Question

104. Giải các phương trình:

a) \(2x^3-x^2-8x+4=0\)

b) \(3x^3+6x^2-75x-150=0\)

c) \(2x^5-3x^4+6x^3-8x^2+3=0\)

Answer 1

b)\(3x^3+6x^2-75x-150=0\Leftrightarrow3\left(x^3+2x^2-25x-50\right)=0\Leftrightarrow x^3+2x^2-25x-50=0\)

<=>\(x^2\left(x+2\right)-25\left(x+2\right)=0\Leftrightarrow\left(x^2-25\right)\left(x+2\right)=0\Leftrightarrow\left(x-5\right)\left(x+5\right)\left(x+2\right)=0\)

<=>x-5=0 hoặc x+5=0 hoặc x+2=0<=>x=5 hoặc x=-5 hoặc x=-2

c)\(2x^5-3x^4+6x^3-8x^2+3=0\Leftrightarrow2x^5+x^4-4x^4-2x^3+8x^3+4x^2-12x^2+3=0\)

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

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

<=>\(\left(2x+1\right)\left(x^4-2x^3+4x^2-6x+3\right)=0\)

<=>\(\left(2x+1\right)\left(x^4-2x^3+x^2+3x^2-6x+3\right)=0\)

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

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

Vì \(x^2\ge0\Rightarrow x^2+3\ge3>0\Rightarrow\orbr{\begin{cases}2x+1=0\\\left(x-1\right)^2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=1\end{cases}}\)

Answer 2

a) 2x³ - x² - 8x + 4 = 0

x².(2x - 1) - 4.(2x - 1) = 0

(x² - 4)(2x - 1) = 0

\(\Rightarrow\orbr{\begin{cases}x^2-4=0\\2x-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x^2=4\\x=\frac{1}{2}\end{cases}}\)

Với x² = 4

=> x = 2 hoặc x = -2

=> x = {-2 ; 2 ; \(\frac{1}{2}\))

Answer 3

a) x²(2x-1) - 4(2x-1) = 0 <=> (2x-1)(x²- 4)=0 <=> x=\(\frac{1}{2}\)hay x=-2 hay x= 2
 

 

Answer 4

Ta có : 2x³ - x² - 8x + 4 = 0 

<=> 2x³ - x² - (8x - 4) = 0 

<=> x²(2x - 1) - 4(2x - 1) = 0

<=> (2x - 1) (x² - 4) = 0

<=> \(\orbr{\begin{cases}2x-1=0\\x^2-4=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}2x=1\\x^2=4\end{cases}}\)

<=> \(\orbr{\begin{cases}x=\frac{1}{2}\\x=-2;2\end{cases}}\)

Answer 5

a) \(2x^3-x^2-8x+4=x^2\left(2x-1\right)-4\left(2x-1\right)=\left(2x-1\right)\left(x^2-4\right)=\left(2x-1\right)\left(x-2\right)\left(x+2\right).\)

\(\)=>\(\orbr{\begin{cases}x-2=0\\x+2=0\end{cases}}\),hoặc 2x-1=0

=>\(x\in\left\{-2;\frac{1}{2};2\right\}\)

b) \(3x^3+6x^2-75x-150=0< =>x^3+2x^2-25x-50=0< =>\)\(x^2\left(x+2\right)-25\left(x+2\right)=\left(x+2\right)\left(x^2-25\right)=\left(x+2\right)\left(x-5\right)\left(x+5\right)\)

\(=>x\in\left\{-2,-5,5\right\}\)

c)