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

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 \(x=1\)

\(ĐK:x\ge\dfrac{2}{3}\)

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

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

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

\(\Leftrightarrow x-2=-\sqrt{3x-2}\)

\(\Leftrightarrow2-x=\sqrt{3x-2}\)

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

\(\Leftrightarrow4-4x+x^2=3x-2\)

\(\Leftrightarrow x^2-7x+6=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=6\end{matrix}\right.\) (vi-et )

Vậy S=\(\left(1;6\right)\)

45. I know how to use this machine, I can help you.

46. excited about the journey.

47. Peter is the best student in my class who can solve this difficult problem.

48. very good at typing.

Em cần giúp điều gì em nhỉ?

Bài 1:

Bài 2:

Bài 3:

Bài 4:

Câu 2:

=>x=1-y và m(1-y)-y=2m

=>x=1-y và m-my-y=2m

=>x=1-y và y(-m-1)=m

=>x=1-y và y=-m/m+1

=>x=1+m/m+1=(m+2)/m+1 và y=-m/m+1

Để x,y nguyên thì m+1+1 chia hết cho m+1 và -m-1+1 chia hết cho m+1

=>\(m+1\in\left\{1;-1\right\}\)

mà m<>0

nên m=-2