Question

Giải phương trình:

\(x^2+2x+7=\left(x+3\right)\sqrt{x^2+5}\)

Answer 1

binhf phương giải pt bậc 4

Answer 2

Đặt \(\sqrt{x^2+5}=a>0\Rightarrow x^2+5=a^2\) pt trở thành:

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

\(\Delta=\left(x+3\right)^2-4\left(2x+2\right)=x^2-2x+1=\left(x-1\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}a=\frac{x+3+x-1}{2}=x+1\\a=\frac{x+3-x+1}{2}=2\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2+5}=x+1\left(x\ge-1\right)\\\sqrt{x^2+5}=2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+5=x^2+2x+1\\x^2+5=4\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x^2=-1\left(vn\right)\end{matrix}\right.\)