Question

\(\sqrt{x-2}=2x+1\)

Answer 1

`\sqrt{x-2}=2x+1`          `ĐK: x >= 2`

`<=>x-2=4x^2+4x+1`

`<=>4x^2+3x+3=0`

`<=>4(x^2+3/4x+3/4)=0`

`<=>(x+3/8)^2+39/64=0`

`<=>(x+3/8)^2=-39/64` (Vô lí vì `(x+3/8)^2 >= 0` mà `-39/64 < 0`)

Vậy ptr vô nghiệm

Answer 2

\(\sqrt{x-2}=2x+1\)

\(x-2=4x^2+4x+1\)

\(x-2-4x^2-4x-1=0\)

\(-3x-3-4x^2=0\)

\(-4x^2-3x-3=0\)

\(4x^2+3x+3=0\)

\(a=4;b=3;c=3\)

\(x=\dfrac{-3\pm\sqrt{3^2-4\times4\times3}}{2\times4}\)

\(x=\dfrac{-3\pm\sqrt{9-48}}{8}\)

\(x=\dfrac{-3\pm\sqrt{-39}}{8}\)

\(x\notin R\)