Question

tính T=\(\sqrt{x^2-16}+\sqrt{x^2-36}biết\sqrt{x^2-16}-\sqrt{x^2-36}=2\)

Answer 1

Ta có: \(\sqrt{x^2-16}-\sqrt{x^2-36}=2\)

\(\Leftrightarrow\left(\sqrt{x^2-16}-\sqrt{x^2-36}\right)\cdot\left(\sqrt{x^2-16}+\sqrt{x^2-36}\right)=2\cdot\left(\sqrt{x^2-16}+\sqrt{x^2-36}\right)\)

\(\Leftrightarrow\left[\left(\sqrt{x^2-16}\right)^2-\left(\sqrt{x^2-36}\right)^2\right]=2\cdot\left(\sqrt{x^2-16}+\sqrt{x^2-36}\right)\)

\(\Leftrightarrow x^2-16-x^2+36=2\cdot\left(\sqrt{x^2-16}+\sqrt{x^2-36}\right)\)

\(\Leftrightarrow20=2\cdot\left(\sqrt{x^2-16}+\sqrt{x^2-36}\right)\)

\(\Leftrightarrow10=\sqrt{x^2-16}+\sqrt{x^2-36}\)

hay \(T=10\)

Vậy \(T=10\).