Question

cho biểu thức P=\(\left(\frac{1}{\sqrt{x}+2}+\frac{1}{\sqrt{x}-2}\right).\frac{\sqrt{x}-2}{\sqrt{x}}\)với x≥0 và x≠4

a)rút gọn P

b)tìm x thục để\(\frac{7P}{3}\)có giá trị nguyên

Answer 1

\(P=\left(\frac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right).\frac{\sqrt{x}-2}{\sqrt{x}}\)

\(=\frac{2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\frac{\left(\sqrt{x}-2\right)}{\sqrt{x}}=\frac{2}{\sqrt{x}+2}\)

b/ \(\frac{7P}{3}=\frac{14}{3\sqrt{x}+6}\) nguyên \(\Rightarrow3\sqrt{x}+6=Ư\left(14\right)=\left\{7;14\right\}\)

\(3\sqrt{x}+6=7\Rightarrow\sqrt{x}=\frac{1}{3}\Rightarrow x=\frac{1}{9}\)

\(3\sqrt{x}+6=14\Rightarrow\sqrt{x}=\frac{8}{3}\Rightarrow x=\frac{64}{9}\)