Question

Giải pt

\(\sqrt{x-1}+9\sqrt{x+1}=4x+9\)

Answer 1

Đk:\(x\ge1\)

\(pt\Leftrightarrow\sqrt{x-1}-\dfrac{1}{2}+9\sqrt{x+1}-\dfrac{27}{2}=4x-5\)

\(\Leftrightarrow\dfrac{x-\dfrac{5}{4}}{\sqrt{x-1}+\dfrac{1}{2}}+\dfrac{9\left(x-\dfrac{5}{4}\right)}{\sqrt{x+1}+\dfrac{3}{2}}=4\left(x-\dfrac{5}{4}\right)\)

\(\Leftrightarrow\left(x-\dfrac{5}{4}\right)\left(\dfrac{1}{\sqrt{x-1}+\dfrac{1}{2}}+\dfrac{9}{\sqrt{x+1}+\dfrac{3}{2}}-4\right)=0\)

Dễ thấy: \(\dfrac{1}{\sqrt{x-1}+\dfrac{1}{2}}+\dfrac{9}{\sqrt{x+1}+\dfrac{3}{2}}-4\) vô nghiệm với \(x\ge1\)

Nên \(x-\dfrac{5}{4}=0\Leftrightarrow x=\dfrac{5}{4}\)