Question

giải phương trình \(\left(x-1\right)\sqrt{2\left(x^2+3\right)}=\left(x-1\right)\left(x-2\right)\)

Answer 1

`(x-1)\sqrt{2(x^2+3)}=(x-1)(x-2)`

`<=>(x-1)(\sqrt{2(x^2+3)}-x+2)=0`

`@TH1:x-1=0<=>x=1`

`@TH2:\sqrt{2(x^2+3)}-x+2=0`

`<=>\sqrt{2x^2+6}=x-2`  , `x >= 2`

`<=>2x^2+6=x^2-4x+4`

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

Ptr có:`\Delta'=2^2-2=2 > 0`

`=>` Ptr có `2` nghiệm pb

`x_1=[-b'+\sqrt{\Delta'}]/a=-2+\sqrt{2}` (ko t/m) 

`x_2=[-b'-\sqrt{\Delta'}]/a=-2-\sqrt{2}` (ko t/m)

Vậy `S={1}`

Answer 2

\(PT\Leftrightarrow\left(x-1\right)\left(\sqrt{2x^2+6}-x+2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=1\left(\text{nhận}\right)\\\sqrt{2x^2+6}=x-2\left(1\right)\end{matrix}\right.\\ \left(1\right)\Leftrightarrow2x^2+6=x^2-4x+4\left(x\ge2\right)\\ \Leftrightarrow x^2+4x+2=0\\ \Rightarrow\left[{}\begin{matrix}x=-2+\sqrt{2}\left(\text{loại}\right)\\x=-2-\sqrt{2}\left(\text{loại}\right)\end{matrix}\right.\)

Vậy \(S=\left\{1\right\}\)