Question

[chi nguyen]

giai pt vo ti sau

\(\sqrt{2x^2+8x+6}-\sqrt{x^2-1}=2x+2\)

Answer 1

[Nguyễn Việt Lâm]

ĐKXĐ: \(x\ge1;x\le-3;x=-1\)

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

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

\(\left(1\right)\Leftrightarrow x+1=0\Rightarrow x=-1\)

\(\left(2\right)\Leftrightarrow\sqrt{2x+6}=\sqrt{x-1}+2\sqrt{x+1}\)

\(\Leftrightarrow2x+6=x-1+4\sqrt{\left(x-1\right)\left(x+1\right)}+4x+4\)

\(\Leftrightarrow4\sqrt{x^2-1}=3-3x\) \(\Leftrightarrow\left\{{}\begin{matrix}3-3x\ge0\\16\left(x^2-1\right)=\left(3-3x\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le1\\7x^2+18x-25=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{-25}{7}\end{matrix}\right.\)

Vậy pt có 3 nghiệm: \(x=-1;1;\dfrac{-25}{7}\)

Answer 2

[chi nguyen]

thank