Question

tìm x

2x²+2²-căn bậc hai 4=x-2

Answer 1

`2x^2 + 2^2 - sqrt{4} = x-2`

`<=> 2x^2 + 4 - 2 = x-2`

`<=> 2x^2 + 4 - 2 - x +2 =0`

`<=> 2x^2 -x + 4 =0`

`=> Delta = (-1)^2 - 4.2.4 = - 31 < 0`

=> pt vô ngiệm 

`x=∅` 

Answer 2

\(2x^2+2^2-\sqrt{4}=x-2\)

\(\Leftrightarrow2x^2+4-2=x-2\)

\(\Leftrightarrow2x^2-x+4=0\)

\(\Leftrightarrow2\left(x^2-\dfrac{1}{2}x+2\right)=0\)

\(\Leftrightarrow x^2-\dfrac{1}{2}x+2=0\)

\(\Leftrightarrow x^2-2.\dfrac{1}{4}x+\left(\dfrac{1}{4}\right)^2-\left(\dfrac{1}{4}\right)^2+2=0\)

\(\Leftrightarrow\left(x-\dfrac{1}{4}\right)^2+\dfrac{31}{16}=0\) ( vô lý )

Vậy pt vô nghiệm