Question

Cho \(\sqrt{16-2x+x^2}-\sqrt{9-2x+x^2}=1\)

Tính A = \(\sqrt{16-2x+x^2}+\sqrt{9-2x+x^2}\)

Answer 1

\(\sqrt{16-2x+x^2}-\sqrt{9-2x+x^2}=1\)

\(\Leftrightarrow\dfrac{\left(\sqrt{16-2x+x^2}-\sqrt{9-2x+x^2}\right)\left(\sqrt{16-2x+x^2}+\sqrt{9-2x+x^2}\right)}{\sqrt{16-2x+x^2}+\sqrt{9-2x+x^2}}=1\)

\(\Leftrightarrow\dfrac{16-2x+x^2-9+2x-x^2}{\sqrt{16-2x+x^2}+\sqrt{9-2x+x^2}}=1\)

\(\Leftrightarrow\dfrac{7}{\sqrt{16-2x+x^2}+\sqrt{9-2x+x^2}}=1\Leftrightarrow\dfrac{7}{A}=1\Rightarrow A=7\)