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

\(a,\dfrac{a\sqrt{a}-8+2a-4\sqrt{a}}{a-4}\left(dk:a\ne4\right)\)

\(=\dfrac{a\sqrt{a}-4\sqrt{a}-8+2a}{a-4}\)

\(=\dfrac{\sqrt{a}\left(a-4\right)+2\left(a-4\right)}{a-4}\)

\(=\dfrac{\left(a-4\right)\left(\sqrt{a}+2\right)}{a-4}\)

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

\(b,\dfrac{12\sqrt{6}}{\sqrt{7+2\sqrt{6}}-\sqrt{7-2\sqrt{6}}}\\ =\dfrac{12\sqrt{6}}{\sqrt{\left(\sqrt{6}+1\right)^2}-\sqrt{\left(\sqrt{6}-1\right)^2}}\\ =\dfrac{12\sqrt{6}}{\left|\sqrt{6}+1\right|-\left|\sqrt{6}-1\right|}\\ =\dfrac{12\sqrt{6}}{\sqrt{6}+1-\sqrt{6}+1}\\ =\dfrac{12\sqrt{6}}{2}\\ =6\sqrt{6}\)

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

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

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

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

\(=-x^2y\)

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

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

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 \(\sqrt{a}\ge0\))  \(\Rightarrow a=\dfrac{\left(\sqrt{6}+1\right)^2}{4}=\dfrac{7+2\sqrt{6}}{4}\left(TM\right)\) 

Vậy...

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: Ta có: \(\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)

\(=\dfrac{15\sqrt{x}-11-\left(3x+7\sqrt{x}-6\right)-\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{15\sqrt{x}-11-3x-7\sqrt{x}+6-2x+2\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{-5\sqrt{x}+2}{\sqrt{x}+3}\)

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

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

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

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

a,ĐKXĐ: tự tìm :v

 \(\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{3+\sqrt{x}}\)

\(=\dfrac{15\sqrt{x}-11}{\left(x+2\sqrt{x}+1\right)-4}+\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{3+\sqrt{x}}\)

\(=\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}+1\right)^2-4}+\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{3+\sqrt{x}}\)

\(=\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\dfrac{3\sqrt{x}-2}{\sqrt{x}-1}+\dfrac{2\sqrt{x}+3}{3+\sqrt{x}}\)

\(=\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\dfrac{\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}+\dfrac{\left(\sqrt{x}-1\right)\left(2\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\dfrac{3x+7\sqrt{x}-6}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}+\dfrac{2x+\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{15\sqrt{x}-11-3x-7\sqrt{x}+6+2x+\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{9\sqrt{x}-x-8}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{\left(9\sqrt{x}-9\right)-\left(x-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{9\left(\sqrt{x}-1\right)-\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{\left(\sqrt{x}-1\right)\left(10-\sqrt{x}\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(\dfrac{10-\sqrt{x}}{\sqrt{x}+3}\)

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

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

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

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

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

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

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

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

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