Question

Cho P= \(\left(\frac{\sqrt{a}}{2}-\frac{1}{2\sqrt{a}}\right)\left(\frac{a-\sqrt{a}}{\sqrt{a}+1}-\frac{a+\sqrt{a}}{\sqrt{a}-1}\right)\)

a) Tìm ĐKXĐ

b) rút gọn P

c) Tìm a để P≥ -2

Answer 1

a) ĐKXĐ: \(a>1\)

b) \(P=\left(\frac{\sqrt{a}}{2}-\frac{1}{2\sqrt{a}}\right)\left(\frac{a-\sqrt{a}}{\sqrt{a}+1}-\frac{a+\sqrt{a}}{\sqrt{a}-1}\right)\)

\(P=\left(\frac{a}{2\sqrt{a}}-\frac{1}{2\sqrt{a}}\right)\left[\frac{\left(a-\sqrt{a}\right)\left(\sqrt{a}-1\right)}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}-\frac{\left(a+\sqrt{a}\right)\left(\sqrt{a}+1\right)}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right]\)

\(P=\frac{a-1}{2\sqrt{a}}\cdot\frac{-2a+a\sqrt{a}+\sqrt{a}-2a-a\sqrt{a}-\sqrt{a}}{a-1}\)

\(P=\frac{-4a}{2\sqrt{a}}=-2\sqrt{a}\)

c) \(P\ge-2\)

\(\Leftrightarrow-2\sqrt{a}\ge-2\)

\(\Leftrightarrow\sqrt{a}\le1\)

\(\Leftrightarrow0\le a\le1\)

Mà \(a>1\)

Vậy \(a\in\varnothing\)