Question

Cho A = \(\frac{2}{\sqrt{x}-2}:\left(\frac{\sqrt{x}}{x-4}+\frac{1}{\sqrt{x}-2}\right)\) với x ≥ 0 và x ≠ 4

1) Rút gọn A

2) Tìm giá trị lớn nhất của A

Answer 1

1) \(A=\frac{2}{\sqrt{x}-2}:\left(\frac{\sqrt{x}}{x-4}+\frac{1}{\sqrt{x}-2}\right)\)

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

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

\(=\frac{2\left(\sqrt{x}+2\right)}{2\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}+2}{\sqrt{x}+1}\)

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

Để A đạt giá trị lớn nhất thì \(\sqrt{x}+1\) nhỏ nhất

Mà \(\sqrt{x}\ge0\Rightarrow\sqrt{x}+1\ge1\Rightarrow A\le2\)

Dấu = xảy ra khi : x = 0

Vậy . . . . . . . . .

Answer 2

a, Ta có : \(A=\frac{2}{\sqrt{x}-2}:\left(\frac{\sqrt{x}}{x-4}+\frac{1}{\sqrt{x}-2}\right)\)

=> \(A=\frac{2}{\sqrt{x}-2}:\left(\frac{\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{1}{\sqrt{x}-2}\right)\)

=> \(A=\frac{2}{\sqrt{x}-2}:\left(\frac{\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right)\)

=> \(A=\frac{2}{\sqrt{x}-2}:\left(\frac{\sqrt{x}+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right)\)

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

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

Answer 3