Question

\(x^2+\sqrt{x+7}=7\)

Answer 1

yêu cầu đề bài đâu a

Answer 2

ĐKXĐ: ...

Đặt \(\sqrt{x+7}=t\ge0\Rightarrow7=t^2-x\)

Pt trở thành:

\(x^2+t=t^2-x\Leftrightarrow x^2-t^2+x+t=0\)

\(\Leftrightarrow\left(x-t\right)\left(x+t\right)+x+t=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+t}=-x\left(x\le0\right)\\\sqrt{x+7}=x+1\left(x\ge-1\right)\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x+7=x^2\\x+7=x^2+2x+1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1-\sqrt{29}}{2}\\x=2\end{matrix}\right.\)