Question

Có BẠN NÀO GIẢI GIÚP MIK CÂU NÀY VỚI
NHA >.............ĐANG CẦN GẤP
  \(p=\frac{x+2}{x\sqrt{x}-1}+\frac{\sqrt{x}+1}{x+\sqrt{x}+1}-\frac{1}{\sqrt{x}-1}\)

 

A.RÚT GỌN

B.TÍNH P KHI \(x=28-6\sqrt{3}\)

C.CHỨNG MINH P<1/3

Answer 1

a) \(P=\frac{x+2}{x\sqrt{x}-1}+\frac{\sqrt{x}+1}{x+\sqrt{x}+1}-\frac{1}{\sqrt{x}-1}\left(ĐKXĐ:1\ne x\ge0\right)\)

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

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

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

b) \(x=28-6\sqrt{3}=\left(3\sqrt{3}-1\right)^2\)thay vào P được : \(P=\frac{3\sqrt{3}-1}{28-6\sqrt{3}+3\sqrt{3}-1+1}=\frac{3\sqrt{3}-1}{28-3\sqrt{3}}\)

c) \(P=\frac{\sqrt{x}}{x+\sqrt{x}+1}=\frac{3\sqrt{x}}{3\left(x+\sqrt{x}+1\right)}=\frac{\left(x+\sqrt{x}+1\right)-\left(x-2\sqrt{x}+1\right)}{3\left(x+\sqrt{x}+1\right)}=-\frac{\left(\sqrt{x}-1\right)^2}{3\left(x+\sqrt{x}+1\right)}+\frac{1}{3}\le\frac{1}{3}\)Vì \(x\ne1\)nên dấu đẳng thức không xảy ra.

Do đó : \(P< \frac{1}{3}\)

Answer 2

ĐKXĐ: \(x\ge0\)

a/ \(P=\frac{x+2}{\sqrt{x}^3-1}+\frac{\sqrt{x}+1}{x+\sqrt{x}+1}-\frac{1}{\sqrt{x}-1}\) \(=\frac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\frac{\sqrt{x}+1}{x+\sqrt{x}+1}-\frac{1}{\sqrt{x}-1}\)

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

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

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

b/ Thay \(x=28-6\sqrt{3}\) vào P ta được: \(P=\frac{\sqrt{28-6\sqrt{3}}}{28-6\sqrt{3}+\sqrt{28-6\sqrt{3}}+1}\)

         \(=\frac{\sqrt{\left(3\sqrt{3}-1\right)^2}}{29-6\sqrt{3}+\sqrt{\left(3\sqrt{3}-1\right)^2}}\) \(=\frac{3\sqrt{3}-1}{29-6\sqrt{3}+3\sqrt{3}-1}=\frac{3\sqrt{3}-1}{28-3\sqrt{3}}\)

c/ \(P< \frac{1}{3}\Leftrightarrow\frac{\sqrt{x}}{x+\sqrt{x}+1}< \frac{1}{3}\)    \(\Leftrightarrow x+\sqrt{x}+1>3\sqrt{x}\)         \(\Leftrightarrow x-2\sqrt{x}+1>0\)

            \(\Leftrightarrow\left(\sqrt{x}-1\right)^2>0\Leftrightarrow\sqrt{x}>1\Leftrightarrow x>1\)

            Vậy x > 1