Question

Answer 1

i: =>2x^2-4x+x-2=0

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

=>x=2 hoặc x=-1/2

m: =>2x^2+5x-2x-5=0

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

=>x=1 hoặc x=-5/2

h: =>(căn x-1)(căn x-4)=0

=>x=1 hoặc x=16

e: =>6x^2-81x+6x-81=0

=>(6x-81)(x+1)=0

=>x=-1 hoặc x=13,5

b: =>x^2-12x+2x-24=0

=>(x-12)(x+2)=0

=>x=12 hoặc x=-2

c: \(\text{Δ}=\left(-2\sqrt{3}\right)^2-4\cdot3\cdot\left(-2\right)=12+24=36>0\)

PT sẽ có hai nghiệm là;

\(\left\{{}\begin{matrix}x_1=\dfrac{2\sqrt{3}-6}{6}\\x_2=\dfrac{2\sqrt{3}+6}{6}\end{matrix}\right.\)

f: =>(x^2-1)(x^2-4)=0

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

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

g: =>3x^4-6x^2+x^2-2=0

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

=>x^2-2=0

=>\(x=\pm\sqrt{2}\)