Question

Bài 34 (trang 19 SGK Toán 9 Tập 1)

Rút gọn các biểu thức sau:

a) $ab^2.\sqrt{\dfrac{3}{a^2b^4}}$ với $a<0$,$b \ne 0$ ;                                               b) $\sqrt{\dfrac{27(a-3)^2}{48}}$ với $a>3$ ;

c) $\sqrt{\dfrac{9+12a+4a^2}{b^2}}$ với $a \ge -1,5$ và $b<0$ ;                         d) $(a-b).\sqrt{\dfrac{ab}{(a-b)^2}}$ với $a<b<0$.

Answer 1

3a2b4=ab2.3a2b4

3a2.b4=ab2.3|a|.|b2|

3(−a).b2 (Do $a<0$ nên $|a|=-a$ và $b\ne 0$ nên $b^2>0$ $\Rightarrow$  $\left|b^2\right|=b^2$)

$=-\sqrt{3}$.

27(a−3)248=9(a−3)216

9.(a−3)216=3.|a−3|4

3(a−3)4. 

(Do $a>3$ nên $|a-3|=a-3$)

9+12a+4a2b2=32+2.3.2a+(2a)2b2

(3+2a)2b2=|3+2a||b|

Answer 2

3a2b4=ab2.3a2b4

3a2.b4=ab2.3|a|.|b2|

3(−a).b2 (Do $a<0$ nên $|a|=-a$ và $b\ne 0$ nên $b^2>0$ $\Rightarrow$  $\left|b^2\right|=b^2$)

$=-\sqrt{3}$.

27(a−3)248=9(a−3)216

9.(a−3)216=3.|a−3|4

3(a−3)4. 

(Do $a>3$ nên $|a-3|=a-3$)

9+12a+4a2b2=32+2.3.2a+(2a)2b2

(3+2a)2b2=|3+2a||b|

Answer 3

3a2b4=ab2.3a2b4

3a2.b4=ab2.3|a|.|b2|

3(−a).b2 (Do $a<0$ nên $|a|=-a$ và $b\ne 0$ nên $b^2>0$ $\Rightarrow$  $\left|b^2\right|=b^2$)

$=-\sqrt{3}$.

27(a−3)248=9(a−3)216

9.(a−3)216=3.|a−3|4

3(a−3)4. 

(Do $a>3$ nên $|a-3|=a-3$)

9+12a+4a2b2=32+2.3.2a+(2a)2b2

(3+2a)2b2=|3+2a||b|

Answer 4

Answer 5

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

b) \(\sqrt{\dfrac{27\left(a-3\right)^2}{48}}=\sqrt{\dfrac{9.\left(a-3\right)^2}{16}}=\dfrac{3\left(a-3\right)}{4}=\dfrac{3a-9}{4}\)

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

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

Answer 6

a) ab^2.\sqrt{\dfrac{3}{a^2b^4}}=ab^2.\dfrac{\sqrt{3}}{\sqrt{a^2b^4}}ab2.a2b43​​=ab2.a2b4​3​​

=ab^2.\dfrac{\sqrt{3}}{\sqrt{a^2}.\sqrt{b^4}}=ab^2.\dfrac{\sqrt{3}}{|a|.|b^2|}=ab2.a2​.b4​3​​=ab2.∣a∣.∣b2∣3​​

=ab^2.\dfrac{\sqrt{3}}{(-a).b^2}=ab2.(−a).b23​​ (Do $a<0$ nên $|a|=-a$ và $b\ne 0$ nên b^2 > 0b2> 0 $\Rightarrow$  \left|b^2\right|=b^2∣∣∣​b2∣∣∣​=b2)

=-\sqrt{3}=−3​.

b) \sqrt{\dfrac{27(a-3)^2}{48}}=\sqrt{\dfrac{9(a-3)^2}{16}}4827(a−3)2​​=169(a−3)2​​

=\dfrac{\sqrt{9}.\sqrt{(a-3)^2}}{\sqrt{16}}=\dfrac{3.|a-3|}{4}=16​9​.(a−3)2​​=43.∣a−3∣​

=\dfrac{3(a-3)}{4}=43(a−3)​. 

(Do $a>3$ nên $|a-3|=a-3$)

c) \sqrt{\dfrac{9+12a+4a^2}{b^2}}=\dfrac{\sqrt{3^2+2.3.2a+(2a)^2}}{\sqrt{b^2}}b29+12a+4a2​​=b2​32+2.3.2a+(2a)2​​

=\dfrac{\sqrt{(3+2a)^2}}{\sqrt{b^2}}=\dfrac{|3+2a|}{|b|}=b2​(3+2a)2​​=∣b∣∣3+2a∣​
=\dfrac{3+2a}{-b}=-\dfrac{2a+3}{b}=−b3+2a​=−b2a+3​.

(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{\dfrac{ab}{(a-b)^2}}=(a-b).\dfrac{\sqrt{ab}}{\sqrt{(a-b)^2}}(a−b).(a−b)2ab​​=(a−b).(a−b)2​ab​​

=(a-b).\dfrac{\sqrt{ab}}{|a-b|}=(a-b).\dfrac{\sqrt{ab}}{-(a-b)}=(a−b).∣a−b∣ab​​=(a−b).−(a−b)ab​​

=-\sqrt{ab}=−ab​.

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

   

Answer 7

a) .

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

ab(a−b)2=(a−b).ab(a−b)2

ab|a−b|=(a−b).ab−(a−b)

$=-\sqrt{ab}$.

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

Answer 8

a) .

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

ab(a−b)2=(a−b).ab(a−b)2

ab|a−b|=(a−b).ab−(a−b)

$=-\sqrt{ab}$.

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

 

Answer 9

 

(vì a < 0 nên |a| = -a, b2 > 0 với mọi b ≠ 0 nên |b2| = b2 )

(vì a > 3 nên |a - 3| = a - 3)

 

Vì b < 0 nên |b| = -b

Vì a ≥ -1,5 nên 3 + 2a ≥ 0. Do đó: |3 + 2a| = 3 + 2a

Vậy:

(vì a < b < 0 và b < 0 nên |a - b| = -(a - b), ab > 0)

 

Answer 10

a, \(ab^2.\sqrt{\dfrac{3}{a^2b^4}}\) với a<0, b≠0
=\(ab^2.\sqrt{\dfrac{3}{\left(ab^2\right)^2}}=ab^2.\sqrt{\dfrac{3}{\left|ab^2\right|}}=ab^2.\dfrac{\sqrt{3}}{-ab^2}=\dfrac{\sqrt{3}}{-1}=-\sqrt{3}\)
b, \(\sqrt{\dfrac{27\left(a-3\right)^2}{48}}\) với a>3
\(=\sqrt{\dfrac{9\left(a-3\right)^2}{16}}=\dfrac{\sqrt{9\left(a-3\right)^2}}{\sqrt{16}}=\dfrac{\sqrt{9}.\sqrt{\left(a-3\right)^2}}{\sqrt{16}}=\dfrac{3.\left|a-3\right|}{4}=\dfrac{3\left(a-3\right)}{4}\)
c, \(\sqrt{\dfrac{9+12a+4a^2}{b^2}}\) với a≥-1,5 và b<0
= \(\dfrac{\sqrt{\left(3+2a\right)^2}}{\sqrt{b^2}}=\dfrac{\left|3+2a\right|}{\left|b^2\right|}=\dfrac{3+2a}{-b}=\dfrac{-3-2a}{b}\)  ( vì a≥-15 và b<0)
d, \(\left(a-b\right).\sqrt{\dfrac{ab}{\left(a-b\right)^2}}\) với a<b<0
<=>\(\left(a-b\right).\dfrac{\sqrt{ab}}{\sqrt{\left(a-b\right)^2}}=\left(a-b\right).\dfrac{\sqrt{ab}}{\left|a-b\right|}=\left(a-b\right).\dfrac{\sqrt{ab}}{-\left(a-b\right)}=-\sqrt{ab}\)
 

 

Answer 11

a) ab^2.\sqrt{\dfrac{3}{a^2b^4}}=ab^2.\dfrac{\sqrt{3}}{\sqrt{a^2b^4}}ab2.a2b43​​=ab2.a2b4​3​​

=ab^2.\dfrac{\sqrt{3}}{\sqrt{a^2}.\sqrt{b^4}}=ab^2.\dfrac{\sqrt{3}}{|a|.|b^2|}=ab2.a2​.b4​3​​=ab2.∣a∣.∣b2∣3​​

=ab^2.\dfrac{\sqrt{3}}{(-a).b^2}=ab2.(−a).b23​​ (Do $a<0$ nên $|a|=-a$ và $b\ne 0$ nên b^2 > 0b2> 0 $\Rightarrow$  \left|b^2\right|=b^2∣∣∣​b2∣∣∣​=b2)

=-\sqrt{3}=−3​.

b) \sqrt{\dfrac{27(a-3)^2}{48}}=\sqrt{\dfrac{9(a-3)^2}{16}}4827(a−3)2​​=169(a−3)2​​

=\dfrac{\sqrt{9}.\sqrt{(a-3)^2}}{\sqrt{16}}=\dfrac{3.|a-3|}{4}=16​9​.(a−3)2​​=43.∣a−3∣​

=\dfrac{3(a-3)}{4}=43(a−3)​. 

(Do $a>3$ nên $|a-3|=a-3$)

c) \sqrt{\dfrac{9+12a+4a^2}{b^2}}=\dfrac{\sqrt{3^2+2.3.2a+(2a)^2}}{\sqrt{b^2}}b29+12a+4a2​​=b2​32+2.3.2a+(2a)2​​

=\dfrac{\sqrt{(3+2a)^2}}{\sqrt{b^2}}=\dfrac{|3+2a|}{|b|}=b2​(3+2a)2​​=∣b∣∣3+2a∣​
=\dfrac{3+2a}{-b}=-\dfrac{2a+3}{b}=−b3+2a​=−b2a+3​.

(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{\dfrac{ab}{(a-b)^2}}=(a-b).\dfrac{\sqrt{ab}}{\sqrt{(a-b)^2}}(a−b).(a−b)2ab​​=(a−b).(a−b)2​ab​​

=(a-b).\dfrac{\sqrt{ab}}{|a-b|}=(a-b).\dfrac{\sqrt{ab}}{-(a-b)}=(a−b).∣a−b∣ab​​=(a−b).−(a−b)ab​​

=-\sqrt{ab}=−ab​.

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

Answer 12

a, = ab² .\(\dfrac{\sqrt{3}}{\sqrt{a^2\left(b^2\right)^2}}\) = ab² . \(\dfrac{\sqrt{3}}{\sqrt{\left(ab^2\right)^2}}\) = ab² . \(\dfrac{\sqrt{3}}{\left|ab^2\right|}\) = ab² . \(\dfrac{\sqrt{3}}{-ab^2}\) ( vì a<0, b²≥0) 

 = \(\dfrac{\sqrt{3}}{-1}\) =\(-\sqrt{3}\)

b,  =    \(\sqrt{\dfrac{9\left(a-3\right)^2}{16}}\) = \(\dfrac{\sqrt{9.\left(a-3\right)^2}}{\sqrt{16}}\) =  \(\dfrac{\sqrt{9}.\sqrt{\left(a-3\right)^2}}{\sqrt{16}}\) =\(\dfrac{3.\left|a-3\right|}{4}\)=\(\dfrac{3.\left(a-3\right)}{4}\) ( Vì a>3)

c, = \(\sqrt{\dfrac{\left(3+2a\right)^2}{b^2}}\)= \(\dfrac{\sqrt{\left(3+2a\right)^2}}{\sqrt{b^2}}\) = \(\dfrac{\left|3+2a\right|}{\left|b\right|}\)=\(\dfrac{3+2a}{-b}\)=\(\dfrac{-\left(3+2a\right)}{b}\)   ( Vì  a≥ -1,5, <0)

d, = (a-b) .\(\dfrac{\sqrt{ab}}{\sqrt{\left(a-b\right)^2}}\)= (a-b) . \(\dfrac{\sqrt{ab}}{\left|a-b\right|}\) = (a - b) .\(\dfrac{\sqrt{a.b}}{-\left(a-b\right)}\)= \(\dfrac{\sqrt{a.b}}{-1}\)= \(-\sqrt{ab}\)   ( Vì a<b <0)

Answer 13

224224

22422|

22 (Do 222)

=−√3.

2216

2√16=3.|a−3|4

=3(a−3)4. 

(Do a>3 nên |a−3|=a−3)

22222

22=|3+2a||b|
=3+2a−b=−2a+3b.

(Do a≥−1,5 ⇒ 3+2a≥0 nên |3+2a|=3+2a và b<0 nên |b|=−b)

22

=(a−b).√ab|a−b|=(a−b).√ab−(a−b)

=−√ab.

(Do a<b<0 nên |a−b|=−(a−b) và ab>0)

Answer 14

Answer 15

Answer 16

a) ab^2.\sqrt{\dfrac{3}{a^2b^4}}=ab^2.\dfrac{\sqrt{3}}{\sqrt{a^2b^4}}ab2.a2b43​​=ab2.a2b4​3​​

=ab^2.\dfrac{\sqrt{3}}{\sqrt{a^2}.\sqrt{b^4}}=ab^2.\dfrac{\sqrt{3}}{|a|.|b^2|}=ab2.a2​.b4​3​​=ab2.∣a∣.∣b2∣3​​

=ab^2.\dfrac{\sqrt{3}}{(-a).b^2}=ab2.(−a).b23​​ (Do $a<0$ nên $|a|=-a$ và $b\ne 0$ nên b^2 > 0b2> 0 $\Rightarrow$  \left|b^2\right|=b^2∣∣∣​b2∣∣∣​=b2)

=-\sqrt{3}=−3​.

b) \sqrt{\dfrac{27(a-3)^2}{48}}=\sqrt{\dfrac{9(a-3)^2}{16}}4827(a−3)2​​=169(a−3)2​​

=\dfrac{\sqrt{9}.\sqrt{(a-3)^2}}{\sqrt{16}}=\dfrac{3.|a-3|}{4}=16​9​.(a−3)2​​=43.∣a−3∣​

=\dfrac{3(a-3)}{4}=43(a−3)​. 

(Do $a>3$ nên $|a-3|=a-3$)

c) \sqrt{\dfrac{9+12a+4a^2}{b^2}}=\dfrac{\sqrt{3^2+2.3.2a+(2a)^2}}{\sqrt{b^2}}b29+12a+4a2​​=b2​32+2.3.2a+(2a)2​​

=\dfrac{\sqrt{(3+2a)^2}}{\sqrt{b^2}}=\dfrac{|3+2a|}{|b|}=b2​(3+2a)2​​=∣b∣∣3+2a∣​
=\dfrac{3+2a}{-b}=-\dfrac{2a+3}{b}=−b3+2a​=−b2a+3​.

(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{\dfrac{ab}{(a-b)^2}}=(a-b).\dfrac{\sqrt{ab}}{\sqrt{(a-b)^2}}(a−b).(a−b)2ab​​=(a−b).(a−b)2​ab​​

=(a-b).\dfrac{\sqrt{ab}}{|a-b|}=(a-b).\dfrac{\sqrt{ab}}{-(a-b)}=(a−b).∣a−b∣ab​​=(a−b).−(a−b)ab​​

=-\sqrt{ab}=−ab​.

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

Answer 17

Answer 18

.

Answer 19

224224

22422|

22 (Do 222)

=−√3.

2216

2√16=3.|a−3|4

=3(a−3)4. 

(Do a>3 nên |a−3|=a−3)

22222

22=|3+2a||b|
=3+2a−b=−2a+3b.

(Do a≥−1,5 ⇒ 3+2a≥0 nên |3+2a|=3+2a và b<0 nên |b|=−b)

22

=(a−b).√ab|a−b|=(a−b).√ab−(a−b)

=−√ab.

(Do a<b<0 nên |a−b|=−(a−b) và ab>0)

Answer 20

(vì a < 0 nên |a| = -a, b2 > 0 với mọi b ≠ 0 nên |b2| = b2 )

(vì a > 3 nên |a - 3| = a - 3

Vì b < 0 nên |b| = -b

Vì a ≥ -1,5 nên 3 + 2a ≥ 0. Do đó: |3 + 2a| = 3 + 2a

Vậy:

(vì a < b < 0 và b < 0 nên |a - b| = -(a - b), ab > 0)

 

Answer 21

a,\(-\sqrt{3}\)

b,\(\dfrac{3\left(a-3\right)}{4}\)

c,\(\dfrac{3+2a}{b}\)

d,\(-\sqrt{ab}\)

Answer 22

a) ab^2.\sqrt{\dfrac{3}{a^2b^4}}=ab^2.\dfrac{\sqrt{3}}{\sqrt{a^2b^4}}ab2.a2b43​​=ab2.a2b4​3​​

=ab^2.\dfrac{\sqrt{3}}{\sqrt{a^2}.\sqrt{b^4}}=ab^2.\dfrac{\sqrt{3}}{|a|.|b^2|}=ab2.a2​.b4​3​​=ab2.∣a∣.∣b2∣3​​

=ab^2.\dfrac{\sqrt{3}}{(-a).b^2}=ab2.(−a).b23​​ (Do $a<0$ nên $|a|=-a$ và $b\ne 0$ nên b^2 > 0b2> 0 $\Rightarrow$  \left|b^2\right|=b^2∣∣∣​b2∣∣∣​=b2)

=-\sqrt{3}=−3​.

b) \sqrt{\dfrac{27(a-3)^2}{48}}=\sqrt{\dfrac{9(a-3)^2}{16}}4827(a−3)2​​=169(a−3)2​​

=\dfrac{\sqrt{9}.\sqrt{(a-3)^2}}{\sqrt{16}}=\dfrac{3.|a-3|}{4}=16​9​.(a−3)2​​=43.∣a−3∣​

=\dfrac{3(a-3)}{4}=43(a−3)​. 

(Do $a>3$ nên $|a-3|=a-3$)

c) \sqrt{\dfrac{9+12a+4a^2}{b^2}}=\dfrac{\sqrt{3^2+2.3.2a+(2a)^2}}{\sqrt{b^2}}b29+12a+4a2​​=b2​32+2.3.2a+(2a)2​​

=\dfrac{\sqrt{(3+2a)^2}}{\sqrt{b^2}}=\dfrac{|3+2a|}{|b|}=b2​(3+2a)2​​=∣b∣∣3+2a∣​
=\dfrac{3+2a}{-b}=-\dfrac{2a+3}{b}=−b3+2a​=−b2a+3​.

(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{\dfrac{ab}{(a-b)^2}}=(a-b).\dfrac{\sqrt{ab}}{\sqrt{(a-b)^2}}(a−b).(a−b)2ab​​=(a−b).(a−b)2​ab​​

=(a-b).\dfrac{\sqrt{ab}}{|a-b|}=(a-b).\dfrac{\sqrt{ab}}{-(a-b)}=(a−b).∣a−b∣ab​​=(a−b).−(a−b)ab​​

=-\sqrt{ab}=−ab​.

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

Answer 23

 

(vì a < 0 nên |a| = -a, b2 > 0 với mọi b ≠ 0 nên |b2| = b2 )

(vì a > 3 nên |a - 3| = a - 3)

   

Vì b < 0 nên |b| = -b

Vì a ≥ -1,5 nên 3 + 2a ≥ 0. Do đó: |3 + 2a| = 3 + 2a

Vậy:

(vì a < b < 0 và b < 0 nên |a - b| = -(a - b), ab > 0)

Answer 24

a) ab² . \(\sqrt{\dfrac{3}{a^2b^4}}\)

= ab² . \(\dfrac{\sqrt{3}}{\sqrt{a^2b^4}}\)

= ab² . \(\dfrac{\sqrt{3}}{\sqrt{a^2}.\sqrt{b^4}}\)

= ab² . \(\dfrac{\sqrt{3}}{\left|a\right|.\left|b^2\right|}\)

= ab² . \(\dfrac{\sqrt{3}}{-a.b^2}\)

= \(\sqrt{3}\)

b) \(\sqrt{\dfrac{27\left(a-3\right)^2}{48}}\)

= \(\sqrt{\dfrac{9\left(a-3\right)^2}{16}}\)

= \(\dfrac{\sqrt{9}.\sqrt{\left(a-3\right)^2}}{\sqrt{16}}\)

= \(\dfrac{3.\left|a-3\right|}{4}\)

= \(\dfrac{3.\left(a-3\right)}{4}\)

c) \(\sqrt{\dfrac{9+12a+4a^2}{b^2}}\)

= \(\dfrac{\sqrt{3^2+2.3.2a+2a^2}}{\sqrt{b^2}}\)

= \(\dfrac{\sqrt{\left(3+2a\right)^2}}{\sqrt{b^2}}\)

= \(\dfrac{\left|3+2a\right|}{\left|b\right|}\)

= \(-\dfrac{2a+3}{b}\)

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

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

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

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

= \(-\sqrt{ab}\)

Answer 25

a)ab\(^2\).\(\sqrt{\dfrac{3}{a^2b^4}}\) =ab\(^2\).\(\dfrac{\sqrt{3}}{\sqrt{a^2b^3}}\)=ab\(^2\).\(\dfrac{\sqrt{3}}{\left(-a\right).b^2}\)doa<0 nên \(\left|a\right|=-a\) và b

\(\ne0\) nên \(b^2>0\)                        b)\(\sqrt{\dfrac{27\left(a-3\right)^2}{48}}=\dfrac{3\left(a-3\right)}{4}\) doa>3 nên \(\left|a-3\right|=a-3\)      C)\(\sqrt{\dfrac{9+12a+4a^4}{b^2}}=-\dfrac{2a+3}{b}\) do a\(\ge-1,5\Rightarrow3-2a\ge0\) nên\(\left|3+2a\right|=3+2a\) và b<0

Answer 26

Rút gọn bt câu a =\(-\sqrt{3}\)                                                                                                                    Rút gọn bt câu b =\(\dfrac{3(a-3)}{4}\)                                                                                                            Rút gọn bt câu c =\(\dfrac{|3+2a|}{|b|}\)                                                                                                              Rút gọn bt câu d =\(\sqrt{ab}\)

Answer 27

a) ab^2.\sqrt{\dfrac{3}{a^2b^4}}=ab^2.\dfrac{\sqrt{3}}{\sqrt{a^2b^4}}ab2.a2b43​​=ab2.a2b4​3​​

=ab^2.\dfrac{\sqrt{3}}{\sqrt{a^2}.\sqrt{b^4}}=ab^2.\dfrac{\sqrt{3}}{|a|.|b^2|}=ab2.a2​.b4​3​​=ab2.∣a∣.∣b2∣3​​

=ab^2.\dfrac{\sqrt{3}}{(-a).b^2}=ab2.(−a).b23​​ (Do $a<0$ nên $|a|=-a$ và $b\ne 0$ nên b^2 > 0b2> 0 $\Rightarrow$  \left|b^2\right|=b^2∣∣∣​b2∣∣∣​=b2)

=-\sqrt{3}=−3​.

b) \sqrt{\dfrac{27(a-3)^2}{48}}=\sqrt{\dfrac{9(a-3)^2}{16}}4827(a−3)2​​=169(a−3)2​​

=\dfrac{\sqrt{9}.\sqrt{(a-3)^2}}{\sqrt{16}}=\dfrac{3.|a-3|}{4}=16​9​.(a−3)2​​=43.∣a−3∣​

=\dfrac{3(a-3)}{4}=43(a−3)​. 

(Do $a>3$ nên $|a-3|=a-3$)

c) \sqrt{\dfrac{9+12a+4a^2}{b^2}}=\dfrac{\sqrt{3^2+2.3.2a+(2a)^2}}{\sqrt{b^2}}b29+12a+4a2​​=b2​32+2.3.2a+(2a)2​​

=\dfrac{\sqrt{(3+2a)^2}}{\sqrt{b^2}}=\dfrac{|3+2a|}{|b|}=b2​(3+2a)2​​=∣b∣∣3+2a∣​
=\dfrac{3+2a}{-b}=-\dfrac{2a+3}{b}=−b3+2a​=−b2a+3​.

(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{\dfrac{ab}{(a-b)^2}}=(a-b).\dfrac{\sqrt{ab}}{\sqrt{(a-b)^2}}(a−b).(a−b)2ab​​=(a−b).(a−b)2​ab​​

=(a-b).\dfrac{\sqrt{ab}}{|a-b|}=(a-b).\dfrac{\sqrt{ab}}{-(a-b)}=(a−b).∣a−b∣ab​​=(a−b).−(a−b)ab​​

=-\sqrt{ab}=−ab​.

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

Answer 28

Answer 29

a) ab^2.\sqrt{\dfrac{3}{a^2b^4}}=ab^2.\dfrac{\sqrt{3}}{\sqrt{a^2b^4}}ab2.a2b43​​=ab2.a2b4​3​​

=ab^2.\dfrac{\sqrt{3}}{\sqrt{a^2}.\sqrt{b^4}}=ab^2.\dfrac{\sqrt{3}}{|a|.|b^2|}=ab2.a2​.b4​3​​=ab2.∣a∣.∣b2∣3​​

=ab^2.\dfrac{\sqrt{3}}{(-a).b^2}=ab2.(−a).b23​​ (Do $a<0$ nên $|a|=-a$ và $b\ne 0$ nên b^2 > 0b2> 0 $\Rightarrow$  \left|b^2\right|=b^2∣∣∣​b2∣∣∣​=b2)

=-\sqrt{3}=−3​.

b) \sqrt{\dfrac{27(a-3)^2}{48}}=\sqrt{\dfrac{9(a-3)^2}{16}}4827(a−3)2​​=169(a−3)2​​

=\dfrac{\sqrt{9}.\sqrt{(a-3)^2}}{\sqrt{16}}=\dfrac{3.|a-3|}{4}=16​9​.(a−3)2​​=43.∣a−3∣​

=\dfrac{3(a-3)}{4}=43(a−3)​. 

(Do $a>3$ nên $|a-3|=a-3$)

c) \sqrt{\dfrac{9+12a+4a^2}{b^2}}=\dfrac{\sqrt{3^2+2.3.2a+(2a)^2}}{\sqrt{b^2}}b29+12a+4a2​​=b2​32+2.3.2a+(2a)2​​

=\dfrac{\sqrt{(3+2a)^2}}{\sqrt{b^2}}=\dfrac{|3+2a|}{|b|}=b2​(3+2a)2​​=∣b∣∣3+2a∣​
=\dfrac{3+2a}{-b}=-\dfrac{2a+3}{b}=−b3+2a​=−b2a+3​.

(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{\dfrac{ab}{(a-b)^2}}=(a-b).\dfrac{\sqrt{ab}}{\sqrt{(a-b)^2}}(a−b).(a−b)2ab​​=(a−b).(a−b)2​ab​​

=(a-b).\dfrac{\sqrt{ab}}{|a-b|}=(a-b).\dfrac{\sqrt{ab}}{-(a-b)}=(a−b).∣a−b∣ab​​=(a−b).−(a−b)ab​​

=-\sqrt{ab}=−ab​.

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

Answer 30

a) .

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

ab(a−b)2=(a−b).ab(a−b)2

ab|a−b|=(a−b).ab−(a−b)

$=-\sqrt{ab}$.

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