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

2: \(A=\dfrac{\sqrt{x}-5}{\sqrt{x}+5}=\dfrac{\sqrt{x}+5-10}{\sqrt{x}+5}\)

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

\(\sqrt{x}+5>=5\forall x\)

=>\(\dfrac{10}{\sqrt{x}+5}< =\dfrac{10}{5}=2\forall x\)

=>\(-\dfrac{10}{\sqrt{x}+5}>=-2\forall x\)

=>\(-\dfrac{10}{\sqrt{x}+5}+1>=-2+1=-1\forall x\)

Dấu '=' xảy ra khi x=0

Vậy: \(A_{min}=-1\) khi x=0

\(A=\dfrac{\sqrt{x}}{\sqrt{x}-5}-\dfrac{10\sqrt{x}}{x-25}-\dfrac{5}{\sqrt{x}+5}\left(x\ge0;x\ne25\right)\)

Để \(A=\dfrac{2\sqrt{x}}{3}\) thì:

\(\dfrac{\sqrt{x}-5}{\sqrt{x}+5}=\dfrac{2\sqrt{x}}{3}\)

\(\Leftrightarrow3\sqrt{x}-15=2x+10\sqrt{x}\)

\(\Leftrightarrow2x+10\sqrt{x}-3\sqrt{x}+15=0\)

\(\Leftrightarrow2x+7\sqrt{x}+15=0\) 

Mà \(2x+7\sqrt{x}+15>0\) (vì \(x\ge0\))

nên không tìm được giá trị nào của \(x\) thoả mãn \(A=\dfrac{2\sqrt{x}}{3}\)

#\(Toru\)

Lời giải:

a. 

$A=\frac{\sqrt{x}(5-\sqrt{x})-(\sqrt{x}+5)(\sqrt{x}+1)}{(\sqrt{x}+5)(5-\sqrt{x})}-\frac{5-9\sqrt{x}}{(\sqrt{x}-5)(\sqrt{x}+5)}$

$=\frac{-2x-10\sqrt{x}}{(\sqrt{x}+5)(5-\sqrt{x})}$

$=\frac{-2\sqrt{x}(\sqrt{x}+5)}{(\sqrt{x}+5)(5-\sqrt{x})}=\frac{2\sqrt{x}}{\sqrt{x}-5}$

b.

$A< 1\Leftrightarrow \frac{2\sqrt{x}}{\sqrt{x}-5}<1$

$\Leftrightarrow \frac{\sqrt{x}+5}{\sqrt{x}-5}<0$

$\Leftrightarrow \sqrt{x}-5<0$

$\Leftrightarrow 0\leq x< 25$

Kết hợp với đkxđ suy ra $0\leq x< 25$

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https://hoc24.vn/cau-hoi/adfracsqrtxsqrtx5-dfracsqrtx15-sqrtx-dfrac5-9sqrtxx-25-voi-xge0xne25rut-gon-a2-tim-tat-ca-cac-gia-tri-cua-x-de-a1.7900547231312

a) ta có : \(\dfrac{x\sqrt{x}-y\sqrt{y}}{\sqrt{x}-\sqrt{y}}=\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)}{\sqrt{x}-\sqrt{y}}=x+\sqrt{xy}+y\)

b) ta có : \(\dfrac{x-\sqrt{3x}+3}{x\sqrt{x}+3\sqrt{3}}=\dfrac{x-\sqrt{3x}+3}{\left(\sqrt{x}+\sqrt{3}\right)\left(x-\sqrt{3x}+3\right)}=\dfrac{1}{\sqrt{x}+\sqrt{3}}\)

c) ta có : \(\dfrac{2}{\sqrt{5}-\sqrt{3}}+\dfrac{3}{\sqrt{6}+\sqrt{3}}\)

\(=\dfrac{2\left(\sqrt{5}+\sqrt{3}\right)}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}+\dfrac{3\left(\sqrt{6}-\sqrt{3}\right)}{\left(\sqrt{6}+\sqrt{3}\right)\left(\sqrt{6}-\sqrt{3}\right)}\)