Question

\(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-5}+2=0\)

\(2x-x^2+\sqrt{6x^2-12x+7}=0\)

\(\left(x+1\right)\left(x+4\right)-3\sqrt{x^2+5x+2}=6\)

\(\frac{1}{2}\sqrt{x-1}-\frac{3}{2}\sqrt{9x-9}+24\sqrt{\frac{x-1}{64}}=-17\)

giúp dùm đi mấy pạn

Answer 1

Answer 2

\(\Leftrightarrow-\left(x^2-2x\right)+\sqrt{6\left(x^2-2x\right)+7}=0\) ĐK \(\sqrt{6x^2-12x+7}\ge0\)

Đặt \(t=x^2-2x\left(t\ge0\right)\Leftrightarrow pt:-t+\sqrt{6t+7}=0\Leftrightarrow\sqrt{6t+7}=t\\ 6t+7-t^2=0\Leftrightarrow\left[\begin{array}{nghiempt}t=7\left(tm\right)\\t=-1\left(ktm\right)\end{array}\right.\)

Với \(t=7\Leftrightarrow x^2-2x-7=0\Leftrightarrow x=1\pm2\sqrt{2}\left(tm\right)\)

Vậy S={​\(1\pm2\sqrt{2}\)}