Question

Cho P= \(\dfrac{x-2\sqrt{x}+22}{\sqrt{x}+3}\)

1) Tính x khi P =4

2)Tìm GTNN của P

3)Tính P khi x= \(3-2\sqrt{2}\)

Answer 1

1) Ta có: P=4

nên \(x-2\sqrt{x}+22=4\sqrt{x}+12\)

\(\Leftrightarrow x-6\sqrt{x}+10=0\)(Vô lý)

3) Thay \(x=3-2\sqrt{2}\) vào P, ta được:

\(P=\dfrac{3-2\sqrt{2}-2\left(\sqrt{2}-1\right)+22}{\sqrt{2}-1+3}\)

\(=\dfrac{3-2\sqrt{2}-2\sqrt{2}+2+22}{2+\sqrt{2}}\)

\(=\dfrac{27-4\sqrt{2}}{2+\sqrt{2}}\)

\(=\dfrac{\left(27-4\sqrt{2}\right)\left(\sqrt{2}-1\right)}{\sqrt{2}}\)

\(=\dfrac{\left(27\sqrt{2}-8\right)\left(\sqrt{2}-1\right)}{2}\)

\(=\dfrac{54-27\sqrt{2}-8\sqrt{2}+8}{2}\)

\(=\dfrac{64-35\sqrt{2}}{2}\)