Question

\(\sqrt{x^2-40x+4000}-\sqrt{x^2+3600}=10\)

Answer 1

TXĐ: D=R

\(\Leftrightarrow\sqrt{x^2-40x+4000}=\sqrt{x^2+3600}+10\)

\(\Leftrightarrow x^2-40x+4000=x^2+3600+20\sqrt{x^2+3600}+100\)

\(\Leftrightarrow15-2x=\sqrt{x^2+3600}\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le\dfrac{15}{2}\\\left(15-2x\right)^2=x^2+3600\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le\dfrac{15}{2}\\3x^2-60x-3375=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-25\\x=45\left(loại\right)\end{matrix}\right.\)