a) \(ĐKXĐ:x>0\)

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

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

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

\(\Leftrightarrow Y=x+\sqrt{x}-2\sqrt{x}-2\)

\(\Leftrightarrow Y=x-\sqrt{x}-2\)

b) Ta có \(Y=x-\sqrt{x}-2=\left(\sqrt{x}-\frac{1}{2}\right)^2-\frac{9}{4}\ge-\frac{9}{4}\)

Dấu "=" xảy ra \(\Leftrightarrow\sqrt{x}-\frac{1}{2}=0\)

\(\Leftrightarrow x=\frac{1}{4}\)

Vậy \(Min_Y=-\frac{9}{4}\Leftrightarrow x=\frac{1}{4}\)

c) Để \(Y-\left|Y\right|=0\)

\(\Leftrightarrow Y=\left|Y\right|\)

\(\Leftrightarrow Y\ge0\)

\(\Leftrightarrow x-\sqrt{x}-2\ge0\)

\(\Leftrightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)\ge0\)

\(\Leftrightarrow\sqrt{x}-2\ge0\) (Vì \(\sqrt{x}+1\ge0\))

\(\Leftrightarrow\sqrt{x}\ge2\)

\(\Leftrightarrow x\ge4\)  (ĐPCM)

ĐKXĐ: x>0

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

Ta có \(Y=x-\sqrt{x}=x-2\sqrt{x}.\frac{1}{2}+\frac{1}{4}-\frac{1}{4}=\left(\sqrt{x}-\frac{1}{2}\right)^2-\frac{1}{4}\ge-\frac{1}{4}\)

Dấu '=' xảy ra khi \(\sqrt{x}-\frac{1}{2}=0\Leftrightarrow x=\frac{1}{4}\)

Vậy GTNN của Y là \(-\frac{1}{4}\)

b) Ta có x>1\(\Leftrightarrow x>\sqrt{x}\Leftrightarrow x-\sqrt{x}>0\)

Ta lại có \(Y-\left|Y\right|=x-\sqrt{x}-\left|x-\sqrt{x}\right|=x-\sqrt{x}-\left(x-\sqrt{x}\right)=0\)

Vậy khi x>1 thì \(Y-\left|Y\right|=0\)

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

\(=\dfrac{\sqrt{x}-\sqrt{y}-\sqrt{x}-\sqrt{y}}{x-y}:\dfrac{2\sqrt{xy}}{x-y}=\dfrac{-2\sqrt{y}}{2\sqrt{xy}}=\dfrac{-1}{\sqrt{x}}=\dfrac{-\sqrt{x}}{x}\)

b, Ta có \(A=\dfrac{-1}{\sqrt{x}}=1\Leftrightarrow\sqrt{x}=-1\left(voli\right)\)

Vậy pt vô nghiệm 

moi tay

giải giùm mình bài 5 với

1,

\(A=\left(\frac{a\sqrt{a}-1}{a-\sqrt{a}}-\frac{a\sqrt{a}+1}{a+\sqrt{a}}\right):\frac{a+2}{a-2}\left(đk:a\ne0;1;2;a\ge0\right)\)

\(=\frac{\left(a\sqrt{a}-1\right)\left(a+\sqrt{a}\right)-\left(a\sqrt{a}+1\right)\left(a-\sqrt{a}\right)}{a^2-a}.\frac{a-2}{a+2}\)

\(=\frac{a^2\sqrt{a}+a^2-a-\sqrt{a}-\left(a^2\sqrt{a}-a^2+a-\sqrt{a}\right)}{a\left(a-1\right)}.\frac{a-2}{a+2}\)

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

Để \(A=1\)\(=>\frac{2a-4}{a+2}=1< =>2a-4-a-2=0< =>a=6\)

2, 

a, Điều kiện xác định của phương trình là \(x\ne4;x\ge0\)

b, Ta có : \(B=\frac{2\sqrt{x}}{x-4}+\frac{1}{\sqrt{x}-2}-\frac{1}{\sqrt{x}+2}\)

\(=\frac{2\sqrt{x}}{x-4}+\frac{\sqrt{x}+2}{x-4}-\frac{\sqrt{x}-2}{x-4}\)

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

c, Với \(x=3+2\sqrt{3}\)thì \(B=\frac{2}{3-2+2\sqrt{3}}=\frac{2}{1+2\sqrt{3}}\)