Question

[~_~ ^~^ ^_^ {_} +_+ -_- ?_?....]

Cho biểu thức A =\(\frac{x+\sqrt{x}+1}{\sqrt{x}+2}\)  và B =\(\frac{2\sqrt{x}}{\sqrt{x}-1}-\frac{x-\sqrt{x}+2}{x-\sqrt{x}}\)  với x > 0; x ≠ 1

1) Tính giá trị của A khi x = 16

2) Chứng minh rằng B = \(\frac{\sqrt{x}+2}{\sqrt{x}}\)

3) Cho P = A.B. So sánh P với 3.

Answer 1

[✰Ťøρ ²⁷ Ťɾїệʉ Vâɳ ŇD✰]

1) Thay x=16 vào A ta có:

A=\(\frac{16+\sqrt{16}+1}{\sqrt{16}+2}\)

A=\(\frac{16+4+1}{4+2}\)

A=\(\frac{21}{6}=\frac{7}{2}\)

Answer 2

[Phạm Thị Thùy Linh]

\(2,\frac{2\sqrt{x}}{\sqrt{x}-1}-\frac{x-\sqrt{x}+2}{x-\sqrt{x}}\)

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

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

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

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

\(3,P=A.B=\frac{x+\sqrt{x}+1}{\sqrt{x}+2}.\frac{\sqrt{x}+2}{\sqrt{x}}=\frac{x+\sqrt{x}+1}{\sqrt{x}}\)

Ta thấy \(\left(\sqrt{x}-1\right)^2>0\Rightarrow x-2\sqrt{x}+1>0\)

\(\Rightarrow x+\sqrt{x}+1>3\sqrt{x}\)

\(\Rightarrow\frac{x+\sqrt{x}+1}{\sqrt{x}}>\frac{3\sqrt{x}}{\sqrt{x}}\Rightarrow\frac{x+\sqrt{x}+1}{\sqrt{x}}>3\left(đpcm\right)\)