\(x^4+2x^3+5x^2+4x-12=0\)

\(\Leftrightarrow\)\(x^4-x^3+3x^3-3x^2+8x^2-8x+12x-12=0\)

\(\Leftrightarrow\)\(x^3\left(x-1\right)+3x^2\left(x-1\right)+8x\left(x-1\right)+12\left(x-1\right)=0\)

\(\Leftrightarrow\)\(\left(x^3+3x^2+8x+12\right)\left(x-1\right)=0\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x-1=0\\x^3+3x^2+8x+12=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x^3+2x^2+x^2+2x+6x+12=0\end{cases}}}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=1\\\left(x+2\right)\left(x^2+x+6\right)=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=1\\x=-2\\x^2+x+6=0\left(1\right)\end{cases}}\)

Giải pt ( 1 ) \(x^2+\frac{1}{2}x.2+\frac{1}{4}+\frac{23}{4}=0\)

\(\Leftrightarrow\)\(\left(x+\frac{1}{2}\right)^2+\frac{23}{4}=0\)suy ra pt ( 1 ) vô nghiệm

Vậy pt có 2 nghiệm là x = 1 ; x = -2

x⁴ + 2x³ + 5x² + 4x - 10 = 0

x⁴ - x³ + 3x³ - 3x² + 8x² - 8x + 12x - 12 = 0

<=> x³(x - 1) + 3x²(x - 1) + 8x(x - 1) + 12(x - 1) = 0

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x^3+3x^2+8x+12=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=1\\x^3+2x^2+x^2+2x+6x+10=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=1\\\left(x+2\right)+\left(x^2+x+6\right)=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=1\\x=-2\\x^2+x+6=0\left(1\right)\end{cases}}\)

Giải (1) \(x^2+\frac{1}{2}x.2+\frac{1}{4}+\frac{23}{4}=0\)

\(\Leftrightarrow\left(x+\frac{1}{2}\right)^2+\frac{23}{4}>0\Rightarrow\text{PT}\left(1\right)\)Vô nghiệm

=> PT có 2 nghiệm: \(\hept{\begin{cases}x=1\\x=-2\end{cases}}\)

Lời giải:

$x^4-2x^3+5x^2-4x-12=0$

$\Leftrightarrow (x^4-2x^3+x^2)+4(x^2-x)-12=0$

$\Leftrightarrow (x^2-x)^2+4(x^2-x)-12=0$

$\Leftrightarrow (x^2-x)^2+4(x^2-x)^2+4-16=0$

$\Leftrightarrow (x^2-x+2)^2-4^2=0$

$\Leftrightarrow (x^2-x+2-4)(x^2-x+2+4)=0$

$\Leftrightarrow (x^2-x-2)(x^2-x+6)=0$

$\Leftrightarrow (x+1)(x-2)(x^2-x+6)=0$

Dễ thấy $x^2-x+6=(x-\frac{1}{2})^2+\frac{23}{4}>0$ nên $(x+1)(x-2)=0$

$\Rightarrow x=-1$ hoặc $x=2$

xem lại câu 1

Đặt x làm thừa số chung là ra đó bạn

phương trình đa thức đối xứng

sorry, e moi hoc lop 7

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) x⁴ + 2x³ + 5x² + 4x - 12 = 0

=> x⁴ - x³+ 3x³- 3x² + 8x² -8x + 12x - 12 = 0

=> x³( x - 1) + 3x²( x - 1) + 8x( x - 1) + 12 ( x - 1 ) = 0

=> ( x - 1)( x³ + 3x² + 8x + 12 ) = 0

=> ( x - 1)( x³ + 2x² + x² + 2x + 6x + 12 ) = 0

=> ( x - 1)[ x²( x + 2) + x( x + 2) + 6( x + 2) ] = 0

=> ( x - 1)( x + 2)( x² + x + 6 ) = 0

Ta thấy : x² + x + 6

= x² + 2.\(\dfrac{1}{2}x+\dfrac{1}{4}-\dfrac{1}{4}+6=\left(x+\dfrac{1}{2}\right)^2+\dfrac{23}{4}\text{≥}\dfrac{23}{4}>0\text{∀}x\)

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

=> x = 1 hoặc x = -2

Vậy,....

b) ( x + 1)³ + ( x - 2)³ = ( 2x - 1)³

=>x³+ 3x² + 3x + 1 + x³ - 6x² + 12x - 8 - ( 8x³ - 12x² + 6x - 1)=0

=> 2x³ - 3x² + 15x - 7 - 8x³ + 12x² - 6x + 1 = 0

=> 9x² - 6x³ + 9x - 6 = 0

=> 9x( x + 1) -6( x³ + 1 ) = 0

=> 9x( x + 1) - 6( x + 1)( x² - x + 1) = 0

=> 3( x + 1)( 3x - 2x² + 2x - 2) = 0

=> 3( x + 1)( - 2x² + 5x - 2) = 0

=> 3( x + 1)( - 2x² + x + 4x - 2) = 0

=> 3( x + 1)[ x( 1 - 2x ) - 2( 1 - 2x ) ] = 0

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

Suy ra :

* x + 1 = 0 => x = -1

* 1 - 2x = 0 => x = \(\dfrac{1}{2}\)

* x - 2 = 0 => x = 2

Vậy,.....

Ta có : 2x⁴ - 5x³ + 4x² -5x +2 =0

<=> ( 2x⁴ +4x² +2) - ( 5x³ + 5x)=0

<=> 2( x⁴+2x²+1) - 5x( x² +1) =0

<=> 2 ( x²+1)² - 5x( x²+1) =0

<=> (x² +1) ( 2( x² +1) -5x ) =0

<=> 2( x² +1) -5x =0           ( vì x²  >_ 0 => x² +1 >0)

<=>2x² +2 -5x =0

<=> 2x² +2 -4x-x =0

<=> (2x²  -4x) +( 2-x) =0

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

<=> (x-2) (2x-1) = 0

<=> x-2 =0 <=> x= 2 hoặc 2x-1 =0 <=> x= 1/2 

vậy x= 2 hoặc x= 1/2 

- học tốt -