Question

Giải phương trình

1) x⁴+x²+6x- 8=0

2) ( x²+x+1) ( x²+x+2)=12

3) (x²+x-2) ( x²+x-3)=12

Answer 1

a/ \(x^4+x^2+6x-8=0\Leftrightarrow\left(x^4-16\right)+\left(x^2-x\right)+\left(2x-2\right)+\left(5x+10\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+2\right)\left(x^2+4\right)+x\left(x-1\right)+2\left(x-1\right)+5\left(x+2\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left[\left(x-2\right)\left(x^2+4\right)+x-1+5\right]=0\)

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

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

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

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

Vậy x = -2; x =1

Answer 2

b/ đặt x² + x + 1 = t    có:

t (t + 1) = 12

<=> t² + t - 12 = 0

<=> (t² - 16) + (t + 4) = 0

<=> (t - 4) (t + 4) + (t + 4) = 0

<=> (t + 4) (t - 4 + 1) = 0

<=> (t + 4) (t - 3) = 0

=> t = -4; t = 3

thay t = x² + x + 1 đc:

      x² + x + 1 = -4          ;          x² + x + 1 = 3

<=> x² + x + 5 = 0                  <=>   x² + x - 2 = 0

 <=> x (loại)                             <=>  (x² - 1) + (x - 1) = 0

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

                                               <=> x = 1; x = -2

c/ đặt x² + x - 2 = a    có:

a (a - 1) = 12

<=> a² - a - 12 = 0

<=> (a² - 16) - (a - 4) = 0

làm tương tự câu b

..........