Question

[Vi Thái Dương]

giúp mình giảii chi tiết với 

Answer 1

[Victorique de Blois]

a, \(P=\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right):\left(\frac{\sqrt{x}-1}{\sqrt{x}}+\frac{1-\sqrt{x}}{x+\sqrt{x}}\right)\)

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

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

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

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

b, \(x=\frac{2}{2+\sqrt{3}}=\frac{2\left(2-\sqrt{3}\right)}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}=4-2\sqrt{3}=\left(\sqrt{3}-1\right)^2\) (thỏa mãn)

thay vào P ta đc : 

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

\(P=\frac{\left(\sqrt{3}-1+1\right)^2}{\sqrt{3}-1}\) vì \(\sqrt{3}-1>0\)

\(P=\frac{\sqrt{3}^2}{\sqrt{3}-1}=\frac{3}{\sqrt{3}-1}\)

c, \(P-2=\frac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}-2=\frac{x+2\sqrt{x}+1-2\sqrt{x}}{\sqrt{x}}=\frac{x+1}{\sqrt{x}}\) mà x > 0

\(\Rightarrow P-2>0\Leftrightarrow P>2\)

d, thay P vào pt ta đc : \(\left(\sqrt{x}+1\right)^2=6\sqrt{x}-3-\sqrt{x-4}\)

\(\Leftrightarrow x+2\sqrt{x}+1=6\sqrt{x}-3-\sqrt{x-4}\)

\(\Leftrightarrow x-4\sqrt{x}+4+\sqrt{x-4}=0\)

\(\Leftrightarrow\left(\sqrt{x}-2\right)^2+\sqrt{x-4}=0\)

mà \(\hept{\begin{cases}\left(\sqrt{x}-2\right)^2\ge0\\\sqrt{x-4}\ge0\end{cases}}\) nên \(\hept{\begin{cases}\sqrt{x}-2=0\\\sqrt{x-4}=0\end{cases}}\Leftrightarrow x=4\left(tm\right)\)

vậy_