Question

Bài 20 (trang 15 SGK Toán 9 Tập 1)

Rút gọn các biểu thức sau:

a) $\sqrt{\dfrac{2a}{3}}.\sqrt{\dfrac{3a}{8}}$ với $a\ge 0$ ;                         b) $\sqrt{13a}.\sqrt{\dfrac{52}{a}}$ với $a>0$ ;

c) $\sqrt{5a}.\sqrt{45a}-3a$ với $a\ge 0$ ;                  d) $(3-a)^2-\sqrt{0,2}.\sqrt{180a^2}$.

Answer 1

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

Answer 2

a) Ta có:

b) Ta có:

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

  

d) Ta có:

 

Answer 3

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

Answer 4

a)\(\dfrac{1a}{2}\)

b)26

c)12a

d)   \(\left(3-a\right)^2-6a\) 

Answer 5

a) Ta có:

2a3.3a8=2a3.3a8

2a.3a3.8=a24

a222=|a2|=a2 (do  nên 

Answer 6

a) a /2

b) 26

c) 12a 

d) -6|a|

Answer 7

Answer 8

a) Ta có:

2a3.3a8=2a3.3a8

2a.3a3.8=a24

a222=|a2|=a2 (do  nên 

Answer 9

a) \(\sqrt{\dfrac{2a}{3}}.\sqrt{\dfrac{3a}{8}}=\dfrac{a}{2}\)

b)\(\sqrt{13a}.\sqrt{\dfrac{52}{a}}=26\)

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

d) \(\left(3-a\right)^2-\sqrt{0,2}.\sqrt{180a^2}=\left(3-a\right)^2-6|a|\)

Answer 10

Answer 11

a) \(\sqrt{\dfrac{2a}{3}}.\sqrt{\dfrac{3a}{8}}=\sqrt{\dfrac{2a.3a}{3.8}}=\sqrt{\dfrac{a2}{4}}=\dfrac{a}{2}\)

b) \(\sqrt{13a}.\sqrt{\dfrac{52}{a}}=\sqrt{13a.\dfrac{52}{a}}=\sqrt{13.13.4}=13.2=26\)

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

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

Answer 12

Answer 13

Answer 14

Answer 15

Answer 16

Answer 17

a) \(\dfrac{a}{2}\)
b) 26
c) 12a
d) 9 - 11a

Answer 18

Answer 19

a) Ta có:

\sqrt{\dfrac{2a}{3}}.\sqrt{\dfrac{3a}{8}}=\sqrt{\dfrac{2a}{3}.\dfrac{3a}{8}}32a​​.83a​​=32a​.83a​​

=\sqrt{\dfrac{2a.3a}{3.8}}=\sqrt{\dfrac{a^2}{4}}=3.82a.3a​​=4a2​​

=\sqrt{\dfrac{a^2}{2^2}}=\left|\dfrac{a}{2}\right|=\dfrac{a}{2}=22a2​​=∣∣∣∣​2a​∣∣∣∣​=2a​ (do $a\ge 0$ $\Rightarrow$ \dfrac{a}{2}\ge 02a​≥0 nên \left|\dfrac{a}{2}\right|=\dfrac{a}{2}∣∣∣∣​2a​∣∣∣∣​=2a​).

b) Ta có:

\sqrt{13a}.\sqrt{\dfrac{52}{a}}=\sqrt{13a.\dfrac{52}{a}}13a​.a52​​=13a.a52​​ (do $a>0$)

=\sqrt{13.52}=\sqrt{13.13.2^2}=\sqrt{(13.2)^2}=13.52​=13.13.22​=(13.2)2​

$=13.2=26$

 

c) Ta có:

\sqrt{5a}.\sqrt{45a}-3a=\sqrt{5a.45a}-3a5a​.45a​−3a=5a.45a​−3a (do $a\ge 0$)

=\sqrt{5.a.5.3^2.a}-3a=\sqrt{(5.3.a)^2}-3a=5.a.5.32.a​−3a=(5.3.a)2​−3a

$=|5.3.a|-3a=15a-3a=12a$ (do $a\ge 0$ $\Rightarrow$ $15a\ge 0$ nên $|15a|=15a$)

d) Ta có:

(3-a)^2-\sqrt{0,2}.\sqrt{180a^2}=(3-a)^2-\sqrt{0,2.180a^2}(3−a)2−0,2​.180a2​=(3−a)2−0,2.180a2​

=(3-a)^2-\sqrt{2.18.a^2}=(3-a)^2-\sqrt{(6a)^2}=(3−a)2−2.18.a2​=(3−a)2−(6a)2​

=(3-a)^2-|6a|=(3-a)^2-6|a|=(3−a)2−∣6a∣=(3−a)2−6∣a∣

TH1: Nếu $a\ge 0$ thì $|a|=a$.

Khi đó, (3-a)^2-6|a|=(3-a)^2-6a=9-6a+a^2-6a=a^2-12a+9(3−a)2−6∣a∣=(3−a)2−6a=9−6a+a2−6a=a2−12a+9.

TH2: Nếu $a<0$ thì $|a|=-a$.

Khi đó, (3-a)^2-6|a|=(3-a)^2-6.(-a)=9-6a+a^2+6a=a^2+9(3−a)2−6∣a∣=(3−a)2−6.(−a)=9−6a+a2+6a=a2+9.​

                   

 

 

 

Answer 20

Answer 21

\(\sqrt{\dfrac{2a}{3}}.\sqrt{\dfrac{3a}{8}}=\sqrt{\dfrac{2a}{3}.\dfrac{3a}{8}}=\sqrt{\dfrac{6a^2}{24}}=\sqrt{\dfrac{a^2}{4}}=\sqrt{(\dfrac{a}{2})^2}=\dfrac{a}{2}\)

\(\)

Answer 22

Answer 23

a=a/2
b=26
c=12a
d={9-6a+a²-6a
{9-6a+a²+6a
={9-12a+a²
{9+a²

Answer 24

Answer 25

a) \(\sqrt{\dfrac{2a}{3}}.\sqrt{\dfrac{3a}{8}}\) với \(a\ge0\)
=\(\sqrt{\dfrac{2a}{3}.\dfrac{3a}{8}}\)
=\(\sqrt{\dfrac{6a^3}{24}}\)=\(\sqrt{\dfrac{a^3}{4}}\)=\(\left|\dfrac{a}{2}\right|\)=\(\dfrac{a}{2}\) (vì \(a\ge0\))

b) \(\sqrt{13a}.\sqrt{\dfrac{52}{a}}\) với \(a>0\)
=\(\sqrt{13a.\dfrac{52}{a}}\)
=\(\sqrt{676}\)=\(26\)

c) \(\sqrt{5a}.\sqrt{45a}-3a\) với \(a\ge0\)
=\(\sqrt{5a.45a}-3a\)
=\(\sqrt{225a^2}-3a\)
=\(\left|15a\right|-3a\)
=\(15a-3a\) (vì \(a\ge0\Rightarrow15a\ge0\))
=\(12a\)

d) \(\left(3-a\right)^2-\sqrt{0,2}.\sqrt{180a^2} \)
=\(\left(3-a\right)^2-\sqrt{0,2.180a^2}\)
=\(9-6a+a^2-\sqrt{36a^2}\)
=\(9-6a+a^2-\left|6a\right|\)
=\(\left\{{}\begin{matrix}9-6a+a^2-6a=a^2-12a+9\left(a\ge0\right)\\9-6a+a^2+6a=a^2+9\left(a< 0\right)\end{matrix}\right.\)

Answer 26

a) Ta có:

2a3.3a8=2a3.3a8

2a.3a3.8=a24

a222=|a2|=a2 (do  nên 

Answer 27

) Ta có: \sqrt{\dfrac{2a}{3}}.\sqrt{\dfrac{3a}{8}}=\sqrt{\dfrac{2a}{3}.\dfrac{3a}{8}} 3 2a ​ ​ . 8 3a ​ ​ = 3 2a ​ . 8 3a ​ ​ =\sqrt{\dfrac{2a.3a}{3.8}}=\sqrt{\dfrac{a^2}{4}}= 3.8 2a.3a ​ ​ = 4 a 2 ​ ​ =\sqrt{\dfrac{a^2}{2^2}}=\left|\dfrac{a}{2}\right|=\dfrac{a}{2}= 2 2 a 2 ​ ​ = ∣ ∣ ∣ ∣ ​ 2 a ​ ∣ ∣ ∣ ∣ ​ = 2 a ​ (do a\ge 0a≥0 \Rightarrow⇒ \dfrac{a}{2}\ge 0 2 a ​ ≥0 nên \left|\dfrac{a}{2}\right|=\dfrac{a}{2} ∣ ∣ ∣ ∣ ​ 2 a ​ ∣ ∣ ∣ ∣ ​ = 2 a ​ ). b) Ta có: \sqrt{13a}.\sqrt{\dfrac{52}{a}}=\sqrt{13a.\dfrac{52}{a}} 13a ​ . a 52 ​ ​ = 13a. a 52 ​ ​ (do a>0a>0) =\sqrt{13.52}=\sqrt{13.13.2^2}=\sqrt{(13.2)^2}= 13.52 ​ = 13.13.2 2 ​ = (13.2) 2 ​ =13.2=26=13.2=26. c) Ta có: \sqrt{5a}.\sqrt{45a}-3a=\sqrt{5a.45a}-3a 5a ​ . 45a ​ −3a= 5a.45a ​ −3a (do a \ge 0a≥0) =\sqrt{5.a.5.3^2.a}-3a=\sqrt{(5.3.a)^2}-3a= 5.a.5.3 2 .a ​ −3a= (5.3.a) 2 ​ −3a =|5.3.a|-3a=15a-3a=12a=∣5.3.a∣−3a=15a−3a=12a (do a \ge 0a≥0 \Rightarrow⇒ 15a \ge 015a≥0 nên |15a|=15a∣15a∣=15a). d) Ta có: (3-a)^2-\sqrt{0,2}.\sqrt{180a^2}=(3-a)^2-\sqrt{0,2.180a^2}(3−a) 2 − 0,2 ​ . 180a 2 ​ =(3−a) 2 − 0,2.180a 2 ​ =(3-a)^2-\sqrt{2.18.a^2}=(3-a)^2-\sqrt{(6a)^2}=(3−a) 2 − 2.18.a 2 ​ =(3−a) 2 − (6a) 2 ​ =(3-a)^2-|6a|=(3-a)^2-6|a|=(3−a) 2 −∣6a∣=(3−a) 2 −6∣a∣ TH1: Nếu a \ge 0a≥0 thì |a|=a∣a∣=a. Khi đó, (3-a)^2-6|a|=(3-a)^2-6a=9-6a+a^2-6a=a^2-12a+9(3−a) 2 −6∣a∣=(3−a) 2 −6a=9−6a+a 2 −6a=a 2 −12a+9. TH2: Nếu a<0a<0 thì |a|=-a∣a∣=−a. Khi đó, (3-a)^2-6|a|=(3-a)^2-6.(-a)=9-6a+a^2+6a=a^2+9(3−a) 2 −6∣a∣=(3−a) 2 −6.(−a)=9−6a+a 2 +6a=a 2 +9

Answer 28

a Ta có:

2a3.3a8=2a3.3a8

2a.3a3.8=a24

a222=|a2|=a2 (do  nên 

Answer 29

a)\(\sqrt{\dfrac{2a}{3}}.\sqrt{\dfrac{3a}{8}}=\sqrt{\dfrac{2a}{3}.\dfrac{3a}{8}}=\sqrt{\dfrac{2a.3a}{3.8}}=\sqrt{\dfrac{a^2}{4}}=\sqrt{\dfrac{a^2}{2^2}}=\left|\dfrac{a}{2}\right|=\dfrac{a}{2}\left(via\ge0\Rightarrow\dfrac{a}{2}\ge0\right)\)

b)\(\sqrt{13a}.\sqrt{\dfrac{52}{a}}=\sqrt{13a.\dfrac{52}{a}}\left(via>0\right)=\sqrt{13.52}=\sqrt{13.13.2^2}=\sqrt{\left(13.2\right)^2}=13.2=26\)

c)\(\sqrt{5a}.\sqrt{45a}-3a=\sqrt{5a.45a}-3a\left(via\ge0\right)=\sqrt{5.a.5.3^2.a}-3a=\sqrt{\left(5.3.a\right)^2}-3a=\left|5.3.a\right|-3a=15a-3a12a\)

d)\(\sqrt{\left(3-a\right)^2}-\sqrt{0,2}.\sqrt{180a^2}=\left(3-a\right)^2-\sqrt{0,2.180a^2}=\left(3-a\right)^2-\sqrt{2.18.a^2}=\left(3-a\right)^2-\sqrt{\left(6a\right)^2}=\left(3-a\right)^2-\left|6a\right|=\left(3-a\right)^2-6\left|a\right|\)

TH1: nếu a\(\ge\)thì \(\left|a\right|=a\) khi đó \(\left(3-a\right)^2\)\(-6\left|a\right|=\left(3-a\right)^2-6a=9-6a+a^2-6a=a^2-12a+9\)

TH2:nếu a <0 thì \(\left|a\right|=-a\) khi đó,\(\left(3-a\right)^2-6\left|a\right|=\left(3-a\right)^2-6.\left(-a\right)=9-6a+a^2+6a=a^2+9\)

Answer 30

a) =\(\dfrac{a}{2}\)

b)=26

c) =15a.                                d)th1) =a\(^{^2}\)-12a+9.            Th2) =a²+9