\sqrt{3x-2}=x-1` `ĐK: x >= 1`
`<=>3x-2=x^2-2x+1`
`<=>x^2-5x+3=0`
Ptr có: `\Delta=(-5)^2-4.3=13 > 0`
`=>` Ptr có `2` nghiệm pb
`x_1=[-b+\sqrt{\Delta}]/[2a]=[5+\sqrt{13}]/2` (t/m)
`x_2=[-b-\sqrt{\Delta}]/[2a]=[5-\sqrt{13}]/2` (ko t/m)
Vậy `S={[5+\sqrt{13}]/2}`
`\sqrt{3x-2}=x-1` `ĐK: x >= 1`
`<=>3x-2=x^2-2x+1`
`<=>x^2-5x+3=0`
Ptr có: `\Delta=(-5)^2-4.3=13 > 0`
`=>` Ptr có `2` nghiệm pb
`x_1=[-b+\sqrt{\Delta}]/[2a]=[5+\sqrt{13}]/2` (t/m)
`x_2=[-b-\sqrt{\Delta}]/[2a]=[5-\sqrt{13}]/2` (ko t/m)
Vậy `S={[5+\sqrt{13}]/2}`
\(đkx\ge\dfrac{2}{3}\\ \left(\sqrt{3x-2}\right)^2=\left(x-1\right)^2\\ 3x-2=x^2-2x+1\\ 3x-2-x^2+2x-1=0\\ -x^2+5x-3=0\\ =>x=\dfrac{5+\sqrt{13}}{2}\left(thoaman\right);x=\dfrac{5-\sqrt{13}}{2}\left(kothoaman\right)\)
\(\sqrt{3x-2}=x-1\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-1\ge0\\3x-2=\left(x-1\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\3x-2=x^2-2x+1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\x^2-5x+3=0\end{matrix}\right.\)
Ta có: \(\Delta=5^2-4.1.3=25-12=13>0\Rightarrow\) Phương trình có hai nghiệm phân biệt là: \(\left[{}\begin{matrix}x=\dfrac{5+\sqrt{13}}{2}\\x=\dfrac{5-\sqrt{13}}{2}\end{matrix}\right.\)
So với điều kiện xác định thì \(x=\dfrac{5+\sqrt{13}}{2}\) thỏa mãn.
Vậy phương trình có nghiệm duy nhất là \(x=\dfrac{5+\sqrt{13}}{2}.\)
\(\sqrt{3x-2}=x-1< =>\left\{{}\begin{matrix}x-1\ge0\\3x-2=\left(x-1\right)^2\end{matrix}\right.\\ < =>\left\{{}\begin{matrix}x\ge1\\x^2-2x+1=3x-2\end{matrix}\right.< =>\left\{{}\begin{matrix}x\ge1\\x^2-5x+3=0\end{matrix}\right.\\ < =>\left\{{}\begin{matrix}x\ge1\\\left[{}\begin{matrix}x=\dfrac{5+\sqrt{13}}{2}\\x=\dfrac{5-\sqrt{13}}{2}\end{matrix}\right.\end{matrix}\right.< =>x=\dfrac{5+\sqrt{13}}{2}\)
