Question

[The Moon]

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Answer 1

[Ami Mizuno]

Ta có: \(x^2-3x+2=\left(1-x\right)\sqrt{3x-2}\) \(\left(x\ge\dfrac{2}{3}\right)\)

\(\Leftrightarrow\left(x-1\right)\left(x-2\right)-\left(1-x\right)\sqrt{3x-2}=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-2+\sqrt{3x-2}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-2+\sqrt{3x-2}=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(TM\right)\\\sqrt{3x-2}=2-x\left(1\right)\end{matrix}\right.\)

Xét (1) ta có: \(\left\{{}\begin{matrix}2-x\ge0\\3x-2=4-4x+x^2\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}2\ge x\\x^2-7x+6=0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x\le2\\\left[{}\begin{matrix}x=6\left(KTM\right)\\x=1\left(TM\right)\end{matrix}\right.\end{matrix}\right.\)

Vậy nghiệm của phương trình là x=1

Answer 2

[Xyz OLM]

ĐKXĐ : x \(\ge\dfrac{2}{3}\)

Ta có \(\left(x-1\right)\left(x-2\right)=\sqrt{3x-2}\left(1-x\right)\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\\sqrt{3x-2}=2-x\end{matrix}\right.\)

Khi x - 1 = 0 <=> x = 1 (tm)

Khi \(\sqrt{3x-2}=2-x\)

<=> \(\left\{{}\begin{matrix}3x-2=x^2-4x+4\\\dfrac{2}{3}\le x\le2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x^2-7x+6=0\\\dfrac{2}{3}\le x\le2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)\left(x-6\right)=0\\\dfrac{2}{3}\le x\le2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=1\\x=6\end{matrix}\right.\\\dfrac{2}{3}\le x\le2\end{matrix}\right.\Leftrightarrow x=1\)

Vậy phương trình 1 nghiêm \(x=1\)