a: ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x\ne1\end{matrix}\right.\)
\(P=\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{2}\)
\(=\left(\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}\right)\cdot\dfrac{2}{\sqrt{x}-1}\)
\(=\dfrac{x+2+\sqrt{x}\left(\sqrt{x}-1\right)-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)^2\cdot\left(x+\sqrt{x}+1\right)}\cdot2\)
\(=\dfrac{x-\sqrt{x}+1+x-\sqrt{x}}{\left(\sqrt{x}-1\right)^2}\cdot\dfrac{2}{x+\sqrt{x}+1}=\dfrac{2}{x+\sqrt{x}+1}\)
b: \(x=7-4\sqrt{3}=\left(2-\sqrt{3}\right)^2\)
Thay \(x=\left(2-\sqrt{3}\right)^2\) vào P, ta được:
\(P=\dfrac{2}{7-4\sqrt{3}+\sqrt{\left(2-\sqrt{3}\right)^2}+1}\)
\(=\dfrac{2}{8-4\sqrt{3}+2-\sqrt{3}}=\dfrac{2}{10-5\sqrt{3}}=\dfrac{4+2\sqrt{3}}{5}\)
c: P>=2/3
=>\(P-\dfrac{2}{3}>=0\)
=>\(\dfrac{2}{x+\sqrt{x}+1}-\dfrac{2}{3}>=0\)
=>\(\dfrac{1}{x+\sqrt{x}+1}-\dfrac{1}{3}>=0\)
=>\(\dfrac{3-x-\sqrt{x}-1}{3\left(x+\sqrt{x}+1\right)}>=0\)
=>\(-x-\sqrt{x}+2>=0\)
=>\(x+\sqrt{x}-2< =0\)
=>\(\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)< =0\)
=>\(\sqrt{x}-1< =0\)
=>0<=x<=1
Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}x>=0\\0< =x< 1\end{matrix}\right.\Leftrightarrow0< =x< 1\)
\(a.ĐK:x\ge0;x\ne1\\ P=\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{2}\\ =\left[\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right]\cdot\dfrac{2}{\sqrt{x}-1}\\ =\dfrac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{2}{\sqrt{x}-1}\\ =\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{2}{\sqrt{x}-1}\\ =\dfrac{2\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)^2\left(x+\sqrt{x}+1\right)}\\ =\dfrac{2}{x+\sqrt{x}+1}\)
b) \(x=7-4\sqrt{3}=2^2-2\cdot2\cdot\sqrt{3}+\left(\sqrt{3}\right)^2=\left(2-\sqrt{3}\right)^2\)
\(P=\dfrac{2}{7-4\sqrt{3}+\sqrt{\left(2-\sqrt{3}\right)^2}+1}=\dfrac{2}{7-4\sqrt{3}+2-\sqrt{3}+1}=\dfrac{2}{10-5\sqrt{3}}\)
\(c.P\ge\dfrac{2}{3}=>\dfrac{2}{x+\sqrt{x}+1}\ge\dfrac{2}{3}\\ =>\dfrac{1}{x+\sqrt{x}+1}\ge\dfrac{1}{3}\\ < =>x+\sqrt{x}+1\le3\\< =>x+\sqrt{x}-2\le0\\ < =>\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)\le0\\ < =>\sqrt{x}-1\le0\\ < =>x\le1\)
Kết hợp với đkxđ thì: 0 <= x < 1
giải giúp mình với ạa , mình cảm ơn.
