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=3

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

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: Sửa đề: căn 6+2căn 5-căn 5

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

b: \(a^3=2-\sqrt{3}+2+\sqrt{3}+3a\)

=>a^3-3a-4=0

=>a^3-3a=4

\(\dfrac{64}{\left(a^2-3\right)^3}-3a=\left(\dfrac{4}{a^2-3}\right)^3-3a\)

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

=4

b)CM: \(ab\sqrt{1+\dfrac{1}{a^2b^2}}-\sqrt{a^2b^2+1}=0\)

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

\(VT=ab\dfrac{\sqrt{a^2b^2+1}}{ab}-\sqrt{a^2b^2+1}\)

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

\(VT=0=VP\)

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) điều kiện xác định : \(a\ge0;a\ne1\)

ta có : \(P=\dfrac{3a+\sqrt{9a}-3}{a+\sqrt{a}-2}-\dfrac{\sqrt{a}-2}{\sqrt{a}-1}+\dfrac{1}{\sqrt{a}+2}-1\)

để \(\left|P\right|=1\Leftrightarrow\left|\dfrac{\sqrt{a}+1}{\sqrt{a}-1}\right|=1\) \(\Leftrightarrow\left[{}\begin{matrix}\dfrac{\sqrt{a}+1}{\sqrt{a}-1}=1\\\dfrac{\sqrt{a}+1}{\sqrt{a}-1}=-1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{\sqrt{a}+1}{\sqrt{a}-1}-1=0\\\dfrac{\sqrt{a}+1}{\sqrt{a}-1}+1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\dfrac{2}{\sqrt{a}-1}=0\\\dfrac{2\sqrt{a}}{\sqrt{a}-1}=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}2=0\left(vôlí\right)\\2\sqrt{a}=0\end{matrix}\right.\Rightarrow a=0\)

vậy \(a=0\)

câu b đề bị sai rồi . thế \(a=1\) vào là bt

Ta có: \(\sqrt{\dfrac{a-1}{a+2}}:\sqrt{\dfrac{a+2}{a^3-3a^2+3a-1}}\)

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

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

`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 : \(a-3< 0\)

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