a) điều kiện xác định : \(a>0\)

ta có : \(D=\dfrac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-\dfrac{2a+\sqrt{a}}{\sqrt{a}}+1\)

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

b) ta có : \(D=2\Leftrightarrow x-\sqrt{x}=2\Leftrightarrow x-\sqrt{x}-2=0\)

\(\Leftrightarrow\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}+1=0\\\sqrt{x}-2=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x\in\varnothing\\x=4\end{matrix}\right.\)

vậy \(x=4\)

c) ta có : \(a>1\Leftrightarrow a-1>0\Leftrightarrow\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)>0\)

\(\Leftrightarrow\sqrt{a}-1>0\Leftrightarrow\sqrt{a}\left(\sqrt{a}-1\right)>0\Leftrightarrow a-\sqrt{a}>0\)

\(\Rightarrow\left|D\right|=\left|a-\sqrt{a}\right|=a-\sqrt{a}=D\) vậy \(D=\left|D\right|\)

d) ta có : \(D=a-\sqrt{a}\Leftrightarrow a-\sqrt{a}-D=0\)

phương trình này luôn có nghiệm \(\Rightarrow\Delta\ge0\)

\(\Leftrightarrow1^2-4\left(-D\right)=4D+1\ge0\Leftrightarrow D\ge\dfrac{-1}{4}\)

\(\Rightarrow D_{min}=\dfrac{-1}{4}\) khi \(\sqrt{a}=\dfrac{-b}{2a}=\dfrac{1}{2}\Leftrightarrow a=\dfrac{1}{4}\)

vậy \(D_{min}=\dfrac{-1}{4}\) khi \(a=\dfrac{1}{4}\)

a) \(P=\dfrac{\sqrt{a}\left[\left(\sqrt{a}\right)^3+1\right]}{a-\sqrt{a}+1}-\dfrac{\sqrt{a}\left(2\sqrt{a}+1\right)}{\sqrt{a}}+1\)

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

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

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

\(P=a-\sqrt{a}\)

b) Với a > 1 thì \(a>\sqrt{a}\) , do đó \(P=a-\sqrt{a}>0\), suy ra \(\left|P\right|=P\)

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

Vậy A nhỏ nhất bằng \(-\dfrac{1}{4}\) khi cà chỉ khi \(\sqrt{a}=\dfrac{1}{2}\) hay \(a=\dfrac{1}{4}\)

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

b: a>1 nên P>0

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

Lời giải:
a.

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

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

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

b.

$A=a-\sqrt{a}=(\sqrt{a}-0,5)^2-0,25\geq -0,25$ với mọi $a>0$

Vậy $A_{\min}=-0,25$ khi $\sqrt{a}-0,5=0$

$\Leftrightarrow a=0,25$

a) Để biểu thức có nghĩa thì \(\dfrac{-a}{3}\ge0\Rightarrow a\le0\)

b) Để biểu thức có nghĩa thì \(\dfrac{1}{a^2}\ge0\) (luôn đúng)

c) Để biểu thức có nghĩa thì \(\dfrac{\left(1-a\right)^3}{a^2}\ge0\Rightarrow\left\{{}\begin{matrix}\left(1-a\right)^3\ge0\\a\ne0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}1-a\ge0\\a\ne0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}a\le1\\a\ne0\end{matrix}\right.\)

d) Để biểu thức có nghĩa thì \(\dfrac{a^2+1}{1-2a}\ge0\Rightarrow1-2a>0\Rightarrow a< \dfrac{1}{2}\)

e) Để biểu thức có nghĩa thì \(a^2-1\ge0\Rightarrow a^2\ge1\Rightarrow\left|a\right|\ge1\)

f) Để biểu thức có nghĩa thì \(\Rightarrow\dfrac{2a-1}{2-a}\ge0\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2a-1\ge0\\2-a>0\end{matrix}\right.\\\left\{{}\begin{matrix}2a-1\le0\\2-a< 0\end{matrix}\right.\end{matrix}\right.\)

\(\left[{}\begin{matrix}\left\{{}\begin{matrix}a\ge\dfrac{1}{2}\\a< 2\end{matrix}\right.\\\left\{{}\begin{matrix}a\le\dfrac{1}{2}\\a>2\end{matrix}\right.\left(l\right)\end{matrix}\right.\Rightarrow\dfrac{1}{2}\le a< 2\)

ĐKXĐ: \(x>0;a\ne9\)

\(A=\left(\dfrac{\sqrt{a}+3}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}+\dfrac{\sqrt{a}-3}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}\right)\left(\dfrac{\sqrt{a}-3}{\sqrt{a}}\right)\)

\(=\left(\dfrac{\sqrt{a}+3+\sqrt{a}-3}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}\right)\left(\dfrac{\sqrt{a}-3}{\sqrt{a}}\right)\)

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

b.

\(A>\dfrac{1}{2}\Rightarrow\dfrac{2}{\sqrt{a}+3}>\dfrac{1}{2}\Rightarrow\sqrt{a}+3< 4\)

\(\Rightarrow\sqrt{a}< 1\Rightarrow a< 1\)

Kết hợp ĐKXĐ \(\Rightarrow0< a< 1\)

ĐKXĐ: \(a>0\)

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

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

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

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

\(\Rightarrow a-\sqrt{a}-2=0\Rightarrow\left[{}\begin{matrix}\sqrt{a}=-1\left(l\right)\\\sqrt{a}=2\end{matrix}\right.\) \(\Rightarrow a=4\)

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

Với \(a>1\Rightarrow\sqrt{a}-1>0\Rightarrow D>0\Rightarrow D=\left|D\right|\)

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

\(\Rightarrow D_{min}=-\frac{1}{4}\) khi \(\sqrt{a}=\frac{1}{2}\Rightarrow a=\frac{1}{4}\)

điều kiện a> 0 

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

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

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

b,  D = 2 => \(a-\sqrt{a}=2\Leftrightarrow a-\sqrt{a}-2=0\)

                                                     \(\Leftrightarrow\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)=0\Leftrightarrow\sqrt{a}-1=0\)( vì a > 0 nên \(\sqrt{a}+1>0\))

                                                        \(\Leftrightarrow a=1\)

c, a > 1 =>  \(\sqrt{a}>1\Rightarrow\sqrt{a}-1>0\)

              \(\Rightarrow D=a-\sqrt{a}=\sqrt{a}\left(\sqrt{a}-1\right)>0\)

            Vậy D = | D |  > 0 

d, \(D=a-\sqrt{a}=a-\sqrt{a}+\frac{1}{4}-\frac{1}{4}=\left(\sqrt{a}-\frac{1}{2}\right)^2-\frac{1}{4}\ge-\frac{1}{4}\)với mọi a > 0 

   vậy Dmin = - 1/4 khi a = 1/4

xin lỗi phàn b anh làm sai. Sửa lại như sau :

b, D = 2 => \(a-\sqrt{a}=2\Rightarrow a-\sqrt{a}-2=0\Leftrightarrow\left(\sqrt{a}-2\right)\left(\sqrt{a}+1\right)=0.\)

                                                                                                    \(\Leftrightarrow\sqrt{a}-2=0\)( vì a > 0, nên căn a + 1 > 0 )

                                                                                                     \(\Leftrightarrow a=4\)