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

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

tth

\(A=\left(\frac{1}{\sqrt{a}-3}+\frac{1}{\sqrt{a}+3}\right)\left(1-\frac{3}{\sqrt{a}}\right)\) \(đk:a>0;a\ne9\)

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

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

\(=\frac{2}{\sqrt{a}+3}\)

\(đk:a>0;a\ne9\)

\(A>\frac{1}{2}=>\frac{2}{\sqrt{a}+3}>\frac{1}{2}\)

\(=>4>\sqrt{a}+3\)

\(< =>\sqrt{a}>1\)

\(< =>a=1\)

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

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

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

          = \(\frac{2}{a-1}\)

b) P nguyên <=> \(\frac{2}{a-1}\)nguyên => 2 \(⋮\)a - 1 

=> ( a- 1 ) = { \(\pm\)1 ; \(\pm\) 2} => a = { -1 ; 0 ; 2 ;3 } 

\(a,ĐKXĐ:\hept{\begin{cases}a>0\\a\ne1\end{cases}}\)

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

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

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

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

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

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

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

\(c,\)Để A= -4 thì 

\(-2\sqrt{a}=-4\Leftrightarrow\sqrt{a}=2\Leftrightarrow a=4\)

Kết bạn với mình nha ....

a: \(A=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}\)

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

b: \(B=\dfrac{2\sqrt{x}-x-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{x+\sqrt{x}+1}{x-1}\)

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

c: \(C=\dfrac{x-9-x+3\sqrt{x}}{x-9}:\left(\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}-2}{\sqrt{x}+3}+\dfrac{x-9}{x+\sqrt{x}-6}\right)\)

\(=\dfrac{3\left(\sqrt{x}-3\right)}{x-9}:\dfrac{9-x+x-4\sqrt{x}+4+x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)

\(=\dfrac{3}{\sqrt{x}+3}\cdot\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}{x-4\sqrt{x}+4}\)

\(=\dfrac{3}{\sqrt{x}-2}\)