-x+1=0\end{cases}}}\)

x4−3x3−2x2+6x+4=0x4−3x3−2x2+6x+4=0

⇔x4−2x3−2x2−x3+2x2+2x−2x2+4x+4=0⇔x4−2x3−2x2−x3+2x2+2x−2x2+4x+4=0

⇔x2(x2−2x−2)−x(x2−2x−2)−2(x2−2x−2)=0⇔x2(x2−2x−2)−x(x2−2x−2)−2(x2−2x−2)=0

⇔(x2−x−2)(x2−2x−2)=0⇔(x2−x−2)(x2−2x−2)=0

⇔(x+1)(x−2)(x−1−√3)(x−1+√3)=0⇔(x+1)(x−2)(x−1−3)(x−1+3)=0

⇔⎡⎢ ⎢ ⎢ ⎢⎣x=−1x=2x=1+√3x=1−√3

tl

x4−3x3−2x2+6x+4=0x4−3x3−2x2+6x+4=0

⇔x4−2x3−2x2−x3+2x2+2x−2x2+4x+4=0⇔x4−2x3−2x2−x3+2x2+2x−2x2+4x+4=0

⇔x2(x2−2x−2)−x(x2−2x−2)−2(x2−2x−2)=0⇔x2(x2−2x−2)−x(x2−2x−2)−2(x2−2x−2)=0

⇔(x2−x−2)(x2−2x−2)=0⇔(x2−x−2)(x2−2x−2)=0

⇔(x+1)(x−2)(x−1−√3)(x−1+√3)=0⇔(x+1)(x−2)(x−1−3)(x−1+3)=0

⇔⎡⎢ ⎢ ⎢ ⎢⎣x=−1x=2x=1+√3x=1−√3

^HT^

Ta thấy x=0 không là nghiệm của phương trình

chia cả 2 vế cho x^2 ta được:

PT <=> x^2-3x-6+3/x+1/(x^2)=0

       <=> (x^2-2+1/(x^2))-3(x-1/x)-4=0

      <=> (x-1/x)^2-3(x-1/x)-4=0

Đặt x-1/x=y

PT <=> y^2-3y-4=0

     <=> y=-4 hoặc y=1

Tại y=-4 , ta có x+1/x+4=0

                       <=> x^2+4x+1=0

                       <=> x=-2+ √3 hoăc x=-2-  √ 3

Tại y=1 ta có x^2-x-1=0

                 <=> x=(1- √  5)/2 hoặc x=(1+  √5)/2