ĐKXĐ: ...

\(A=\left(\frac{\sqrt{x}\left(\sqrt{x}-5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}-1\right):\left(\frac{25-x+\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+5\right)}-\frac{\sqrt{x}+3}{\sqrt{x}+5}\right)\)

\(=\left(\frac{\sqrt{x}}{\sqrt{x}+5}-\frac{\sqrt{x}+5}{\sqrt{x}+5}\right):\left(\frac{25-x+x-25}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+5\right)}-\frac{\sqrt{x}+3}{\sqrt{x}+5}\right)\)

\(=\frac{-5}{\left(\sqrt{x}+5\right)}.\frac{\left(\sqrt{x}+5\right)}{-\left(\sqrt{x}+3\right)}=\frac{5}{\sqrt{x}+3}\)

b/ \(B=\frac{x+16}{\sqrt{x}+3}=\sqrt{x}-3+\frac{25}{\sqrt{x}+3}=\sqrt{x}+3+\frac{25}{\sqrt{x}+3}-6\)

\(\Rightarrow B\ge2\sqrt{\frac{\left(\sqrt{x}+3\right).25}{\sqrt{x}+3}}-6=4\)

\(B_{min}=4\) khi \(\left(\sqrt{x}+3\right)^2=25\Rightarrow x=4\)

ĐKXĐ : \(x\ge0,x\ne25,x\ne9\)

a) \(A=\left(\frac{x-5\sqrt{x}}{x-25}-1\right):\left(\frac{25-x}{x+2\sqrt{x}-15}-\frac{\sqrt{x}+3}{\sqrt{x}+5}+\frac{\sqrt{x}-5}{\sqrt{x}-3}\right)\)

\(=\left(\frac{x-5\sqrt{x}-\left(x-25\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\right):\left(\frac{-\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}-\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}+\frac{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\right)\)

\(=\frac{-5\left(\sqrt{x}-5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}:\left(\frac{25-x-x+9+x-25}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\right)\)

\(=-\frac{5}{\sqrt{x}+5}:\frac{-\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+5\right)}=\frac{-5}{\sqrt{x}+5}.\left(\frac{-\left(\sqrt{x}+5\right)}{\sqrt{x}+3}\right)=\frac{5}{\sqrt{x}+3}\)

b) \(A< 1\Rightarrow\frac{5}{\sqrt{x}+3}< 1\Rightarrow\sqrt{x}+3>5\Rightarrow\sqrt{x}>2\Rightarrow x>4\)

Chú ý kết hợp với điều kiện xác định.

\(1,\frac{\sqrt{x}+1}{\sqrt{x}-3}=\frac{\sqrt{x}-3+4}{\sqrt{x}-3}=1+\frac{4}{\sqrt{x}-3}\)

Để \(\frac{\sqrt{x}+1}{\sqrt{x}-3}\in Z\Rightarrow\frac{4}{\sqrt{x}-3}\in Z\)

\(\Rightarrow\sqrt{x}-3\in\left(1;4;-1;-4\right)\)

\(\Rightarrow\sqrt{x}\in\left(4;7;2;-1\right)\)

\(\Rightarrow\sqrt{x}=4\Leftrightarrow x=2\)

\(4,A=x+\sqrt{x}+1\)

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

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

\(\Rightarrow A\ge\frac{3}{4}.\left(\sqrt{x}+\frac{1}{2}\right)^2\ge0\)

Dấu "=" xảy ra khi :

\(\sqrt{x}+\frac{1}{2}=0\Leftrightarrow\sqrt{x}=-\frac{1}{2}\)

Vậy Min A = 3/4 khi căn x = -1/2

Câu 1:

\(\frac{A}{B}\ge\frac{x}{4}+5\Leftrightarrow\frac{\sqrt{x}+4}{\sqrt{x}-1}:\frac{1}{\sqrt{x}-1}\ge\frac{x}{4}+5\)

\(\Rightarrow\sqrt{x}+4\ge\frac{x}{4}+5\Rightarrow x-4\sqrt{x}+4\le0\)

\(\Rightarrow\left(\sqrt{x}-2\right)^2\le0\Rightarrow\sqrt{x}-2=0\Rightarrow x=4\)

Câu 2:

Bạn coi lại đề, biểu thức B không hợp lý

M = \(\frac{2\sqrt{x}-9x}{x-5\sqrt{x}+6}-\frac{\sqrt{x}+3}{\sqrt{x}-2}-\frac{2\sqrt{x}+1}{3-\sqrt{x}}\)

    =\(\frac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\frac{\left(\sqrt{x}+3\right)\left(3-\sqrt{x}\right)+\left(\sqrt{x}-2\right)\left(2\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(3-\sqrt{x}\right)}\)

    =\(\frac{2\sqrt{x}-9}{x-5\sqrt{x}+6}+\frac{9-x+2x-3\sqrt{x}}{x-5\sqrt{x}+6}\)

    =\(\frac{x-\sqrt{x}}{x-5\sqrt{x}+6}\)

ĐKXĐ :\(\left\{{}\begin{matrix}x\ge0\\\sqrt{x}-3\ne0\\\sqrt{x}+5\ne0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x\ge0\\\sqrt{x}\ne3\\\sqrt{x}\ne-5\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x\ge0\\x\ne9\end{matrix}\right.\)

- Ta có : \(\left(\frac{x-5\sqrt{x}}{25}-1\right):\left(\frac{25-x}{x+2\sqrt{x}-15}-\frac{\sqrt{x}+3}{\sqrt{x}+5}+\frac{\sqrt{x}-5}{\sqrt{x}-3}\right)\)

\(=\left(\frac{x-5\sqrt{x}-25}{25}\right):\left(\frac{25-x}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}-\frac{\sqrt{x}+3}{\sqrt{x}+5}+\frac{\sqrt{x}-5}{\sqrt{x}-3}\right)\)

\(=\left(\frac{x-5\sqrt{x}-25}{25}\right):\left(\frac{25-x}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}-\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}+\frac{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+5\right)}\right)\)

\(=\left(\frac{x-5\sqrt{x}-25}{25}\right):\left(\frac{25-x}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}-\frac{x-9}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}+\frac{x-25}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+5\right)}\right)\)

\(=\left(\frac{x-5\sqrt{x}-25}{25}\right):\left(\frac{25-x-x+9+x-25}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\right)\)

\(=\left(\frac{x-5\sqrt{x}-25}{25}\right):\left(\frac{-x+9}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\right)\)

\(=\left(\frac{x-5\sqrt{x}-25}{25}\right):\left(\frac{-\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\right)\)

\(=\left(\frac{x-5\sqrt{x}-25}{25}\right)\left(\frac{\sqrt{x}+5}{-\sqrt{x}-3}\right)\)

\(=\frac{\left(x-5\sqrt{x}-25\right)\left(\sqrt{x}+5\right)}{-25\left(\sqrt{x}+3\right)}=\frac{x\sqrt{x}+5x-5x-25\sqrt{x}-25\sqrt{x}-125}{-25\left(\sqrt{x}+3\right)}\)

\(=\frac{x\sqrt{x}-125-50\sqrt{x}}{-25\left(\sqrt{x}+3\right)}\)