Question

tìm x biết :

a,\(\sqrt{x^2-2x}+1=5\)

b, \(\sqrt{x^2-2x}+1=x-1\)

Answer 1

a)

\(\sqrt{x^2-2x}+1=5\)(với \(\left[{}\begin{matrix}x\ge2\\x\le0\end{matrix}\right.\))

\(\Leftrightarrow x^2-2x-16=0\)

\(\Leftrightarrow\left(x-1\right)^2=17\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\sqrt{17}+1\\x=-\sqrt{17}+1\end{matrix}\right.\)(t/m)

Vậy.........

b)

\(\sqrt{x^2-2x}+1=x-1\)(với \(\left[{}\begin{matrix}x\ge2\\x\le0\end{matrix}\right.\))

\(\Leftrightarrow x^2-2x=x^2-4x+4\)

\(\Leftrightarrow2x=4\Leftrightarrow x=2\)(t/m)

Vậy..........