Vì x>0; y>0

Nên áp dụng BĐT Cô-si ta có: \(x+y\ge2\sqrt{xy}\)

\(\Rightarrow\)\(\frac{1}{x}+\frac{1}{y}\ge2\sqrt{\frac{1}{x}.\frac{1}{y}}=2\sqrt{\frac{1}{xy}}\)

Mà \(\frac{1}{x}+\frac{1}{y}=\frac{1}{2}\)

Nên \(\frac{1}{2}\ge2.\frac{1}{\sqrt{xy}}\Rightarrow\frac{1}{4}\ge\frac{1}{\sqrt{xy}}\)

\(\Rightarrow4\le\sqrt{xy}\) (C)

Ta có: \(\sqrt{x}+\sqrt{y}\ge2\sqrt{\sqrt{xy}}\)

Thế (C) vào ta được: \(\sqrt{x}+\sqrt{y}\ge2\sqrt{4}=4\)

Dấu "=" xảy ra <=> x = y

Vậy A_Min = 4 khi và chỉ khi x = y

\(\frac{1}{x}+\frac{1}{y}>=\frac{4}{x+y}\Rightarrow\frac{1}{2}>=\frac{4}{x+y}\Rightarrow x+y>=8\left(1\right)\)(bđt svacxo)

\(\frac{1}{x}+\frac{1}{y}>=2\sqrt{\frac{1}{x}\cdot\frac{1}{y}}=\frac{2}{\sqrt{xy}}\Rightarrow\frac{1}{2}>=\frac{2}{\sqrt{xy}}\Rightarrow\sqrt{xy}>=4\Rightarrow2\sqrt{xy}>=8\left(2\right)\)(bđt cosi)

từ \(\left(1\right);\left(2\right)\Rightarrow x+2\sqrt{xy}+y>=8+8=16\Rightarrow\left(\sqrt{x}+\sqrt{y}\right)^2>=16\)

mà \(\sqrt{x}>0;\sqrt{y}>0\Rightarrow\sqrt{x}+\sqrt{y}>=4\)

dấu = xảy ra khi x=y=4

vậy min A là 4 khi x=y=4

\(A=\frac{x-1+3\sqrt{x-1}+2}{x-1+4\sqrt{x-1}+3}=\frac{\left(\sqrt{x-1}+1\right)\left(\sqrt{x-1}+2\right)}{\left(\sqrt{x-1}+1\right)\left(\sqrt{x-1}+3\right)}=\frac{\sqrt{x-1}+2}{\sqrt{x-1}+3}\ge\frac{2}{3}\)

GTNN của A = \(\frac{2}{3}\)khi x = 1

\(P=\frac{x^2+\sqrt{x}}{x-\sqrt{x}+1}+1-\frac{2x+\sqrt{x}}{\sqrt{x}}\left(đkxđ\Leftrightarrow x\ge0\right).\)

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

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

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

\(P=x-\sqrt{x}-1=\sqrt{x}^2-2.\sqrt{x}.\frac{1}{2}+\frac{1}{4}-\frac{1}{4}-1\)

\(=\left(\sqrt{x}-\frac{1}{2}\right)^2-\frac{5}{4}\)

\(\Rightarrow P_{min}=-\frac{5}{4}\Leftrightarrow\left(\sqrt{x}-\frac{1}{2}\right)^2=0\)

\(\Rightarrow\sqrt{x}=\frac{1}{2}\Rightarrow x=\frac{1}{4}\)

Bạn bình phương lên là tính đc GTLN đó

cảm ơn bạn

Ta có : 

\(B=\frac{\sqrt{x}+1}{\sqrt{x}-1}=1+\frac{2}{\sqrt{x}-1}\)

\(\Rightarrow B\sqrt{x}=\sqrt{x}+\frac{2.\sqrt{x}}{\sqrt{x}-1}\)

\(\Rightarrow B\sqrt{x}=\left(\sqrt{x}-1+\frac{2}{\sqrt{x}-1}\right)+3\)

\(\Rightarrow B\sqrt{x}\ge2\sqrt{\left(\sqrt{x}-1\right).\frac{2}{\sqrt{x}-1}}+3\)

\(\Rightarrow B\sqrt{x}\ge2\sqrt{2}+3\)

\(P=\left(\frac{\sqrt{x}}{\sqrt{x}-2}+\frac{4\sqrt{x}-3}{2\sqrt{x}-x}\right):\)\(\left(\frac{\sqrt{x}+2}{\sqrt{x}}-\frac{\sqrt{x}-4}{\sqrt{x}-2}\right)\)

\(=\left(\frac{\sqrt{x}}{\sqrt{x}-2}-\frac{4\sqrt{x}-3}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\)\(:\left(\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)-\sqrt{x}\left(\sqrt{x}-4\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\)

\(=\frac{x-4\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-2\right)}:\frac{x-4-x+4\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

\(=\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}.\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{4\left(\sqrt{x}-1\right)}\)

\(=\frac{\sqrt{x}-3}{4}\)

\(b,\)Để \(P>0\Rightarrow\frac{\sqrt{x}-3}{4}>0\)

Mà \(4>0\Rightarrow\sqrt{x}-3>0\Rightarrow\sqrt{x}>3\Rightarrow x>9\)

\(c,\sqrt{P}_{min}=0\Rightarrow\frac{\sqrt{x}-3}{4}=0\)

\(\Leftrightarrow\sqrt{x}-3=0\Rightarrow\sqrt{x}=3\Rightarrow x=9\)

thank