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

Bài 1: 

a: Ta có: \(x^2-2\sqrt{5}x+5=0\)

\(\Leftrightarrow x-\sqrt{5}=0\)

hay \(x=\sqrt{5}\)

b: Ta có: \(\sqrt{x+3}=1\)

\(\Leftrightarrow x+3=1\)

hay x=-2

a. ĐKXĐ: \(x>0\)

\(P=\left(\dfrac{1}{\sqrt{x}}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right):\dfrac{\sqrt{x}}{x+\sqrt{x}}\)

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

b. Để \(P=-1\) thÌ  \(\dfrac{x+\sqrt{x}+1}{\sqrt{x}}=-1\) 

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

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

\(\Leftrightarrow\left(\sqrt{x}+1\right)^2=0\)

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

\(\Leftrightarrow\sqrt{x}=-1\) ( vô lý )

Vậy không có x thỏa mãn ycbt

c. Ta có \(x=\dfrac{8}{\sqrt{5}-1}-\dfrac{8}{\sqrt{5}+1}=\dfrac{8\sqrt{5}+8-8\sqrt{5}+8}{5-1}=\dfrac{16}{4}=4\)

Thay x=4 vào P, ta được

\(P=\dfrac{4+\sqrt{4}+1}{\sqrt{4}}=\dfrac{4+2+1}{2}=\dfrac{7}{2}\)

d. \(P=\dfrac{x+\sqrt{x}+1}{\sqrt{x}}\) \(\Rightarrow P-3=\dfrac{x+\sqrt{x}+1}{\sqrt{x}}-3\)

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

Mà \(\left\{{}\begin{matrix}\left(\sqrt{x}-1\right)^2\ge0\\\sqrt{x}>0\end{matrix}\right.\) \(\Rightarrow P-3\ge0\Rightarrow P\ge3\)

Dấu "=" xảy ra khi \(\left(\sqrt{x}-1\right)^2=0\Leftrightarrow\sqrt{x}-1=0\Leftrightarrow\sqrt{x}=1\Leftrightarrow x=1\)

Vậy \(P_{min}=3\) khi \(x=1\)

a: Ta có: \(N=\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}\)

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

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

a: Ta có: \(A=\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}\)

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

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

Lời giải:
a.

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

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

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

b.

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

Vậy $A_{\min}=\frac{3}{4}$ khi $\sqrt{x}=\frac{1}{2}\Leftrightarrow x=\frac{1}{4}$

Với \(x< 9\) biểu thức này chỉ có max, ko có min

Để có min thì cần \(x>9\)

\(P=\dfrac{x-5}{\sqrt{x}-3}=\dfrac{x-9+4}{\sqrt{x}-3}=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)+4}{\sqrt{x}-3}\)

\(P=\sqrt{x}+3+\dfrac{4}{\sqrt{x}-3}=\sqrt{x}-3+\dfrac{4}{\sqrt{x}-3}+6\)

\(P\ge2\sqrt{\left(\sqrt{x}-3\right).\dfrac{4}{\sqrt{x}-3}}+6=10\)

\(P_{min}=10\) khi \(\sqrt{x}-3=\dfrac{4}{\sqrt{x}-3}\Leftrightarrow\sqrt{x}-3=2\Rightarrow x=25\)

Nếu chưa học BĐT Cô-si như cách làm trên thì:

\(P=\dfrac{x-5}{\sqrt{x}-3}=\dfrac{x-10\sqrt{x}+25+10\sqrt{x}-30}{\sqrt{x}-3}\)

\(P=\dfrac{\left(\sqrt{x}-5\right)^2+10\left(\sqrt{x}-3\right)}{\sqrt{x}-3}=\dfrac{\left(\sqrt{x}-5\right)^2}{\sqrt{x}-3}+10\)

Do \(x>9\Rightarrow\sqrt{x}>3\Rightarrow\sqrt{x}-3>0\Rightarrow\dfrac{\left(\sqrt{x}-5\right)^2}{\sqrt{x}-3}>0\)

\(\Rightarrow P\ge10\)

a: Thay x=9 vào A, ta được:

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

\(B=\dfrac{3\sqrt{x}-15+20-2\sqrt{x}}{x-25}=\dfrac{\sqrt{x}+5}{x-25}=\dfrac{1}{\sqrt{x}-5}\)

b: Để \(A=B\cdot\left|x-4\right|\) thì \(\left|x-4\right|=\dfrac{A}{B}=\dfrac{\sqrt{x}+2}{\sqrt{x}-5}:\dfrac{1}{\sqrt{x}-5}=\sqrt{x}+2\)

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

\(\Leftrightarrow x-\sqrt{x}-6=0\)

=>x=9

\(M=\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}\left(x\ge0,x\ne1\right)\)

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

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

2) Thay x=9 vào M đã rút gọn ta được:

\(M=\dfrac{\sqrt{9}-1}{9+\sqrt{9}+1}=\dfrac{2}{13}\)

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

\(\Leftrightarrow x.M+\sqrt{x}\left(M-1\right)+1+M=0\) (*)

Tại x=0 pt (*) <=> M=-1  (1)

Tại x khác 0, coi pt (*) là pt bậc 2 ẩn \(\sqrt{x}\)

Pt (*) có nghiệm không âm <=> \(\left\{{}\begin{matrix}\Delta\ge0\\S\ge0\\P\ge0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}-3M^2-6M+1\ge0\\\dfrac{1-M}{M}\ge0\\\dfrac{1+M}{M}\ge0\end{matrix}\right.\)

\(\Rightarrow0< M\le\dfrac{-3+2\sqrt{3}}{3}\) (2)

Từ (1) (2)=>  \(M_{min}=-1\) <=> x=0