Question

\(\sqrt{x^2-6x+9}+2x=4\)

Giải pt

Answer 1

\(\sqrt{x^2-6x+9}+2x=4\)

\(\Leftrightarrow\sqrt{\left(x-3\right)^2}=4-2x\)

\(\Leftrightarrow\left|x-3\right|=4-2x\)

\(\left|x-3\right|=\left\{{}\begin{matrix}4-2xkhix\ge2\\-4+2xkhix< 2\end{matrix}\right.\)

Với \(x\ge2\Rightarrow x-3=4-2x\Rightarrow3x=7\Rightarrow x=\dfrac{7}{3}\left(tm\right)\)

Với \(x< 2\Rightarrow x-3=-4+2x\Rightarrow-x=-1\Rightarrow x=1\left(tm\right)\)

Vậy \(S=\left\{-1;\dfrac{7}{3}\right\}\)

Answer 2

ĐKXĐ: `x\inRR`

`pt<=>sqrt(x^2-6x+9)=4-2x`

`<=>sqrt((x-3)^2)=4-2x`

`<=>|x-3|=4-2x(**)`

Ta thấy rằng `VT(**)>=0AAx\inRR` nên `4-2x>=0<=>x<=2`

Khi đó `|x-3|=3-x`

Suy ra `3-x=4-2x`

`<=>x=1(TM)`

Vậy `S={1}`