Question

tính A nếu \(x\ge\sqrt{2}\)

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

Answer 1

Sửa đề:

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

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

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

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

\(A=\left|\sqrt{x^2-1}+1\right|-\left|\sqrt{x^2-1}-1\right|\)

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

\(A=2\)