Question

Giải phương trình:

a)\(\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\sqrt{9x-18}+6\sqrt{\dfrac{x-2}{81}}=4\)

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

Answer 1

a)ĐK \(x\ge2\)

\(\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\sqrt{9\left(x-2\right)}+6\dfrac{\sqrt{x-2}}{\sqrt{81}}=4\)

\(\Leftrightarrow\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}.3\sqrt{x-2}+6\dfrac{\sqrt{x-2}}{9}=4\)

\(\Leftrightarrow\dfrac{1}{3}\sqrt{x-2}-2\sqrt{x-2}+\dfrac{2}{3}\sqrt{x-2}=4\)

\(\Leftrightarrow-\sqrt{x-2}=4\left(vl\right)\)

Answer 2

b) \(\sqrt{x-1-2\sqrt{x-1}+1}=\sqrt{x-1}\) (ĐK \(x\ge1\)

\(\Leftrightarrow\sqrt{\left(\sqrt{x-1}-1\right)^2}=\sqrt{x-1}\)

\(\Leftrightarrow\left|\sqrt{x-1}-1\right|=\sqrt{x-1}\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}-1=\sqrt{x-1}\\1-\sqrt{x-1}=\sqrt{x-1}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}-1=0\left(vl\right)\\2\sqrt{x-1}=1\end{matrix}\right.\)

\(\Leftrightarrow\sqrt{x-1}=\dfrac{1}{2}\)

\(\Leftrightarrow x-1=\dfrac{1}{4}\)

\(\Leftrightarrow x=\dfrac{5}{4}\)