1)Để căn có nghĩa \(\Leftrightarrow\dfrac{-a}{3}\ge0\Leftrightarrow a\le0\)

Vậy...

2)Để căn có nghĩa \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{a^2+1}{1-3a}\ge0\\1-3a\ne0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}1-3a>0\left(vìa^2+1>0\right)\\1-3a\ne0\end{matrix}\right.\)

\(\Leftrightarrow1-3a>0\Leftrightarrow3a< 1\Leftrightarrow a< \dfrac{1}{3}\)

Vậy...

3)Để căn có nghĩa 

\(\Leftrightarrow a^2-6a+10\ge0\Leftrightarrow\left(a^2-6a+9\right)+1\ge0\Leftrightarrow\left(a-3\right)^2+1\ge0\left(lđ;\forall a\right)\)

Vậy căn luôn có nghĩa với mọi a

4)Để căn có nghĩa \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{a-1}{a+2}\ge0\\a+2\ne0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}\left\{{}\begin{matrix}a-1\ge0\\a+2>0\end{matrix}\right.\\\left\{{}\begin{matrix}a-1\le0\\a+2< 0\end{matrix}\right.\end{matrix}\right.\\a+2\ne0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}a\ge1\\a>-2\end{matrix}\right.\\\left\{{}\begin{matrix}a\le1\\a< -2\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}a\ge1\\a< -2\end{matrix}\right.\)

Vậy...

a) Ta có: \(A=\dfrac{a^2-1}{3}\cdot\sqrt{\dfrac{9}{\left(1-a\right)^2}}\)

\(=\dfrac{\left(a+1\right)\cdot\left(a-1\right)}{3}\cdot\dfrac{3}{\left|1-a\right|}\)

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

=-a-1

b) Ta có: \(B=\sqrt{\left(3a-5\right)^2}-2a+4\)

\(=\left|3a-5\right|-2a+4\)

\(=5-3a-2a+4\)

=9-5a

c) Ta có: \(C=4a-3-\sqrt{\left(2a-1\right)^2}\)

\(=4a-3-\left|2a-1\right|\)

\(=4a-3-2a+1\)

\(=2a-2\)

d) Ta có: \(D=\dfrac{a-2}{4}\cdot\sqrt{\dfrac{16a^4}{\left(a-2\right)^2}}\)

\(=\dfrac{a-2}{4}\cdot\dfrac{4a^2}{\left|a-2\right|}\)

\(=\dfrac{a^2\left(a-2\right)}{-\left(a-2\right)}\)

\(=-a^2\)

1) \(ĐK:3-2a>0\Leftrightarrow a< \dfrac{3}{2}\)

2) \(ĐK:2x-5< 0\Leftrightarrow x< \dfrac{5}{2}\)

3) \(ĐK:3-5a< 0\Leftrightarrow a>\dfrac{3}{5}\)

4) \(ĐK:a< 0\)

5) \(ĐK:-a\ge0\Leftrightarrow a\le0\)

1: \(P=\dfrac{3a+3\sqrt{a}-3-a+1-a+4}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}\)

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

2: Để P nguyên thì \(\sqrt{a}-1+2⋮\sqrt{a}-1\)

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

hay \(a\in\left\{4;0;9\right\}\)

a=3

a =4 .bạn xem MÌNH trả lời câu hỏi của NGUYỄN THỊ DIỆP

a) \(\sqrt{\dfrac{2a}{3}}\cdot\sqrt{\dfrac{3a}{8}}\)

\(=\sqrt{\dfrac{2a\cdot3a}{3\cdot8}}\)

\(=\sqrt{\dfrac{6a^2}{24}}\)

\(=\sqrt{\dfrac{a^2}{4}}\)

\(=\dfrac{\sqrt{a^2}}{\sqrt{4}}\)

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

b) \(\sqrt{3a}\cdot\sqrt{\dfrac{52}{a}}\)

\(=\sqrt{3a\cdot\dfrac{52}{a}}\) 

\(=\sqrt{3\cdot52}\)

\(=\sqrt{13\cdot3\cdot4}\)

\(=2\sqrt{39}\)

c) \(2y^2\cdot\sqrt{\dfrac{x^4}{4y^2}}\)

\(=2y^2\cdot\dfrac{\sqrt{\left(x^2\right)^2}}{\sqrt{\left(2y\right)^2}}\)

\(=2y^2\cdot\dfrac{x^2}{-2y}\)

\(=\dfrac{2y^2\cdot x^2}{-2y}\)

\(=-x^2y\)

`M=sqrt{(3a-1)^2}+2a-3`

`=|3a-1|+2a-3`

`=3a-1+2a-3(do \ a>=1/3)`

`=5a-4`

`N=sqrt{(4-a)^2}-a+5`

`=|4-a|-a+5`

`=a-4-a+5(do \ a>4)`

`=1`

`I=sqrt{(3-2a)^2}+2-7`

`=|3-2a|-5`

`=3-2a-5(do \ a<3/2)`

`=-2-2a`

`K=(a^2-9)/4*sqrt{4/(a-2)^2}`

`=(a^2-9)/4*|2/(a-2)|`

`=(a^2-9)/(2|a-2|)`

Nếu `3>a>2=>|a-2|=a-2`

`=>K=(a^2-9)/(2(a-2))`

Nếu `a<2=>|a-2|=2-a`

`=>K=(a^2-9)/(2(2-a))`

\(M=\left|3a-1\right|+2a-3\)

Mà \(a-\dfrac{1}{3}\ge0\)

\(\Rightarrow M=3a-1+2a-3=5a-4\)

\(N=\left|4-a\right|-a+5\)

Mà \(4-a< 0\)

\(\Rightarrow N=a-4-a+5=1\)

\(I=\left|3-2a\right|-5\)

Mà \(a-\dfrac{3}{2}< 0\)

\(\Rightarrow I=3-2a-5=-2a-2\)

K, Ta có : \(a-3< 0\)

\(\Rightarrow K=\dfrac{2\left(a^2-9\right)}{4\left|a-2\right|}=\dfrac{\left(a-3\right)\left(a+3\right)}{\left|2a-4\right|}\)
 

\(A=\left|a-3\right|-3a=3-a-3a=3-4a\)

\(B=4a+3-\left|2a-1\right|=4a+3-2a+1=2a+4\)

\(C=\dfrac{4}{a^2-4}\left|a-2\right|=\dfrac{-4\left(a-2\right)}{\left(a-2\right)\left(a+2\right)}=\dfrac{-4}{a+2}\)

\(D=\dfrac{a^2-9}{12}:\sqrt{\dfrac{\left(a+3\right)^2}{16}}=\dfrac{a^2-9}{12}:\dfrac{\left|a+3\right|}{4}=\dfrac{\left(a-3\right)\left(a+3\right).4}{-12\left(a+3\right)}=\dfrac{3-a}{3}\)

\(A=\sqrt{\left(a-3\right)^2}-3a\)

=3-a-3a

=3-4a

a: \(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{4}+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

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

b: \(\sqrt{\dfrac{2a}{3}}\cdot\sqrt{\dfrac{3a}{8}}=\sqrt{\dfrac{6a^2}{24}}=\sqrt{\dfrac{a^2}{4}}=\dfrac{a}{2}\)

c: \(\sqrt{5a\cdot45a}-3a=-15a-3a=-18a\)