Question

Giải phương trình:
`x(3-\sqrt{3x-1})=\sqrt{3x^2+2x-1}-x\sqrt{x+1}+1`

Chú Lâm cíu cháu :<

Answer 1

ĐKXĐ: ...

\(\Leftrightarrow3x-1-x\sqrt{3x-1}+x\sqrt{x+1}-\sqrt{\left(x+1\right)\left(3x-1\right)}=0\)

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

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

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

\(\Leftrightarrow...\)

Answer 2

ĐKXĐ: x \(\ge\)\(\dfrac{1}{3}\)

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

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

  \(\Leftrightarrow\)\(\left[{}\begin{matrix}\sqrt{x+1}=\sqrt{3x-1}\\1=\sqrt{3x-1}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{2}{3}\end{matrix}\right.\)(t/m x \(\ge\)\(\dfrac{1}{3}\))

Vậy.....................

Answer 3

 

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

ta có:\(x\left(3-\sqrt{3x-1}\right)\)

=\(3x-x\sqrt{3x-1}\)

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

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

Ta có \(\sqrt{3x^2+2x-1}-x\sqrt{x+1}+1\)

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

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

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

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

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

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

⇔​​\(\sqrt{3x-1}=\sqrt{x+1}\)

⇔​\(3x-1=x+1\)

⇔\(2x=2\)

⇔x=1(N)

​Vậy x=1