Question

[JLEIZ]

GIÚP MÌNH VS HUHU

Answer 1

[Nguyễn Lê Phước Thịnh]

a: ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x\ne1\end{matrix}\right.\)

\(P=\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{2}\)

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

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

\(=\dfrac{1-\sqrt{x}+x-\sqrt{x}}{\left(\sqrt{x}-1\right)^2}\cdot\dfrac{2}{x+\sqrt{x}+1}\)

\(=\dfrac{2}{x+\sqrt{x}+1}\)

b: \(x=7-4\sqrt{3}=\left(2-\sqrt{3}\right)^2\)

Khi \(x=\left(2-\sqrt{3}\right)^2\) thì 

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

\(P=\dfrac{2}{7-4\sqrt{3}+2-\sqrt{3}+1}\)

\(=\dfrac{2}{10-5\sqrt{3}}=\dfrac{4+2\sqrt{3}}{5}\)

c: P>=2/3

=>P-2/3>=0

=>\(\dfrac{2}{x+\sqrt{x}+1}-\dfrac{2}{3}>=0\)

=>\(\dfrac{1}{x+\sqrt{x}+1}-\dfrac{1}{3}>=0\)

=>\(\dfrac{3-x-\sqrt{x}-1}{3\left(x+\sqrt{x}+1\right)}>=0\)

=>\(-x-\sqrt{x}+2>=0\)

=>\(x+\sqrt{x}-2< =0\)

=>\(\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)< =0\)

=>\(\sqrt{x}-1< =0\)

=>0<=x<=1

Kết hợp ĐKXĐ, ta được: 0<=x<1