a/ \(\sqrt{4a^4-12a^2+9}-\sqrt{a^4-8a^2+16}\)

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

= \(|2a^2-3|-|a^2-4|\)

= \(2a^2-3+a^2-4\)
= \(3a^2-7\)

Thay a=\(\sqrt{3}\).Ta có:

\(3.\left(\sqrt{3}\right)^2-7\)

= 3.3-7=2

b/ \(\sqrt{10a^2-12a\sqrt{10}+36}\)

= \(\sqrt{\left(a\sqrt{10}\right)^2-2.a\sqrt{10}.6+6^2}\)

= \(\sqrt{\left(a\sqrt{10}-6\right)^2}\)

= \(|a\sqrt{10}-6|\)

=  \(-a\sqrt{10}+6\)

Thay  a= \(\sqrt{\frac{5}{2}}-\sqrt{\frac{2}{5}}\)=\(\frac{3}{\sqrt{10}}\),Ta có:

\(-\frac{3}{\sqrt{10}}.\sqrt{10}+6\)

= -3+6 =3

\(\sqrt{9a^2-12a+4}-9a+1\)

=\(\sqrt{\left(3a\right)^2-2.3a.2+2^2}-9a+1\)

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

=\(|3a-2|-9a+1\)

=\(3a-2-9a+1\)

=\(-6a-1\)

Thay \(a=\frac{1}{3}\)ta có:

\(-6.\frac{1}{3}-1\)

= \(-3\)

mình nhầm khúc này nha

=\(|3a-2|-9a+1\)

=\(2-3a-9a+1\)

=\(3-12a\)

Thay a=1/3. ta có

\(3-12.\frac{1}{3}\)

=\(3-4\)

=\(-1\)

ukm cảm ơn bạn 

\(1) \sqrt{9a^2.b^2}\)=3ab

\(2) \sqrt{3a}.\sqrt{27a}=\sqrt{3a}.3\sqrt{3a}=9a\)

\(3) \sqrt{3a^5}.12a=12\sqrt{3a^7}\)

\(4) \sqrt{5a}.\sqrt{45a}-3a=15a-3a=12a\)

\(5) \sqrt{3+\sqrt{a}}.\sqrt{3-\sqrt{a}}=\sqrt{(3+\sqrt{a}).(3-\sqrt{a})} =\sqrt{9-a} \)

\(6) \sqrt{3+\sqrt{5}}.\sqrt{3\sqrt{5}} =\sqrt{\sqrt{3\sqrt{5}}.(3+\sqrt{5})} =\sqrt{9+\sqrt{15}}\)

1) \(\sqrt{9a^2b^2}=3ab\)

2) \(\sqrt{3a}\cdot\sqrt{27a}=9a\)

4) \(\sqrt{5a}\cdot\sqrt{45a}-3a=15a-3a=12a\)

a) \(\sqrt{9a^4}=\sqrt{\left(3a^2\right)^2}=\left|3a^2\right|=3a^2\)

b) \(2\sqrt{a^2}-5a=2\left|a\right|-5a=-2a-5a=-7a\)

c) \(\sqrt{16\left(1+4x+4x^2\right)}=\sqrt{\left[4\left(1+2x\right)\right]^2}=\left|4\left(1+2x\right)\right|=4\left(1+2x\right)\)

a) \(3\sqrt{a^2-4a+4}=3\sqrt{\left(a-2\right)^2}=3\left|a-2\right|=3\left(a-2\right)\) (vì \(a\ge2\))

b) \(2\sqrt{9a^2+12a+4}=2\sqrt{\left(3a+2\right)^2}=2\left|3a+2\right|=2\left(-3a-2\right)=-2\left(3a+2\right)\) (vì \(a< -\frac{2}{3}\))

Với a = -9 C không có nghĩa 

Đẫ bảo là a = -9 thì biểu thức không tính đc mở máy tính ra BẤm  thử mà xem 

a) Ta có: \(2\sqrt{3a}-\sqrt{12a^3}-5\cdot\sqrt{\frac{a}{3}}-\frac{1}{4}\cdot\sqrt{27a}\)

\(=2\sqrt{3a}-2a\sqrt{3a}-\frac{5\sqrt{a}}{\sqrt{3}}-\frac{1}{4}\cdot3\sqrt{3a}\)

\(=2\sqrt{3a}-\frac{3}{4}\sqrt{3a}-2a\sqrt{3a}-\frac{5\sqrt{a}}{\sqrt{3}}\)

\(=\frac{5}{4}\sqrt{3a}-2a\sqrt{3a}-5\sqrt{3a}\cdot\frac{1}{3}\)

\(=\frac{5}{4}\sqrt{3a}-\frac{5}{3}\sqrt{3a}-2a\sqrt{3a}\)

\(=\frac{-5}{12}\sqrt{3a}-2a\sqrt{3a}\)

b) Ta có: \(2a\sqrt{b+a}+\left(a+b\right)\cdot\sqrt{\frac{1}{a+b}}-\sqrt{a^3+a^2b}\)

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

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

\(=\left(a+1\right)\cdot\sqrt{a+b}\)

c) Ta có: \(2\sqrt{a}+5\sqrt{\frac{a}{9}}-a\sqrt{\frac{16}{a}}\cdot\sqrt{a^3}\)

\(=2\sqrt{a}+5\cdot\frac{\sqrt{a}}{3}-4a^2\)

\(=\frac{11}{3}\sqrt{a}-4a^2\)