Question

Tìm x

d, \(\sqrt{x-2\sqrt{x-1}=\sqrt{x-1}-1}\)

e, \(\sqrt{1-12x+36x^2}=5\)

g, \(\sqrt{23+8\sqrt{7}}-\sqrt{7}=4\)

Answer 1

- Đề sai nhiều vậy sửa lại đi bạn ;-;

Answer 2

e) Ta có: \(\sqrt{1-12x+36x^2}=5\)

\(\Leftrightarrow\left|6x-1\right|=5\)

\(\Leftrightarrow\left[{}\begin{matrix}6x-1=5\\6x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}6x=6\\6x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{2}{3}\end{matrix}\right.\)

Vậy: \(S=\left\{1;-\dfrac{2}{3}\right\}\)