Question

Bài 47 (trang 27 SGK Toán 9 Tập 1)

Rút gọn:

a) $\dfrac{2}{x^{2}-y^{2}} \sqrt{\dfrac{3(x+y)^{2}}{2}}$ với $x \ge 0, y \ge 0$ và $x \ne y$ ;

b) $\dfrac{2}{2 a-1} \sqrt{5 a^{2}\left(1-4 a+4 a^{2}\right)}$ với $a>0,5$.

Answer 1

a) Ta có : Vì \(x\ge0\)và \(y\ge0\)nên \(x+y\ge0\)\(\Leftrightarrow\left|x+y\right|=x+y\)

\(\frac{2}{x^2-y^2}\sqrt{\frac{3\left(x+y\right)^2}{2}}\)

\(=\frac{2}{x^2-y^2}\sqrt{\frac{3}{2}.\left(x+y\right)^2}\)

\(=\frac{2}{x^2-y^2}.\sqrt{\frac{3}{2}}.\left|x+y\right|\)

\(=\frac{2}{\left(x-y\right)\left(x+y\right)}.\sqrt{\frac{3}{2}}.\left(x+y\right)\)

\(=\frac{2}{x-y}.\sqrt{\frac{3}{2}}\)

\(=\frac{1}{x-y}.2.\sqrt{\frac{3}{2}}\)

\(=\frac{1}{x-y}.\sqrt{\frac{2^2.3}{2}}\)

\(=\frac{1}{x-y}.\sqrt{6}=\frac{\sqrt{6}}{x-y}\)

Answer 2

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

do \(x\ge0;y\ge0\)

\(=\frac{2\sqrt{3}}{\sqrt{2}\left(x-y\right)}=\frac{2\sqrt{6}}{2\left(x-y\right)}=\frac{\sqrt{6}}{x-y}\)

Answer 3

b) Ta có :

\(\frac{2}{2a-1}\sqrt{5a^2\left(1-4a+4a^2\right)}\)

\(=\frac{2}{2a-1}\sqrt{5a^2\left(1-2.2a+2^2.a^2\right)}\)

\(=\frac{2}{2a-1}\sqrt{5a^2\left[1^2-2.1.2a+\left(2a\right)^2\right]}\)

\(=\frac{2}{2a-1}\sqrt{5a^2\left(1-2a\right)^2}\)

\(=\frac{2}{2a-1}.\sqrt{5}.\sqrt{a^2}.\sqrt{\left(1-2a\right)^2}\)

\(=\frac{2}{2a-1}\sqrt{5}.\left|a\right|.\left|1-2a\right|\)

Vì a > 0,5 <=> a > 0 <=> | a | = a

Vì a > 0,5 <=> 2a > 2 . 0,5 <=> 2a > 1 hay 1 < 2a

<=> 1 - 2a < 0 <=> | 1 - 2a | = - ( 1 - 2a )

= -1 + 2a = 2a - 1

Thay vào trên ta được : 

\(=\frac{2}{2a-1}\sqrt{5}.\left|a\right|.\left|1-2a\right|=\frac{2}{2a-1}\sqrt{5}.a.\left(2a-1\right)=2a\sqrt{5}\)

Vậy : \(\frac{2}{2a-1}\sqrt{5a^2\left(1-4a+4a^2\right)}=2a\sqrt{5}\)

Answer 4

b, \(\frac{2}{2a-1}\sqrt{5a^2\left(1-4a+4a^2\right)}\)

\(=\frac{2}{2a-1}\sqrt{5a^2\left(1-2a\right)^2}=\frac{2}{2a-1}\sqrt{5a^2\left(2a-1\right)^2}\)

\(=\frac{2\sqrt{5}a\left(2a-1\right)}{2a-1}=2\sqrt{5}a\)

Answer 5

2x2−y23(x+y)22

2x2−y232.(x+y)2

2x2−y2(x+y)2.32

2x2−y2.|x+y|32

|x+y|x2−y23.222

x+y(x−y)(x+y).6 (Do $x\ge 0,y\ge 0$ $\Rightarrow$ $x+y\ge 0$ nên $|x+y|=x+y$)

6x−y.

22a−15a2(1−4a+4a2)

22a−15a2(1−2.1.2a+(2a)2)

22a−15a2(1−2a)2

22a−15.a2.(1−2a)2

22a−15.|a|.|1−2a|

22a−15.a.(2a−1) (Do $a>0,5$ $\Rightarrow$ $2a-1>0$ nên $|a|=a$ và $|1-2a|=2a-1$)

$=2a\sqrt{5}$.

Answer 6

a) \(\dfrac{2}{x^2-y^2}\sqrt{\dfrac{3\left(x+y\right)^2}{2}}=\dfrac{2}{\left(x+y\right)\left(x-y\right)}.\dfrac{\sqrt{3}.\left(x+y\right)}{\sqrt{2}}=\dfrac{\sqrt{6}}{x-y}\)

b) \(\dfrac{2}{2a-1}\sqrt{5a^2\left(1-4a+4a^2\right)}=\dfrac{2}{2a-1}.\sqrt{5}a.\left(2a-1\right)=2\sqrt{5}a\)

Answer 7

a)\(\dfrac{\sqrt{6}}{x-y}\)
b)\(2a\sqrt{5}\)

Answer 8

Answer 9

Answer 10

Answer 11

Answer 12

a, \(\dfrac{2}{x^2-y^2}\sqrt{\dfrac{3(x+y)^2}{2}với}x\ge0,y\ge0\)\(vàx\ne y\)

 \(=\dfrac{\sqrt{6}}{(x-y)}\)

b, \(=2a\sqrt{5}\)

Answer 13

a) \dfrac{2}{x^2-y^2}\sqrt{\dfrac{3(x+y)^2}{2}}x2−y22​23(x+y)2​​

=\dfrac{2}{x^2-y^2}\sqrt{\dfrac{3}{2}.(x+y)^2}=x2−y22​23​.(x+y)2​

=\dfrac{2}{x^2-y^2}\sqrt{(x+y)^2}.\sqrt{\dfrac{3}{2}}=x2−y22​(x+y)2​.23​​

=\dfrac{2}{x^2-y^2}.|x+y|\sqrt{\dfrac{3}{2}}=x2−y22​.∣x+y∣23​​

=\dfrac{|x+y|}{x^{2}-y^{2}}\sqrt{\dfrac{3.2^2}{2}}=x2−y2∣x+y∣​23.22​​

=\dfrac{x+y}{(x-y)(x+y)}.\sqrt{6}=(x−y)(x+y)x+y​.6​ (Do $x\ge 0,y\ge 0$ $\Rightarrow$ $x+y\ge 0$ nên $|x+y|=x+y$)

=\dfrac{\sqrt{6}}{x-y}=x−y6​​.

b) \dfrac{2}{2 a-1} \sqrt{5a^2(1-4 a+4 a^2)}2a−1

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Answer 14

a căn 6 trên (x-y)

b 2a căn 5

Answer 15

Answer 16

Answer 17

b)\(\dfrac{2}{2a-1}\sqrt{5a^2\left(1-4a+4a^2\right)}=\dfrac{2\left|a\right|}{2a-1}\sqrt{5\left(1-2.2a+\left(2a\right)^2\right)}=\dfrac{2a}{2a-1}\sqrt{5\left(1-2a\right)^2}=\dfrac{2a\left|1-2a\right|}{2a-1}\sqrt{5}=\dfrac{2a\left(2a-1\right)}{2a-1}\sqrt{5}=2a\sqrt{5}\)

a) \(\dfrac{2}{x^2-y^2}\sqrt{\dfrac{3\left(x+y\right)^2}{2}}=\dfrac{\left|x+y\right|}{x^2-y^2}\sqrt{\dfrac{3.2^2}{2}}=\dfrac{\left(x+y\right)}{\left(x-y\right)\left(x+y\right)}\sqrt{6}=\dfrac{1}{x-y}\sqrt{6}\)

Answer 18

a)\(\dfrac{2}{x^2-y^2}\sqrt{\dfrac{3\left(x+y\right)^2}{2}}=\dfrac{\left|x+y\right|}{x^2-y^2}\sqrt{\dfrac{3.2^2}{2}}=\dfrac{x+y}{\left(x-y\right)\left(x+y\right)}\sqrt{6}=\dfrac{1}{x-y}\sqrt{6}\)                                 b)\(\dfrac{2}{2a-1}\sqrt{5a^2\left(1-4a+4a^2\right)}=\dfrac{2\left|a\right|}{2a-1}\sqrt{5.\left(1-2a\right)^2}=\dfrac{2a\left|1-2a\right|}{2a-1}\sqrt{5}=2a\sqrt{5}\)

Answer 19

2a căn5

Answer 20

a)\(\dfrac{2}{x^2-y^2}\sqrt{\dfrac{3\left(x+y\right)^2}{2}}\)=\(\dfrac{2}{x^2-y^2}\sqrt{\dfrac{3}{2}\left(x+y\right)^2}\) =\(\dfrac{2}{x^2-y^2}\sqrt{\left(x+y\right)^2}.\sqrt{\dfrac{3}{2}}\)=\(\dfrac{2}{x^2-y^2}.\left|x+y\right|\sqrt{\dfrac{3}{2}}\)=                            \(\dfrac{\left|x+y\right|}{x^2-y^2}.\sqrt{\dfrac{3.2^2}{2}}=\dfrac{x+y}{\left(x-y\right)\left(x+y\right)}.\sqrt{6}\)  (do x\(\ge\)0 y\(\ge\)0 \(\Rightarrow x+y\ge\)nên\(\left|x+y\right|=x+y\))                                   b)\(\dfrac{2}{2a-1}\sqrt{5a^2\left(1-4a+4a^2\right)}\)=\(\dfrac{2}{2a-1}\sqrt{5a^2\left(1-2.1.2a+\left(2a\right)^2\right)}\)

Answer 21

   

 

Answer 22

a)\(\dfrac{\sqrt{6}}{x-y}\)

b)\(2a\sqrt{5}\)

Answer 23

a= 2/(x-y) . (x+y) .\(\sqrt{ }\)3(x+y)^2.2/2.2  = 2/(x+y).(x-y) . \(\sqrt{ }\)6(x+y)^2/\(\sqrt{ }\)4  = 2/(x+y). (x-y) . \(\sqrt{ }\)6(x+y)/2    = \(\sqrt{ }\)6/x-y                                                                                                                                                                                                                                                                                                   b = 2/2a-1 . \(\sqrt{ }\)5a^2 . (1-2a)^2                                                                                                          = 2/2a-1 . \(\sqrt{ }\)5 . \(\sqrt{ }\)a^2 . \(\sqrt{ }\)(1-2a)^2                                                                                                = 2/2a-1 . \(\sqrt{ }\)5 . a . (2a-1)                                                                                                              = 2\(\sqrt{ }\)5-a

Answer 24

a)\(\dfrac{2}{x^2-y^2}\cdot\sqrt{\dfrac{3\cdot\left(x+y\right)^2}{2}}\)     (voi \(x\ge0;x\ne y\) )                                                                              =\(\dfrac{2}{x^2-y^2}\cdot\dfrac{\sqrt{3}\cdot\sqrt{\left(x+y\right)^2}}{\sqrt{2}}\)                                                                                                               =\(\dfrac{2}{\left(x-y\right)\cdot\left(x+y\right)}\cdot\dfrac{\sqrt{3}\cdot\left|x+y\right|}{\sqrt{2}}\)                                                                                              =\(\dfrac{2\sqrt{3}}{\sqrt{2}\cdot\left(x-y\right)}\)                                                                                                                              =\(\sqrt{\dfrac{12}{2}}\cdot\dfrac{1}{x-y}\)

    =\(\sqrt{6}\cdot\dfrac{1}{x-y}\)

b)\(\dfrac{2}{2a-1}\cdot\sqrt{5a^2\cdot\left(1-4a+4a^2\right)}\)  ( Voi  a>0,5 )

= \(\dfrac{2}{2a-1}\cdot\sqrt{5a^2\cdot\left(1-2a\right)^2}\)

 =\(\dfrac{2}{2a-1}\cdot\sqrt{5a^2}\cdot\left|1-2a\right|\)

=\(\dfrac{2}{2a-1}\cdot\sqrt{5a^2}\cdot\left(2a-1\right)\)

= \(2\sqrt{5a^2}\)

= \(2a\sqrt{5}\)

Answer 25

a) 2x2−y23(x+y)22\dfrac{2}{x^2-y^2}\sqrt{\dfrac{3(x+y)^2}{2}}x2−y22​23(x+y)2​

=2x2−y232.(x+y)2=\dfrac{2}{x^2-y^2}\sqrt{\dfrac{3}{2}.(x+y)^2}=x2−y22​23​.(x+y)2

=2x2−y2(x+y)2.32=\dfrac{2}{x^2-y^2}\sqrt{(x+y)^2}.\sqrt{\dfrac{3}{2}}=x2−y22​(x+y)2

​.23​

​

=2x2−y2.∣x+y∣32=\dfrac{2}{x^2-y^2}.|x+y|\sqrt{\dfrac{3}{2}}=x2−y22​.∣x+y∣23​

​

=∣x+y∣x2−y23.222=\dfrac{|x+y|}{x^{2}-y^{2}}\sqrt{\dfrac{3.2^2}{2}}=x2−y2∣x+y∣​23.22​

​

=x+y(x−y)(x+y).6=\dfrac{x+y}{(x-y)(x+y)}.\sqrt{6}=(x−y)(x+y)x+y​.6

​ (Do $x\ge 0,y\ge 0$ $\Rightarrow$ $x+y\ge 0$ nên $|x+y|=x+y$)

=6x−y=\dfrac{\sqrt{6}}{x-y}=x−y6

​​.

b) 22a−15a2(1−4a+4a2)\dfrac{2}{2 a-1} \sqrt{5a^2(1-4 a+4 a^2)}2a−12​5a2(1−4a+4a2)

=22a−15a2(1−2.1.2a+(2a)2)=\dfrac{2}{2 a-1} \sqrt{5a^2(1-2.1.2a+(2a)^2)}=2a−12​5a2(1−2.1.2a+(2a)2)

​

=22a−15a2(1−2a)2=\dfrac{2}{2 a-1} \sqrt{5a^2(1-2a)^2}=2a−12​5a2(1−2a)2

​

=22a−15.a2.(1−2a)2=\dfrac{2}{2 a-1} \sqrt{5}.\sqrt{a^2}.\sqrt{(1-2a)^2}=2a−12​5

​.a2​.(1−2a)2

​

=22a−15.∣a∣.∣1−2a∣=\dfrac{2}{2 a-1} \sqrt{5}.|a|.|1-2a|=2a−12​5

​.∣a∣.∣1−2a∣

=22a−15.a.(2a−1)=\dfrac{2}{2 a-1} \sqrt{5}.a.(2a-1)=2a−12

​.

Answer 26

a) \dfrac{2}{x^2-y^2}\sqrt{\dfrac{3(x+y)^2}{2}}x2−y22​23(x+y)2​​

=\dfrac{2}{x^2-y^2}\sqrt{\dfrac{3}{2}.(x+y)^2}=x2−y22​23​.(x+y)2​

=\dfrac{2}{x^2-y^2}\sqrt{(x+y)^2}.\sqrt{\dfrac{3}{2}}=x2−y22​(x+y)2​.23​​

=\dfrac{2}{x^2-y^2}.|x+y|\sqrt{\dfrac{3}{2}}=x2−y22​.∣x+y∣23​​

=\dfrac{|x+y|}{x^{2}-y^{2}}\sqrt{\dfrac{3.2^2}{2}}=x2−y2∣x+y∣​23.22​​

=\dfrac{x+y}{(x-y)(x+y)}.\sqrt{6}=(x−y)(x+y)x+y​.6​ (Do $x\ge 0,y\ge 0$ $\Rightarrow$ $x+y\ge 0$ nên $|x+y|=x+y$)

=\dfrac{\sqrt{6}}{x-y}=x−y6​​.

b) \dfrac{2}{2 a-1} \sqrt{5a^2(1-4 a+4 a^2)}2

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Answer 27

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Answer 28

a) \(\dfrac{2}{x^2-y^2}\)\(\sqrt{\dfrac{3\left(x+y\right)^2}{2}}\) 

= \(\dfrac{2}{\left(x-y\right).\left(x+y\right)}\)\(\sqrt{\dfrac{3}{2}.\left(x+y\right)^2}\)

 

= \(\dfrac{2}{\left(x+y\right).\left(x-y\right)}\).|x+y| \(\sqrt{\dfrac{3}{2}}\)

= \(\dfrac{x+y}{\left(x-y\right).\left(x+y\right)}\) .\(\sqrt{\dfrac{3.2^2}{2}}\)

= \(\dfrac{\sqrt{6}}{x-y}\)

b) \(\dfrac{2}{2a-1}\) \(\sqrt{5a^2\left(1-4a+4a^2\right)}\)

= \(\dfrac{2}{2a-1}\)\(\sqrt{5a^2\left(1-2a\right)^2}\)

=\(\dfrac{2}{2a-1}\)\(\sqrt{5}\).\(\sqrt{a^2}\).\(\sqrt{\left(1-2a\right)^2}\)

= \(\dfrac{2}{2a-1}\)\(\sqrt{5}\). |a| . |1-2a|

= \(\dfrac{2}{2a-1}\)\(\sqrt{5}\) . |a| .|1-2a|

= \(\dfrac{2}{2a-1}\)\(\sqrt{5}\).a.( 2a-1)          ( vì a > 0,5 )

= 2a\(\sqrt{5}\)

= 

Answer 29

Answer 30

2x2−y23(x+y)22

2x2−y232.(x+y)2

2x2−y2(x+y)2.32

2x2−y2.|x+y|32

|x+y|x2−y23.222

x+y(x−y)(x+y).6 (Do $x\ge 0,y\ge 0$ $\Rightarrow$ $x+y\ge 0$ nên $|x+y|=x+y$)

6x−y.

22a−15a2(1−4a+4a2)

22a−15a2(1−2.1.2a+(2a)2)

22a−15a2(1−2a)2

22a−15.a2.(1−2a)2

22a−15.|a|.|1−2a|

22a−15.a.(2a−1) (Do $a>0,5$ $\Rightarrow$ $2a-1>0$ nên $|a|=a$ và $|1-2a|=2a-1$)

$=2a\sqrt{5}$.