Question

Giải phương trình sau:

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

Answer 1

\(\sqrt{x-1+2\sqrt{x-2}}+x+1=5\sqrt{x-2}\)(ĐK: \(x\ge2\))

\(\Leftrightarrow\sqrt{x-2+2\sqrt{x-2}+1}+x+1=5\sqrt{x-2}\)

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

\(\Leftrightarrow\sqrt{x-2}+1+x+1=5\sqrt{x-2}\)

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

\(\Leftrightarrow x^2+4x+4=16\left(x-2\right)\) (bình phương cả 2 vế)

\(\Leftrightarrow x^2-12x+36=0\)

\(\Leftrightarrow\left(x-6\right)^2=0\Leftrightarrow x=6\left(TM\right)\)

Vậy PT có nghiệm là \(x=6\)