Question

cho P=\(\dfrac{1}{\sqrt{x}+2}-\dfrac{5}{x-\sqrt{x}-6}-\dfrac{\sqrt{x}-2}{3-\sqrt{x}}\)

a) Rút gọn P

b) tìm giá trị lớn nhất của P

Answer 1

b) Với \(x\ge0;x\ne9\)

P = \(\dfrac{\sqrt{x}+4}{\sqrt{x}+2}\) = 1+\(\dfrac{2}{\sqrt{x}+2}\)

Vì x\(\ge0\) \(\Rightarrow\sqrt{x}\ge0\Rightarrow\sqrt{x}+2\ge2\Rightarrow\dfrac{2}{\sqrt{x}+2}\le1\)

\(\Rightarrow P=1+\dfrac{2}{\sqrt{x}+2}\le2\)

Dấu "=" xảy ra \(\Leftrightarrow x=0\) ( Thỏa mãn ĐKXĐ )

Vậy GTLN của P=2 \(\Leftrightarrow x=0\)

Answer 2

a) \(P=\dfrac{1}{\sqrt{x}+2}-\dfrac{5}{x-\sqrt{x}-6}-\dfrac{\sqrt{x}-2}{3-\sqrt{x}}\) với \(x\ge0;x\ne9\)

\(\Rightarrow P=\dfrac{\sqrt{x}-3}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}-\dfrac{5}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}+\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\)=\(\dfrac{\sqrt{x}-3-5+\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\)

= \(\dfrac{x+\sqrt{x}-12}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\)

= \(\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+4\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\)

= \(\dfrac{\sqrt{x}+4}{\sqrt{x}+2}\)