sqrt(x)<x

sqrt(x)=x voi x=1,x=0

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

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

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

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

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:

     \(H-\sqrt{H}=\sqrt{H}\left(\sqrt{H}-1\right)< 0\)

 \(\Rightarrow H< \sqrt{H}\)

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

\(\Rightarrow H\le\sqrt{H};bằng.khi.x=y=1\)

\(M=\dfrac{\sqrt{x}+1-1}{\sqrt{x}+1}=1-\dfrac{1}{\sqrt{x}+1}< =1\)

=>M<=căn M

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 \(P< \dfrac{1}{3}\)

Có \(A=\dfrac{\sqrt{x}-5}{\sqrt{x}+5}=1-\dfrac{10}{\sqrt{x}+5}\)

Dễ thấy \(\dfrac{10}{\sqrt{x}+5}>0\forall x\Rightarrow A=1-\dfrac{10}{\sqrt{x}+5}< 1\)

=> A < 2