Bài 2 :
a. \(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}+2=0\\ \sqrt{x-1}+\sqrt{4\left(x-1\right)}-\sqrt{25\left(x-1\right)}+2=0\\ \sqrt{x-1}+2\sqrt{x-1}-5\sqrt{x-1}+2=0\\ \sqrt{x-1}=0\\ x=1\)
b.\(\dfrac{1}{2}\sqrt{x-1}-\dfrac{3}{2}\sqrt{9x-9}+24\sqrt{\dfrac{x-1}{64}}=-17\\ \dfrac{1}{2}\sqrt{x-1}-\dfrac{3}{2}\sqrt{9\left(x-1\right)}+24\sqrt{x-1}:\sqrt{64}=-17\\ \dfrac{1}{2}\sqrt{x-1}-\dfrac{3}{2}.3\sqrt{x-1}+3\sqrt{x-1}=-17\\ \dfrac{1}{2}-\dfrac{9}{2}+3.\sqrt{x-1}=-17\\ -\sqrt{x-1}=-17\\ \sqrt{x-1}=17\\ x-1=17\\ x=18\)
a) Điều kiện: \(\left\{{}\begin{matrix}x-1\ge0\\4x-4\ge0\\25x-25\ge0\end{matrix}\right.\Leftrightarrow x\ge1\)
\(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}+2=0\)
\(\Leftrightarrow\sqrt{x-1}+\sqrt{4\left(x-1\right)}-\sqrt{25\left(x-1\right)}+2=0\)
\(\Leftrightarrow\sqrt{x-1}+2\sqrt{x-1}-5\sqrt{x-1}+2=0\)
\(\Leftrightarrow-2\sqrt{x-1}+2=0\)
\(\Leftrightarrow\sqrt{x-1}=1\)
\(\Leftrightarrow x-1=1\)
\(\Leftrightarrow x=2\) (nhận)
Vậy \(x=2.\)
b) Điều kiện: \(\left\{{}\begin{matrix}x-1\ge0\\9x-9\ge0\\\dfrac{x-1}{64}\ge0\end{matrix}\right.\Leftrightarrow x\ge1\)
\(\dfrac{1}{2}\sqrt{x-1}-\dfrac{3}{2}\sqrt{9x-9}+24\sqrt{\dfrac{x-1}{64}}=-17\)
\(\Leftrightarrow\dfrac{1}{2}\sqrt{x-1}-\dfrac{3}{2}\sqrt{9\left(x-1\right)}+24\sqrt{\dfrac{x-1}{64}}=-17\)
\(\Leftrightarrow\dfrac{1}{2}\sqrt{x-1}-\dfrac{3}{2}.3\sqrt{x-1}+24.\dfrac{\sqrt{x-1}}{8}=-17\)
\(\Leftrightarrow\dfrac{1}{2}\sqrt{x-1}-\dfrac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\)
\(\Leftrightarrow-\sqrt{x-1}=-17\)
\(\Leftrightarrow\sqrt{x-1}=17\)
\(\Leftrightarrow x-1=289\)
\(\Leftrightarrow x=290\) (nhận)
Vậy \(x=290.\)
