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a) Ta có: \(P=\left(\dfrac{\sqrt{x}}{x\sqrt{x}-1}+\dfrac{1}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}\)
\(=\dfrac{\sqrt{x}+x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{x+\sqrt{x}+1}{\sqrt{x}+1}\)
\(=\dfrac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Áp dụng Cô-si:
\(x+y\ge2\sqrt{xy}\)
Do đó:
\(H\le\dfrac{\sqrt{xy}}{2\sqrt{xy}-\sqrt{xy}}=1\)
Mà \(x>y\) nên dấu "=" không xảy ra
\(\Rightarrow H< 1\)
Kết hợp dữ kiện đề bài, ta được:
\(\Rightarrow0< H< 1\)
\(\Rightarrow\sqrt{H}< 1\)
Xét:
\(H-\sqrt{H}=\sqrt{H}\left(\sqrt{H}-1\right)< 0\)
\(\Rightarrow H< \sqrt{H}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có
\(x+y\ge2\sqrt{xy}\\ \Leftrightarrow x+y\ge\sqrt{xy}+\sqrt{xy}\\ \Leftrightarrow x+y-\sqrt{xy}\ge\sqrt{xy}\\ \Rightarrow\dfrac{\sqrt{xy}}{yx-\sqrt{xy}+y}\)
Có mẫu luôn lớn hơn hoặc = tử số
Bằng khi x = y = 1
\(\Rightarrow H\le\sqrt{H};bằng.khi.x=y=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(M=\dfrac{\sqrt{x}+1-1}{\sqrt{x}+1}=1-\dfrac{1}{\sqrt{x}+1}< =1\)
=>M<=căn M
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(P=\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{x+2+x-1-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\)
\(P-\dfrac{1}{3}=\dfrac{\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{1}{3}=-\dfrac{x-2\sqrt{x}+1}{x+\sqrt{x}+1}=-\dfrac{\left(\sqrt{x}-1\right)^2}{x+\sqrt{x}+1}\le0;\forall x\ge0\)
\(\Rightarrow P\le\dfrac{1}{3}\)
Dấu "=" xảy ra khi \(x=1\) ko thỏa mãn ĐKXĐ nên \(P< \dfrac{1}{3}\)
\(\sqrt{x}< x\)
vì \(\left(\sqrt{x}\right)^2=x\)với \(\forall\)\(x\ge0\)
học tốt
Vì: \(x\ge0\) nên \(\sqrt{x}\ge0\)
+) \(\sqrt{x}=x\Leftrightarrow x=x^2\Leftrightarrow x-x^2=0\Leftrightarrow x\left(1-x\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
+) \(\sqrt{x}< x\Leftrightarrow x< x^2\Leftrightarrow x-x^2< 0\Leftrightarrow x\left(1-x\right)< 0\Leftrightarrow x>1\)
+) \(\sqrt{x}>x\Leftrightarrow x>x^2\Leftrightarrow x-x^2>0\Leftrightarrow x\left(1-x\right)>0\Leftrightarrow0< x< 1\)
Vậy: Nếu \(x=0\) thì \(x=1\) hoặc \(\sqrt{x}=x\)
Nếu \(x>1\) thì \(\sqrt{x}< x\)
Nếu \(0< x< 1\) thì \(\sqrt{x}>x\)
=.= hok tốt!!