a) ĐS: ; b) ĐS: 26; c) ĐS: 12a

d) - = - 6a + 9 -

= - 6a + 9 - = - 6a + 9 - 6│a│.

Khi a ≥ 0 thì │a│= a.

Do đó - = - 6a + 9 -6a = - 12a + 9.

Khi a < 0 thì │a│= a.

Do đó - = - 6a + 9 + 6a = + 9.

a, \(\sqrt{\frac{2a}{3}}.\sqrt{\frac{3a}{8}}=\sqrt{\frac{6a^2}{24}}=\sqrt{\frac{a^2}{4}}=\left|\frac{a}{2}\right|=\frac{a}{2}\)

do \(a\ge0\)

b, \(\sqrt{13a}.\sqrt{\frac{52}{a}}=\sqrt{\frac{676a}{a}}=\sqrt{676}=26\)

c, \(\sqrt{5a}.\sqrt{45a}-3a=\sqrt{225a^2}-3a=\left|15a\right|-3a\)

\(=15a-3a=12a\)do a > 0 

d, \(=\left(3-a\right)^2-\sqrt{0,2}.\sqrt{180a^2}\)

\(=\left(3-a\right)^2-\sqrt{36a^2}=\left(3-a\right)^2-\left|6a\right|\)

Với \(a\ge0\Rightarrow\left(3-a\right)^2-6a=a^2-6a+9-6a=a^2-12a+9\)

Với \(a< 0\Rightarrow\left(3-a\right)^2+6a=a^2-6a+9+6a=a^2+9\)

a) Ta có:

b) Ta có:

c) Do a ≥ 0 nên bài toán luôn xác định. Ta có:

d) Ta có:

b) \(\sqrt{13a}\).\(\sqrt{\frac{52}{a}}\)=\(\sqrt{13a.\frac{52}{a}}\)=\(\sqrt{13.13.2.2}\)=13.2=26

\(A=a+2\sqrt{a}-3\sqrt{a}-6-a-2\sqrt{a}-1+3\sqrt{a}\)

\(A=-7\)

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

\(=a-3\sqrt{a}+2\sqrt{a}-6-a-2\sqrt{a}-1+3\sqrt{a}\)

\(=-7\)

a) Ta có: \(\sqrt{27\cdot48\left(1-a^2\right)}\)

\(=\sqrt{3^4\cdot4^2\cdot\left(1-a^2\right)}\)

\(=36\sqrt{1-a^2}\)

c) Ta có: \(\sqrt{5a}\cdot\sqrt{45a}-3a\)

\(=15a-3a=12a\)

b) Ta có: \(B=\dfrac{1}{a-b}\cdot\sqrt{a^4\cdot\left(a-b\right)^2}\)

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

\(=a^2\)

d) Ta có: \(D=\left(3-a\right)^2-\sqrt{0.2}\cdot\sqrt{180a^2}\)

\(=a^2-6a+9-\sqrt{36a^2}\)

\(=a^2-6a+9-\left|6a\right|\)

\(=\left[{}\begin{matrix}a^2-6a+9-6a\left(a\ge0\right)\\a^2-6a+9+6a\left(a< 0\right)\end{matrix}\right.\)

\(=\left[{}\begin{matrix}a^2-12a+9\\a^2+9\end{matrix}\right.\)

\(A=9.4\left|1-a\right|=36\left(a-1\right)\) (a>1)

\(B=\dfrac{a^2\left|a-b\right|}{a-b}=\dfrac{a^2\left(a-b\right)}{a-b}=a^2\) (a>b)

\(C=5.3\left|a\right|-3a=15a-3a=12a\)

\(D=9-6a+a^2-6\left|a\right|=\left[{}\begin{matrix}a^2-12a+9\left(a\ge0\right)\\a^2+9\left(a< 0\right)\end{matrix}\right.\) 

1) \(\sqrt{\frac{24}{3}}\cdot\sqrt{\frac{3a}{8}}=\sqrt{\frac{72a}{24}}=\sqrt{3a}\)

2) \(\sqrt{13a}\cdot\sqrt{\frac{52}{a}}=\sqrt{\frac{13a\cdot52}{a}}=\sqrt{676}=26\)

3) \(\sqrt{5a}\cdot\sqrt{45a}-3a=\sqrt{225a^2}-3a=15a-3a=12a\)

4) \(\left(3-a\right)^2-\sqrt{0,2}\cdot\sqrt{180a^2}=a^2-6a+9-\sqrt{36a^2}=a^2-6a+9-6a=a^2-12a+9\)

a: \(\left(1-\sqrt{x}\right)\left(1+\sqrt{x}+x\right)-\sqrt{x^3}\)

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

\(=1-2x\sqrt{x}\)

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

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

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

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

\(=\left(3-a\right)^2-\sqrt{0,2.180a^2}=9-6a+a^2-\sqrt{36a^2}=9-6a+a^2-6.lal\)

a.

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

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

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

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

b.

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

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

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

\(\rightarrow A=\dfrac{-2}{\sqrt{2}-1+1}=\dfrac{-1}{\sqrt{2}}=\sqrt{2}\)

=>\(A=\left(\dfrac{\sqrt{a}-1}{\sqrt{a}}\right).\left(\dfrac{\sqrt{a}+1+\sqrt{a}-1}{a-1}\right)\left(a>0,a\ne1\right)\)

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

b, \(a=3-2\sqrt{2}=\left(\sqrt{2}-1\right)^2\) thế vào A

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