Question

[BiBo MoMo]

giải phương trình

\(x^2+2x+7=3\sqrt{\left(x^2+1\right)\left(x+3\right)}\)

Answer 1

[Tạ Đức Hoàng Anh]

Ta có: \(x^2+2x+7=3\sqrt{\left(x^2+1\right)\left(x+3\right)}\)               \(\left(ĐK:x\ge-3\right)\)

    \(\Leftrightarrow4x^2+8x+28=12\sqrt{\left(x^2+1\right)\left(x+3\right)}\)

    \(\Leftrightarrow4x^2+4-12\sqrt{\left(x^2+1\right)\left(x+3\right)}+9x+27=x+3\)

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

    \(\Leftrightarrow2\sqrt{x^2+1}-3\sqrt{x+3}=\sqrt{x+3}\)

    \(\Leftrightarrow2\sqrt{x^2+1}=4\sqrt{x+3}\)

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

    \(\Leftrightarrow x^2+1=4.\left(x+3\right)\)

    \(\Leftrightarrow x^2-4x-11=0\)

    \(\Leftrightarrow\left(x^2-4x+4\right)-15=0\)

    \(\Leftrightarrow\left(x-2\right)^2=15\)

    \(\Leftrightarrow x-2=\pm\sqrt{15}\)

 +  \(x-2=\sqrt{15}\)\(\Leftrightarrow\)\(x=2+\sqrt{15}\)\(\left(TM\right)\)

 +  \(x-2=-\sqrt{15}\)\(\Leftrightarrow\)\(x=2-\sqrt{15}\)\(\left(TM\right)\)

Vậy \(S=\left\{2+\sqrt{15};2-\sqrt{15}\right\}\)

Answer 2

[Nguyễn Thùy Trang ( team ASL)]

Đặt \(\hept{\begin{cases}\sqrt{x^2+1}=a\\\sqrt{x+3}=b\end{cases}\Rightarrow\hept{\begin{cases}x^2+1=a^2\\x+3=b^2\end{cases}\left(a,b>0\right)}}\)

Có phương trình mới : \(a^2+2b^2=3ab\)

\(\Leftrightarrow\left(a-b\right)\left(a-2b\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}a=b\\a=2b\end{cases}}\)

Với \(a=b\Leftrightarrow\sqrt{x^2+1}=\sqrt{x+3}\)

\(\Leftrightarrow x^2+1=x+3\)

\(\Leftrightarrow x^2-x-2=0\Leftrightarrow\orbr{\begin{cases}x=2\\x=-1\end{cases}}\)thỏa mãn

Với \(a=2b\Leftrightarrow\sqrt{x^2+1}=2\sqrt{x+3}\)

\(\Leftrightarrow x^2+1=4x+12\)

\(\Leftrightarrow x^2-4x-11=0\Leftrightarrow\orbr{\begin{cases}x=2+\sqrt{15}\\x=2-\sqrt{15}\end{cases}}\)thỏa mãn