Question

Tìm x

\(2x^3-x^2+3x+6=0\)

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

Answer 1

2x^3-x^2+3x+6=0

<=> 2x^3+2x^2-3x^2-3x+6x+6=0

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

<=> 2x^2-3x+6 =0 hoặc x=-1

2x^2-3x+6=0

<=>2(x^2-3x/2+9/16+39/16)=0

<=>2(x-3/4)^2+39/8=0 ( vô lí )

vậy x=-1

Answer 2

câu b

đặt x^2+x=a

pt <=> a(a+1)=6

<=>a^2+a-6=0

<=>a=-3 hoặc a=2

th1 a=-3

<=> x^2+x+3=0

<=>x^2+x+1/4+11/4=0

<=>(x+1/2)^2+11/4=0( vô lí )

th2 a=2

=> x^2+x-2=0

=> x=1 hoặc x=-2

vậy x=1 hoăc x=-2

Answer 3

\(a,2x^3-x^2+3x+6=0\)