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 \(\Rightarrow\dfrac{2a-1}{2-a}\ge0\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2a-1\ge0\\2-a>0\end{matrix}\right.\\\left\{{}\begin{matrix}2a-1\le0\\2-a< 0\end{matrix}\right.\end{matrix}\right.\)

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

1) Để biểu thức có nghĩa thì \(a^2+2a-3\ge0\)

\(\Leftrightarrow\left(a+3\right)\left(a-1\right)\ge0\)

\(\Leftrightarrow\left[{}\begin{matrix}a-1\ge0\\a+3\le0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a\ge1\\a\le-3\end{matrix}\right.\)

2) Để biểu thức có nghĩa thì \(\left\{{}\begin{matrix}a-1\ge0\\a\ne0\end{matrix}\right.\Leftrightarrow a\ge1\)

3) Để biểu thức có nghĩa thì \(a>0\)

4) Để biểu thức có nghĩa thì \(\left\{{}\begin{matrix}a\ne-\dfrac{1}{2}\\\left[{}\begin{matrix}a-1\ge0\\2a+1< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a\ne-\dfrac{1}{2}\\\left[{}\begin{matrix}a\ge1\\a< -\dfrac{1}{2}\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a\ge1\\a< -\dfrac{1}{2}\end{matrix}\right.\)

1) Để biểu thức có nghĩa  \(\Rightarrow a^2+2a-3\ge0\Rightarrow\left(a-1\right)\left(a+3\right)\ge0\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}a-1\ge0\\a+3\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}a-1\le0\\a+3\le0\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}a\ge1\\a\le-3\end{matrix}\right.\)

2) Để biểu thức có nghĩa \(\Rightarrow\dfrac{\left(a-1\right)^3}{a^2}\ge0\Rightarrow\left\{{}\begin{matrix}\left(a-1\right)^3\ge0\\a\ne0\end{matrix}\right.\Rightarrow a\ge1\)

3) Để biểu thức có nghĩa \(\Rightarrow\dfrac{a^2+1}{2a}\ge0\Rightarrow2a>0\Rightarrow a>0\)

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

\(\Rightarrow\left[{}\begin{matrix}a\ge1\\a< -\dfrac{1}{2}\end{matrix}\right.\)

a) \(A=\left(\dfrac{2a+1}{\sqrt{a^3}-1}-\dfrac{\sqrt{a}}{a+\sqrt{a}+1}\right)\left(\dfrac{1+\sqrt{a^3}}{1+\sqrt{a}}-\sqrt{a}\right)\left(đk:a\ge0,a\ne1\right)\)

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

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

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

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

b) \(A=\sqrt{a}-1=6\)

\(\Leftrightarrow\sqrt{a}=7\Leftrightarrow a=49\)

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

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

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

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

c) Để A=2 thì \(a-\sqrt{a}-2=0\)

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

\(\Leftrightarrow a=4\)

\(P=\left(\dfrac{3\sqrt{a}}{a+\sqrt{ab}+\sqrt{b}}-\dfrac{3a}{a\sqrt{a}-b\sqrt{b}}+\dfrac{1}{\sqrt{a}-\sqrt{b}}\right):\dfrac{\left(a-1\right)\left(\sqrt{a}-\sqrt{b}\right)}{2a+2\sqrt{ab}+2b}\left(đk:a\ne b,a\ge0,b\ge0\right)\)

\(=\dfrac{3a-3\sqrt{ab}-3a+a+\sqrt{ab}+b}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+\sqrt{b}\right)}.\dfrac{2\left(a+\sqrt{ab}+b\right)}{\left(a-1\right)\left(\sqrt{a}-\sqrt{b}\right)}\)

\(=\dfrac{a-2\sqrt{ab}+b}{\sqrt{a}-\sqrt{b}}.\dfrac{2}{\left(a-1\right)\left(\sqrt{a}-\sqrt{b}\right)}=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2.2}{\left(\sqrt{a}-\sqrt{b}\right)^2\left(a-1\right)}=\dfrac{2}{a-1}\in Z\)

\(\Rightarrow a-1\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\)

Do \(a\ge0\)

\(\Rightarrow a\in\left\{0;2;3\right\}\)

Ta có: \(P=\left(\dfrac{3\sqrt{a}}{a+\sqrt{ab}+b}-\dfrac{3a}{a\sqrt{a}-b\sqrt{b}}+\dfrac{1}{\sqrt{a}-\sqrt{b}}\right):\left(\dfrac{\left(a-1\right)\left(\sqrt{a}-\sqrt{b}\right)}{2a+2\sqrt{ab}+2b}\right)\)

\(=\dfrac{3a-3\sqrt{ab}-3a+a+\sqrt{ab}+b}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}\cdot\dfrac{2\left(a+\sqrt{ab}+b\right)}{\left(a-1\right)\left(\sqrt{a}-\sqrt{b}\right)}\)

\(=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\left(\sqrt{a}-\sqrt{b}\right)^2}\cdot\dfrac{2}{a-1}\)

\(=\dfrac{2}{a-1}\)

Để P là số nguyên thì \(a-1\in\left\{1;-1;2;-2\right\}\)

hay \(a\in\left\{2;0;3\right\}\)

a ĐKXĐ \(a\ge0,a\ne\dfrac{1}{4},a\ne1\)

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

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

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

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

b Xét hiệu \(P-\dfrac{2}{3}=\dfrac{a+1}{a+\sqrt{a}+1}-\dfrac{2}{3}=\dfrac{3a+3-2a-2\sqrt{a}-2}{a+\sqrt{a}+1}=\dfrac{a-2\sqrt{a}+1}{a+\sqrt{a}+1}=\dfrac{\left(\sqrt{a}-1\right)^2}{a+\sqrt{a}+\dfrac{1}{4}+\dfrac{3}{4}}=\dfrac{\left(\sqrt{a}-1\right)^2}{\left(\sqrt{a}+\dfrac{1}{2}\right)^2+\dfrac{3}{4}}>0\) \(\Rightarrow P>\dfrac{2}{3}\) 

c Ta có \(P=\dfrac{\sqrt{6}}{\sqrt{6}+1}\Rightarrow\dfrac{a+1}{a+\sqrt{a}+1}=\dfrac{\sqrt{6}}{\sqrt{6}+1}\) \(\Rightarrow\left(a+1\right)\left(\sqrt{6}+1\right)=\sqrt{6}\left(a+\sqrt{a}+1\right)\Leftrightarrow a\sqrt{6}+a+\sqrt{6}+1=a\sqrt{6}+\sqrt{6a}+\sqrt{6}\)

\(\Leftrightarrow a-\sqrt{6a}+1=0\Leftrightarrow a-\sqrt{6a}+\dfrac{6}{4}-\dfrac{2}{4}=0\Leftrightarrow\left(\sqrt{a}-\dfrac{\sqrt{6}}{2}\right)^2=\dfrac{1}{2}\)

\(\Rightarrow\left[{}\begin{matrix}\sqrt{a}=\dfrac{\sqrt{6}+1}{2}\\\sqrt{a}=\dfrac{1-\sqrt{6}}{2}\left(L\right)\end{matrix}\right.\) (Do \(\sqrt{a}\ge0\))  \(\Rightarrow a=\dfrac{\left(\sqrt{6}+1\right)^2}{4}=\dfrac{7+2\sqrt{6}}{4}\left(TM\right)\) 

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