a) \(\frac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-\frac{2a+\sqrt{a}}{\sqrt{a}}+1\)

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

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

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

b) \(\frac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-2\sqrt{a}=2\)

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

\(\Leftrightarrow a^2-2\sqrt{a}.a+2a-\sqrt{a}-2a=2a-2\sqrt{a}+2\)

\(\Leftrightarrow a^2-2\sqrt{a}.a+2a-\sqrt{a}-2a=-2\sqrt{a}+2\)

\(\Leftrightarrow-2\sqrt{a}.a+2a-\sqrt{a}-2a=-2\sqrt{a}+2-a^2\)

\(\Leftrightarrow-2\sqrt{a}.a-\sqrt{a}=-2\sqrt{a}+2-a^2\)

\(\Leftrightarrow-2a\sqrt{a}+\sqrt{a}=2-a^2\)

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

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

\(\Leftrightarrow4a^3-4a^2+a=4-4a^2+a^4\)

\(\Leftrightarrow\orbr{\begin{cases}a=4\left(\text{thỏa mãn}\right)\\a=1\left(\text{loại}\right)\end{cases}}\)

=> a = 4

Cách ngắn hơn :

\(đkxđ\Leftrightarrow x\ge0\)

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

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

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

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

\(=a+\sqrt{a}-2\sqrt{a}=a-\sqrt{a}\)

\(b,A=2\Rightarrow a-\sqrt{a}=2\)

\(\Rightarrow a-\sqrt{a}-2=0\)

\(\Rightarrow a+\sqrt{a}-2\sqrt{a}-2=0\)

\(\Rightarrow\sqrt{a}\left(\sqrt{a}+1\right)-2\left(\sqrt{a}+1\right)=0\)

\(\Rightarrow\left(\sqrt{a}-2\right)\left(\sqrt{a}+1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}\sqrt{a}=2\\\sqrt{a}=-1\end{cases}\Rightarrow\orbr{\begin{cases}a=4\\a\in\varnothing\end{cases}}}\)

\(\Rightarrow a=4\)

\(c,A=a-\sqrt{a}=\sqrt{a}^2-2.\sqrt{a}.\frac{1}{2}+\frac{1}{4}-\frac{1}{4}\)

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

\(\Rightarrow A_{min}=-\frac{1}{4}\Leftrightarrow\left(\sqrt{a}-\frac{1}{2}\right)^2=0\)

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

Vậy với \(a=\frac{1}{4}\)thì A có giá trị nhỏ nhất là \(-\frac{1}{4}\)

Làm màu quá :))

a) \(A=\frac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-\frac{2a+\sqrt{a}}{\sqrt{a}}+1\left(ĐK:a\ge0\right)\)

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

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

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

b) Để A=2 \(\Leftrightarrow a-\sqrt{a}=2\)

                  \(\Leftrightarrow a-\sqrt{a}-2=0\)

                   \(\Leftrightarrow\left(\sqrt{a}+1\right)\left(\sqrt{a}-2\right)=0\)

                   \(\Leftrightarrow\sqrt{a}-2=0\left(Vì\sqrt{a}+1\ne0\right)\)

                   \(\Leftrightarrow a=4\) (TM)

Vậy a=4 thì A=2

c) \(A=a-\sqrt{a}=a-\sqrt{a}+\frac{1}{4}-\frac{1}{4}=\left(\sqrt{a}-\frac{1}{2}\right)^2-\frac{1}{4}\)

Vì: \(\left(\sqrt{a}-\frac{1}{2}\right)^2\ge0\)

=> \(\left(\sqrt{a}-\frac{1}{2}\right)^2-\frac{1}{4}\ge-\frac{1}{4}\)

Vậy GTNN của A là \(-\frac{1}{4}\) khi \(a=\frac{1}{4}\)

Bài 1

a) \(P=\frac{3a+\sqrt{9a}-3}{a+\sqrt{a}-2}-\frac{\sqrt{a}+1}{\sqrt{a}+2}+\frac{\sqrt{a}-2}{1-\sqrt{a}}\)    (ĐK : x\(\ge0\) ; x\(\ne\) 1)

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

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

         \(=\frac{3a+\sqrt{9a}-3-a+1-a+4}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}\)

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

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

         \(=\frac{\sqrt{a}+1}{\sqrt{a}-1}\)

b) \(P=\frac{\sqrt{a}+1}{\sqrt{a}-1}=\frac{\sqrt{a}-1+2}{\sqrt{a}-1}=1+\frac{2}{\sqrt{a}-1}\)

Vậy để P là số nguyên thì: \(\sqrt{a}-1\inƯ\left(2\right)\)

Mà Ư(2)={-1;1;2;-1}

=> \(\sqrt{a}-1\in\left\{1;-1;2;-2\right\}\)

Ta có bảng sau:

vậy a={0;4;9} thì P nguyên

Bài 2

  \(P=\frac{\sqrt{a+4\sqrt{a-4}}+\sqrt{a-4\sqrt{a-4}}}{\sqrt{1-\frac{8}{a}+\frac{16}{a^2}}}\)(ĐK:a\(\ge\)8)

      \(=\frac{\sqrt{\left(a-4\right)+4\sqrt{a-4}+4}+\sqrt{\left(a-4\right)-4\sqrt{a-4}+4}}{\sqrt{\left(1-\frac{4}{a}\right)^2}}\)

     \(=\frac{\sqrt{\left(\sqrt{a-4}+2\right)^2}+\sqrt{\left(\sqrt{a-4}-2\right)^2}}{1-\frac{4}{a}}\)

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

     \(=2\sqrt{a-4}\cdot\frac{a}{a-4}\)

     \(=\frac{2a}{\sqrt{a-4}}\)