a: \(=4\left|a-3\right|=4\left(a-3\right)=4a-12\)

b: \(=9\cdot\left|a-9\right|=9\left(9-a\right)=81-9a\)

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

d: \(=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}{a-b}\)

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

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

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

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

\(=-a^2\)

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

= \(\sqrt{\dfrac{6+2\sqrt{5}}{4x^2}}-\sqrt{\dfrac{6-2\sqrt{5}}{4}}=\sqrt{\dfrac{5+2\sqrt{5}+1}{4x^2}}-\sqrt{\dfrac{5-2\sqrt{5}+1}{4}}\) = \(\sqrt{\dfrac{\left(\sqrt{5}+1\right)^2}{\left(2x\right)^2}}-\sqrt{\dfrac{\left(\sqrt{5}-1\right)^2}{2^2}}=\dfrac{\left|\sqrt{5}+1\right|}{\left|2x\right|}-\dfrac{\left|\sqrt{5}-1\right|}{2}=\dfrac{\sqrt{5}+1}{2x}-\dfrac{\sqrt{5}-1}{2}\)

Thay x = 1 vào biểu thức \(\dfrac{\sqrt{5}+1}{2x}-\dfrac{\sqrt{5}-1}{2}\) ta được :

\(\dfrac{\sqrt{5}+1}{2}-\dfrac{\sqrt{5}-1}{2}=\dfrac{\sqrt{5}+1-\sqrt{5}+1}{2}=1\)

Vậy tại x =1 thì giá trị của biểu thức \(\sqrt{\dfrac{3+\sqrt{5}}{2x^2}}-\sqrt{\dfrac{3-\sqrt{5}}{2}}\) là bằng 1

b) \(\dfrac{\sqrt{a^3+4a^2+4a}}{\sqrt{a\left(a^2-2ab+b^2\right)}}-\dfrac{\sqrt{b^3-4b^2+4b}}{\sqrt{b\left(a^2-2ab+b^2\right)}}+ab\)

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

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

Thay a = 4 và b = 3 vào biểu thức a - b +ab ta được :

4 - 3 + 4.3 = 13

Vậy tại a = 4 ; b = 3 thì giá trị của biểu thức \(\dfrac{\sqrt{a^3+4a^2+4a}}{\sqrt{a\left(a^2-2ab+b^2\right)}}-\dfrac{\sqrt{b^3-4b^2+4b}}{\sqrt{b\left(a^2-2ab+b^2\right)}}+ab\) là bằng 13

c) \(ab^2.\sqrt{\dfrac{4}{a^2b^4}}+ab=ab^2.\dfrac{2}{ab^2}+ab=2+ab\)

Thay a = 1 và b = -2 vào BT : 2 + ab ta được :

2 + 1.(-2) = 2 + (-2) = 0

Vậy tại a = 1 ; b = -2 thì giá trị của biểu thức \(ab^2.\sqrt{\dfrac{4}{a^2b^4}}+ab\) là bằng 0

d) \(\dfrac{a+b}{b^2}.\sqrt{\dfrac{a^2b^2}{a^2+2ab+b^2}}\) = \(\dfrac{a+b}{b^2}.\dfrac{\sqrt{a^2b^2}}{\sqrt{a^2+2ab+b^2}}=\dfrac{a+b}{b^2}.\dfrac{ab}{a+b}=\dfrac{ab}{b^2}\)

Thay a = 1 ; b =2 vào BT : \(\dfrac{ab}{b^2}\) ta được : \(\dfrac{1.2}{2^2}=\dfrac{1}{2}\)

Vậy tại a =1 ; b =2 GT của BT : \(\dfrac{a+b}{b^2}.\sqrt{\dfrac{a^2b^2}{a^2+2ab+b^2}}\) là \(\dfrac{1}{2}\)

@phynit

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

b) \(\left(a-b\right).\sqrt{\dfrac{ab}{\left(a-b\right)^2}}\)

\(\Leftrightarrow\left(a-b\right).\dfrac{\left|ab\right|}{\left|a-b\right|}=-ab\)

a) $a{b}^2.\sqrt{\frac{3}{a^2{b}^4}}=a{b}^2.\frac{\sqrt{3}}{\sqrt{a^2{b}^4}}$

$=a{b}^2.\frac{\sqrt{3}}{\sqrt{a^2}.\sqrt{b^4}}=a{b}^2.\frac{\sqrt{3}}{|a|.|b^2|}$

$=a{b}^2.\frac{\sqrt{3}}{(-a).b^2}$ (Do $a<0$ nên $|a|=-a$ và $b\ne 0$ nên $b^2>0$ $\Rightarrow$  $\left|b^2\right|=b^2$)

$=-\sqrt{3}$.

b) $\sqrt{\frac{27(a-3{)}^2}{48}}=\sqrt{\frac{9(a-3{)}^2}{16}}$

$=\frac{\sqrt{9}.\sqrt{(a-3{)}^2}}{\sqrt{16}}=\frac{3.|a-3|}{4}$

$=\frac{3(a-3)}{4}$. 

(Do $a>3$ nên $|a-3|=a-3$)

c) $\sqrt{\frac{9+12a+4a^2}{b^2}}=\frac{\sqrt{3^2+\mathrm{2.3.2}a+(2a{)}^2}}{\sqrt{b^2}}$

$=\frac{\sqrt{(3+2a{)}^2}}{\sqrt{b^2}}=\frac{|3+2a|}{|b|}$
$=\frac{3+2a}{-b}=-\frac{2a+3}{b}$.

(Do $a\ge -1,5$ $\Rightarrow$ $3+2a\ge 0$ nên $|3+2a|=3+2a$ và $b<0$ nên $|b|=-b$)

d) $(a-b).\sqrt{\frac{ab}{(a-b{)}^2}}=(a-b).\frac{\sqrt{ab}}{\sqrt{(a-b{)}^2}}$

$=(a-b).\frac{\sqrt{ab}}{|a-b|}=(a-b).\frac{\sqrt{ab}}{-(a-b)}$

$=-\sqrt{ab}$.

(Do $a<b<0$ nên $|a-b|=-(a-b)$ và $ab>0$)

a) $a{b}^2.\sqrt{\frac{3}{a^2{b}^4}}=a{b}^2.\frac{\sqrt{3}}{\sqrt{a^2{b}^4}}$

$=a{b}^2.\frac{\sqrt{3}}{\sqrt{a^2}.\sqrt{b^4}}=a{b}^2.\frac{\sqrt{3}}{|a|.|b^2|}$

$=a{b}^2.\frac{\sqrt{3}}{(-a).b^2}$ (Do $a<0$ nên $|a|=-a$ và $b\ne 0$ nên $b^2>0$ $\Rightarrow$  $\left|b^2\right|=b^2$)

$=-\sqrt{3}$.

b) $\sqrt{\frac{27(a-3{)}^2}{48}}=\sqrt{\frac{9(a-3{)}^2}{16}}$

$=\frac{\sqrt{9}.\sqrt{(a-3{)}^2}}{\sqrt{16}}=\frac{3.|a-3|}{4}$

$=\frac{3(a-3)}{4}$. 

(Do $a>3$ nên $|a-3|=a-3$)

c) $\sqrt{\frac{9+12a+4a^2}{b^2}}=\frac{\sqrt{3^2+\mathrm{2.3.2}a+(2a{)}^2}}{\sqrt{b^2}}$

$=\frac{\sqrt{(3+2a{)}^2}}{\sqrt{b^2}}=\frac{|3+2a|}{|b|}$
$=\frac{3+2a}{-b}=-\frac{2a+3}{b}$.

(Do $a\ge -1,5$ $\Rightarrow$ $3+2a\ge 0$ nên $|3+2a|=3+2a$ và $b<0$ nên $|b|=-b$)

d) $(a-b).\sqrt{\frac{ab}{(a-b{)}^2}}=(a-b).\frac{\sqrt{ab}}{\sqrt{(a-b{)}^2}}$

$=(a-b).\frac{\sqrt{ab}}{|a-b|}=(a-b).\frac{\sqrt{ab}}{-(a-b)}$

$=-\sqrt{ab}$.

(Do $a<b<0$ nên $|a-b|=-(a-b)$ và $ab>0$)

Với `a > 0,b >= 0` có:

`Bth=[a\sqrt{b}+b]/[a-b] . \sqrt{[b(a+b-2\sqrt{ab})]/[a^2+2a\sqrt{b}+b]} . (\sqrt{a}+\sqrt{b})`

 `=[\sqrt{b}(a+\sqrt{b})]/[a-b].\sqrt{[b(\sqrt{a}-\sqrt{b})^2]/[(a+\sqrt{b})^2]}.(\sqrt{a}+\sqrt{b})`

`=[\sqrt{b}(a+\sqrt{b})|\sqrt{a}-\sqrt{b}|.\sqrt{b}.(\sqrt{a}+\sqrt{b})]/[(a-b)(a+\sqrt{b})]`

`=[b|\sqrt{a}-\sqrt{b}|]/[\sqrt{a}-\sqrt{b}]`

`={(b\text{ nếu }\sqrt{a} >= \sqrt{b}),(-b\text{ nếu }\sqrt{a} < \sqrt{b}):}`

ko biet

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

\(b,A=\dfrac{\sqrt{a}}{\sqrt{a}-5}-\dfrac{10\sqrt{a}}{a-25}-\dfrac{5}{\sqrt{a}+5}\)

\(\Rightarrow A=\dfrac{\sqrt{a}\left(\sqrt{a}+5\right)}{\left(\sqrt{a}-5\right)\left(\sqrt{a}+5\right)}-\dfrac{10\sqrt{a}}{\left(\sqrt{a}-5\right)\left(\sqrt{a}+5\right)}-\dfrac{5\left(\sqrt{a}-5\right)}{\left(\sqrt{a}-5\right)\left(\sqrt{a}+5\right)}\)

\(\Rightarrow A=\dfrac{a+5\sqrt{a}}{\left(\sqrt{a}-5\right)\left(\sqrt{a}+5\right)}-\dfrac{10\sqrt{a}}{\left(\sqrt{a}-5\right)\left(\sqrt{a}+5\right)}-\dfrac{5\sqrt{a}-25}{\left(\sqrt{a}-5\right)\left(\sqrt{a}+5\right)}\)

\(\Rightarrow A=\dfrac{a+5\sqrt{a}-10\sqrt{a}-5\sqrt{a}+25}{\left(\sqrt{a}-5\right)\left(\sqrt{a}+5\right)}\)

\(\Rightarrow A=\dfrac{a-10\sqrt{a}+25}{\left(\sqrt{a}-5\right)\left(\sqrt{a}+5\right)}\)

\(\Rightarrow A=\dfrac{\left(\sqrt{a}-5\right)^2}{\left(\sqrt{a}-5\right)\left(\sqrt{a}+5\right)}\)

\(\Rightarrow A=\dfrac{\sqrt{a}-5}{\sqrt{a}+5}\)

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

b: \(A=\dfrac{a+5\sqrt{a}-10\sqrt{a}-5\sqrt{a}+25}{\left(\sqrt{a}-5\right)\left(\sqrt{a}+5\right)}=\dfrac{\left(\sqrt{a}-5\right)^2}{a-25}=\dfrac{\sqrt{a}-5}{\sqrt{a}+5}\)