a) x = 2 7                         b) x = 2.

c) x = 2                          d) x = 1.

a) \(x^3-x^2+3x-3>0\)

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

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

Mà: \(x^2+3>0\forall x\) 

\(\Leftrightarrow x-1>0\)

\(\Leftrightarrow x>1\)

b) \(x^3+x^2+9x+9< 0\)

\(\Leftrightarrow x^2\left(x+1\right)+9\left(x+1\right)< 0\)

\(\Leftrightarrow\left(x^2+9\right)\left(x+1\right)< 0\)

Mà: \(x^2+9>0\forall x\)

\(\Leftrightarrow x+1< 0\)

\(\Leftrightarrow x< -1\)

d) \(4x^3-14x^2+6x-21< 0\)

\(\Leftrightarrow2x^2\left(2x-7\right)+3\left(2x-7\right)< 0\)

\(\Leftrightarrow\left(2x^2+3\right)\left(2x-7\right)< 0\)

Mà: \(2x^2+3>0\forall x\)

\(\Leftrightarrow2x-7< 0\)

\(\Leftrightarrow2x< 7\)

\(\Leftrightarrow x< \dfrac{7}{2}\)

d) \(x^2\left(2x^2+3\right)+2x^2>-3\)

\(\Leftrightarrow2x^4+3x^2+2x^2+3>0\)

\(\Leftrightarrow2x^4+5x^2+3>0\)

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

Mà: 

\(x^2+1>0\forall x\)

\(2x^2+3>0\forall x\)

\(\Rightarrow x\in R\)

a: =>x^2(x-1)+3(x-1)>0

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

=>x-1>0

=>x>1

b: =>x^2(x+1)+9(x+1)<0

=>(x+1)(x^2+9)<0

=>x+1<0

=>x<-1

c: 4x^3-14x^2+6x-21<0

=>2x^2(2x-7)+3(2x-7)<0

=>2x-7<0

=>x<7/2

d: =>x^2(2x^2+3)+2x^2+3>0

=>(2x^2+3)(x^2+1)>0(luôn đúng)

Bài 2 

P(x) + Q(x) =  x³ – 6x + 2 + 2x² - 4x³ + x - 5 =  - 3x³ + 2x² – 5x - 3 

P(x) - Q(x) = x³ – 6x + 2 - 2x² + 4x³ - x + 5 = 5x³ − 2x² − 7x+7

Bai 3

a)(x-8)(x³+8)=0

=>x-8=0 hoac x³+8=0

=>x   =8 hoac x³    =-8

=>x   =8 hoac x     =-2

Vậy x=8 hoặc x=-2

b)(4x-3)-(x+5)=3(10-x)

=>4x-3-x-5=30-3x

=>4x-x+3x=30+3+5

=>x(4-1+3)=38

=>6x         =38

=>x           =\(\dfrac{38}{6}\)

=>x           =\(\dfrac{19}{3}\)

Vậy x=\(\dfrac{19}{3}\)

a) 

\(x^3+\left(x-5\right)\left(x+8\right)=2x^2-37\\ \Leftrightarrow x^3+x^2+3x-40=2x^2-37\\ \Leftrightarrow x^3-x^2+3x-3=0\\ \Leftrightarrow x^2\left(x-3\right)+3\left(x-3\right)=0\\ \Leftrightarrow\left(x^2+3\right)\left(x-3\right)=0\)

Vì \(x^2+3\ge3>0\Rightarrow x-3=0\\ \Leftrightarrow x=3\)

b)

\(x\left(x-1\right)\left(x+1\right)\left(x+2\right)=24\\ \Leftrightarrow\left[x\left(x+1\right)\right]\left[\left(x-1\right)\left(x+2\right)\right]=24\\ \Leftrightarrow\left(x^2+x\right)\left(x^2+x-2\right)=24\)

Đặt \(x^2+x=y\)

\(\Rightarrow y\left(y-2\right)=24\\ \Leftrightarrow y^2-2y+1=25\\ \Leftrightarrow\left(y-1\right)^2=25\\ \Leftrightarrow\left[{}\begin{matrix}y-1=5\\y-1=-5\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}y=6\\y=-4\end{matrix}\right.\)

Nếu y = 6 

\(\Rightarrow x^2+x=6\\ \Leftrightarrow x^2+x-6=0\\ \Leftrightarrow x^2+2x-3x-6=0\\ \Leftrightarrow x\left(x+2\right)-3\left(x+2\right)=0\\ \Leftrightarrow\left(x-3\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)

Nếu y = -4

\(\Rightarrow x^2+x=-4\\ \Leftrightarrow x^2+x+\dfrac{1}{4}=-4+\dfrac{1}{4}\\ \Leftrightarrow\left(x+\dfrac{1}{2}\right)^2=-\dfrac{15}{4}\)

Mà \(\left(x+\dfrac{1}{.2}\right)^2\ge0>-\dfrac{15}{4}\)

`=> Loại`

c) Vế còn lại là bao nhiêu?

1: \(=x^2+1\)

3: \(=\left(x-y-z\right)^2\)

11: \(2x^2-12xy+18y^2\)

\(=2\left(x^2-6xy+9y^2\right)\)

\(=2\left(x-3y\right)^2\)

12: \(\left(x^2+x\right)^2+3\left(x^2+x\right)+2\)

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

a: =>2x^2+9x-6x-27=0

=>x(2x+9)-3(2x+9)=0

=>(2x+9)(x-3)=0

=>x=3 hoặc x=-9/2

b: =>-10x^2+6x-5x+3=0

=>-2x(5x-3)-(5x-3)=0

=>(5x-3)(-2x-1)=0

=>x=-1/2 hoặc x=5/3

c: =>-x^3+2x^2-x^2+4=0

=>-x^2(x-2)-(x-2)(x+2)=0

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

=>x-2=0

=>x=2

d: =>(x^3+8)-4x(x+2)=0

=>(x+2)(x^2-2x+4)-4x(x+2)=0

=>(x+2)(x^2-6x+4)=0

=>x=-2 hoặc \(x=3\pm\sqrt{5}\)

a) x = 0                 b) x = - 1 3

c) x = 28 15               d) x = -82.

a) Ta có: \(1+6x-6x^2-x^3\)

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

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

b:Ta có: \(x^3-4x^2+8x-8\)

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

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

c: Ta có: \(x^3+2x^2+2x+1\)

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

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

d: Ta có: \(8x^3-12x^2+6x-1\)

\(=\left(2x\right)^3-3\cdot\left(2x\right)^2\cdot1+3\cdot2x\cdot1^2-1^3\)

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