Question

[Quoc Tran Anh Le]

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Đề số 2 trong bộ đề Toán chuyên, gửi tới các bạn.

Answer 1

[Hồng Phúc]

I.1.

ĐK:  \(x\in R\)

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

\(\Leftrightarrow2x^2+6x+2=2\left(x+3\right)\sqrt{x^2+1}\)

\(\Leftrightarrow x^2+1+x^2+6x+9-2\left(x+3\right)\sqrt{x^2+1}=8\)

\(\Leftrightarrow\left(x+3-\sqrt{x^2+1}\right)^2=8\)

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

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

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

\(\Leftrightarrow\left\{{}\begin{matrix}x+3-2\sqrt{2}\ge0\\x^2+1=x^2+2\left(3-2\sqrt{2}\right)x+17-12\sqrt{2}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge2\sqrt{2}-3\\2\left(3-2\sqrt{2}\right)x=12\sqrt{2}-16\end{matrix}\right.\)

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

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

\(\Leftrightarrow\left\{{}\begin{matrix}x+3+2\sqrt{2}\ge0\\x^2+1=x^2+2\left(3+2\sqrt{2}\right)x+17+12\sqrt{2}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-3-2\sqrt{2}\\2\left(3+2\sqrt{2}\right)x=-16-12\sqrt{2}\end{matrix}\right.\)

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

Vậy phương trình có nghiệm \(x=\pm2\sqrt{2}\)

Answer 2

[Nguyễn Ngọc Lộc]

Câu 1 :

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

- Đặt \(\sqrt{x^2+1}=a\left(a\ge0\right)\)

PT TT : \(a^2+3x=a\left(x+3\right)\)

\(\Leftrightarrow a^2-ax-3a+3x=0\)

\(\Leftrightarrow a^2-a\left(x+3\right)+3x=0\)

Có : \(\Delta=b^2-4ac=\left(a+3\right)^2-4.3a=a^2+6a+9-12a\)

\(=a^2-6a+9=\left(a-3\right)^2\ge0\forall a\)

TH1 : \(\Delta=0\Rightarrow a=3\left(TM\right)\)

\(\Rightarrow\sqrt{x^2+1}=3\)

\(\Rightarrow x=\pm2\sqrt{2}\)

TH2 : \(\Delta>0\)

=> Pt có 2 nghiệm phân biệt :\(\left\{{}\begin{matrix}a=\dfrac{x+3+\sqrt{\left(x-3\right)^2}}{2}\\a=\dfrac{x+3-\sqrt{\left(x-3\right)^2}}{2}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\sqrt{x^2+1}=\dfrac{x+3+\left|x-3\right|}{2}\\\sqrt{x^2+1}=\dfrac{x+3-\left|x-3\right|}{2}\end{matrix}\right.\)

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

\(\Rightarrow\left\{{}\begin{matrix}x^2+1=9\\x^2+1=x^2\end{matrix}\right.\)

\(\Rightarrow x=\pm2\sqrt{2}\)

Vậy phương trình có tập nghiệm là \(S=\left\{\pm2\sqrt{2}\right\}\)