Question

Giải PT

13\(\sqrt{x-1}\) + 9\(\sqrt{x+1}\) = 16x

Answer 1

x>=1

\(\Leftrightarrow16x-13\sqrt{x-1}-9\sqrt{x+1}=0\)

\(\Leftrightarrow13\left(x-1-\sqrt{x-1}+\dfrac{1}{4}\right)+3\left(x+1-3\sqrt{x+1}+\dfrac{9}{4}\right)=0\)

\(\Leftrightarrow13\left(\sqrt{x-1}-\dfrac{1}{2}\right)^2+3\left(\sqrt{x+1}-\dfrac{3}{2}\right)^2=0\)

\(\left\{{}\begin{matrix}\sqrt{x-1}=\dfrac{1}{2}\\\sqrt{x+1}=\dfrac{3}{2}\end{matrix}\right.\)

x=5/4(tm)