Question

Bài 76 (trang 41 SGK Toán 9 Tập 1)

Cho biểu thức

$Q=\dfrac{a}{\sqrt{a^{2}-b^{2}}}-\left(1+\dfrac{a}{\sqrt{a^{2}-b^{2}}}\right): \dfrac{b}{a-\sqrt{a^{2}-b^{2}}}$ với $a>b>0$.

a) Rút gọn $Q$.

b) Xác định giá trị của $Q$ khi $a=3 b$.

Answer 1

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Answer 2

a) a √ a 2 − b 2 − ( 1 + a √ a 2 − b 2 ) : b a − √ a 2 − b 2 = a √ a 2 − b 2 − a + √ a 2 − b 2 √ a 2 − b 2 ⋅ a − √ a 2 − b 2 b = a √ a 2 − b 2 − a 2 − ( √ a 2 − b 2 ) 2 b √ a 2 − b 2 = a √ a 2 − b 2 − a 2 − ( a 2 − b 2 ) b √ a 2 − b 2 = a √ a 2 − b 2 − b 2 b ⋅ √ a 2 − b 2 = a √ a 2 − b 2 − b √ a 2 − b 2 = a − b √ a 2 − b 2 = √ a − b ⋅ √ a − b √ a − b ⋅ √ a + b (do a > b > 0 )$ = √ a − b √ a + b Vậy Q = √ a − b √ a + b . b) Thay a = 3 b vào Q = √ a − b √ a + b , ta được: Q = √ 3 b − b √ 3 b + b = √ 2 b √ 4 b = √ 2 b √ 2 ⋅ √ 2 b = 1 √ 2 = √ 2 2 .

Answer 3

a) \dfrac{a}{\sqrt{a^{2}-b^{2}}}-\left(1+\dfrac{a}{\sqrt{a^{2}-b^{2}}}\right): \dfrac{b}{a-\sqrt{a^{2}-b^{2}}}a2−b2​a​−(1+a2−b2​a​):a−a2−b2​b​

=\dfrac{a}{\sqrt{a^{2}-b^{2}}}-\dfrac{a+\sqrt{a^{2}-b^{2}}}{\sqrt{a^{2}-b^{2}}} \cdot \dfrac{a-\sqrt{a^{2}-b^{2}}}{b}=a2−b2​a​−a2−b2​a+a2−b2​​⋅ba−a2−b2​​

=\dfrac{a}{\sqrt{a^{2}-b^{2}}}-\dfrac{a^{2}-\left(\sqrt{a^{2}-b^{2}}\right)^{2}}{b \sqrt{a^{2}-b^{2}}}=a2−b2​a​−ba2−b2​a2−(a2−b2​)2​

=\dfrac{a}{\sqrt{a^{2}-b^{2}}}-\dfrac{a^{2}-\left(a^{2}-b^{2}\right)}{b \sqrt{a^{2}-b^{2}}}=a2−b2​a​−ba2−b2​a2−(a2−b2)​

=\dfrac{a}{\sqrt{a^{2}-b^{2}}}-\dfrac{b^{2}}{b \cdot \sqrt{a^{2}-b^{2}}}=a2−b2​a​−b⋅a2−b2​b2​

=\dfrac{a}{\sqrt{a^{2}-b^{2}}}-\dfrac{b}{\sqrt{a^{2}-b^{2}}}=a2−b2​a​−a2−b2​b​

=\dfrac{a-b}{\sqrt{a^{2}-b^{2}}}=a2−b2​a−b​

=\dfrac{\sqrt{a-b} \cdot \sqrt{a-b}}{\sqrt{a-b} \cdot \sqrt{a+b}}=a−b​⋅a+b​a−b​⋅a−b​​ (do $a>b>0$)$

=\dfrac{\sqrt{a-b}}{\sqrt{a+b}}=a+b​a−b​​

Vậy Q=\dfrac{\sqrt{a-b}}{\sqrt{a+b}}Q=a+b​a−b​​.
b) Thay $a=3b$ vào Q=\dfrac{\sqrt{a-b}}{\sqrt{a+b}}Q=a+b​a−b​​, ta được:

Q=\dfrac{\sqrt{3 b-b}}{\sqrt{3 b+b}}=\dfrac{\sqrt{2 b}}{\sqrt{4 b}}Q=3b+b​3b−b​​=4b​2b​​

=\dfrac{\sqrt{2 b}}{\sqrt{2} \cdot \sqrt{2 b}}=\dfrac{1}{\sqrt{2}}=\dfrac{\sqrt{2}}{2}=2​⋅2b​2b​​=2​1​=22​​.

Answer 4

Answer 5

a) \dfrac{a}{\sqrt{a^{2}-b^{2}}}-\left(1+\dfrac{a}{\sqrt{a^{2}-b^{2}}}\right): \dfrac{b}{a-\sqrt{a^{2}-b^{2}}}a2−b2​a​−(1+a2−b2​a​):a−a2−b2​b​

=\dfrac{a}{\sqrt{a^{2}-b^{2}}}-\dfrac{a+\sqrt{a^{2}-b^{2}}}{\sqrt{a^{2}-b^{2}}} \cdot \dfrac{a-\sqrt{a^{2}-b^{2}}}{b}=a2−b2​a​−a2−b2​a+a2−b2​​⋅ba−a2−b2​​

=\dfrac{a}{\sqrt{a^{2}-b^{2}}}-\dfrac{a^{2}-\left(\sqrt{a^{2}-b^{2}}\right)^{2}}{b \sqrt{a^{2}-b^{2}}}=a2−b2​a​−ba2−b2​a2−(a2−b2​)2​

=\dfrac{a}{\sqrt{a^{2}-b^{2}}}-\dfrac{a^{2}-\left(a^{2}-b^{2}\right)}{b \sqrt{a^{2}-b^{2}}}=a2−b2​a​−ba2−b2​a2−(a2−b2)​

=\dfrac{a}{\sqrt{a^{2}-b^{2}}}-\dfrac{b^{2}}{b \cdot \sqrt{a^{2}-b^{2}}}=a2−b2​a​−b⋅a2−b2​b2​

=\dfrac{a}{\sqrt{a^{2}-b^{2}}}-\dfrac{b}{\sqrt{a^{2}-b^{2}}}=a2−b2​a​−a2−b2​b​

=\dfrac{a-b}{\sqrt{a^{2}-b^{2}}}=a2−b2​a−b​

=\dfrac{\sqrt{a-b} \cdot \sqrt{a-b}}{\sqrt{a-b} \cdot \sqrt{a+b}}=a−b​⋅a+b​a−b​⋅a−b​​ (do $a>b>0$)$

=\dfrac{\sqrt{a-b}}{\sqrt{a+b}}=a+b​a−b​​

Vậy Q=\dfrac{\sqrt{a-b}}{\sqrt{a+b}}Q=a+b​a−b​​.
b) Thay $a=3b$ vào Q=\dfrac{\sqrt{a-b}}{\sqrt{a+b}}Q=a+b​a−b​​, ta được:

Q=\dfrac{\sqrt{3 b-b}}{\sqrt{3 b+b}}=\dfrac{\sqrt{2 b}}{\sqrt{4 b}}Q=3b+b​3b−b​​=4b​2b​​

=\dfrac{\sqrt{2 b}}{\sqrt{2} \cdot \sqrt{2 b}}=\dfrac{1}{\sqrt{2}}=\dfrac{\sqrt{2}}{2}=2​⋅2b​2b​​=2​1​=22​​.

Answer 6

Answer 7

Answer 8

máy em bị lỗi

Answer 9

a) \dfrac{a}{\sqrt{a^{2}-b^{2}}}-\left(1+\dfrac{a}{\sqrt{a^{2}-b^{2}}}\right): \dfrac{b}{a-\sqrt{a^{2}-b^{2}}}a2−b2​a​−(1+a2−b2​a​):a−a2−b2​b​

=\dfrac{a}{\sqrt{a^{2}-b^{2}}}-\dfrac{a+\sqrt{a^{2}-b^{2}}}{\sqrt{a^{2}-b^{2}}} \cdot \dfrac{a-\sqrt{a^{2}-b^{2}}}{b}=a2−b2​a​−a2−b2​a+a2−b2​​⋅ba−a2−b2​​

=\dfrac{a}{\sqrt{a^{2}-b^{2}}}-\dfrac{a^{2}-\left(\sqrt{a^{2}-b^{2}}\right)^{2}}{b \sqrt{a^{2}-b^{2}}}=a2−b2​a​−ba2−b2​a2−(a2−b2​)2​

=\dfrac{a}{\sqrt{a^{2}-b^{2}}}-\dfrac{a^{2}-\left(a^{2}-b^{2}\right)}{b \sqrt{a^{2}-b^{2}}}=a2−b2​a​−ba2−b2​a2−(a2−b2)​

=\dfrac{a}{\sqrt{a^{2}-b^{2}}}-\dfrac{b^{2}}{b \cdot \sqrt{a^{2}-b^{2}}}=a2−b2​a​−b⋅a2−b2​b2​

=\dfrac{a}{\sqrt{a^{2}-b^{2}}}-\dfrac{b}{\sqrt{a^{2}-b^{2}}}=a2−b2​a​−a2−b2​b​

=\dfrac{a-b}{\sqrt{a^{2}-b^{2}}}=a2−b2​a−b​

=\dfrac{\sqrt{a-b} \cdot \sqrt{a-b}}{\sqrt{a-b} \cdot \sqrt{a+b}}=a−b​⋅a+b​a−b​⋅a−b​​ (do $a>b>0$)$

=\dfrac{\sqrt{a-b}}{\sqrt{a+b}}=a+b​a−b​​

Vậy Q=\dfrac{\sqrt{a-b}}{\sqrt{a+b}}Q=a+b​a−b​​.
b) Thay $a=3b$ vào Q=\dfrac{\sqrt{a-b}}{\sqrt{a+b}}Q=a+b​a−b​​, ta được:

Q=\dfrac{\sqrt{3 b-b}}{\sqrt{3 b+b}}=\dfrac{\sqrt{2 b}}{\sqrt{4 b}}Q=3b+b​3b−b​​=4b​2b​​

=\dfrac{\sqrt{2 b}}{\sqrt{2} \cdot \sqrt{2 b}}=\dfrac{1}{\sqrt{2}}=\dfrac{\sqrt{2}}{2}=2​⋅2b​2b​​=2​1​=22​​.

Answer 10

ee

Answer 11

 

 

Answer 12

Answer 13

 

aa2−b2−(1+aa2−b2):ba−a2−b2

aa2−b2−a+a2−b2a2−b2⋅a−a2−b2b

aa2−b2−a2−(a2−b2)2ba2−b2

aa2−b2−a2−(a2−b2)ba2−b2

aa2−b2−b2b⋅a2−b2

aa2−b2−ba2−b2

a−ba2−b2

a−b⋅a−ba−b⋅a+b (do $a>b>0$)$

a−ba+b

a−ba+b.
b) Thay , ta được:

3b−b3b+b=2b4b

 

2b2⋅2b=12=22.

Answer 14

a)\(\dfrac{\sqrt{a-b}}{\sqrt{a+b}}\)
b)\(\dfrac{\sqrt{2}}{2}\)

Answer 15

aa2−b2−(1+aa2−b2):ba−a2−b2

aa2−b2−a+a2−b2a2−b2⋅a−a2−b2b

 

aa2−b2−a2−(a2−b2)2ba2−b2

 

aa2−b2−a2−(a2−b2)ba2−b2

 

aa2−b2−b2b⋅a2−b2

 

aa2−b2−ba2−b2

 

a−ba2−b2

 

a−b⋅a−ba−b⋅a+b (do $a>b>0$)

 

a−ba+b

 

a−ba+b.

b) Thay , ta được:

 

3b−b3b+b=2b4b

 

2b2⋅2b=12=22.

 

Answer 16

a) 

aa2−b2−(1+aa2−b2):ba−a2−b2

 

aa2−b2−a+a2−b2a2−b2⋅a−a2−b2b

 

aa2−b2−a2−(a2−b2)2ba2−b2

 

aa2−b2−a2−(a2−b2)ba2−b2

 

aa2−b2−b2b⋅a2−b2

 

aa2−b2−ba2−b2

 

a−ba2−b2

 

a−b⋅a−ba−b⋅a+b (do $a>b>0$)$

 

a−ba+b

 

a−ba+b.

b)

a−ba+b, ta được:

 

3b−b3b+b=2b4b

 

2b2⋅2b=12=22.

Answer 17

Answer 18

 

Answer 19

a) \dfrac{a}{\sqrt{a^{2}-b^{2}}}-\left(1+\dfrac{a}{\sqrt{a^{2}-b^{2}}}\right): \dfrac{b}{a-\sqrt{a^{2}-b^{2}}}a2−b2​a​−(1+a2−b2​a​):a−a2−b2​b​

=\dfrac{a}{\sqrt{a^{2}-b^{2}}}-\dfrac{a+\sqrt{a^{2}-b^{2}}}{\sqrt{a^{2}-b^{2}}} \cdot \dfrac{a-\sqrt{a^{2}-b^{2}}}{b}=a2−b2​a​−a2−b2​a+a2−b2​​⋅ba−a2−b2​​

=\dfrac{a}{\sqrt{a^{2}-b^{2}}}-\dfrac{a^{2}-\left(\sqrt{a^{2}-b^{2}}\right)^{2}}{b \sqrt{a^{2}-b^{2}}}=a2−b2​a​−ba2−b2​a2−(a2−b2​)2​

=\dfrac{a}{\sqrt{a^{2}-b^{2}}}-\dfrac{a^{2}-\left(a^{2}-b^{2}\right)}{b \sqrt{a^{2}-b^{2}}}=a2−b2​a​−ba2−b2​a2−(a2−b2)​

=\dfrac{a}{\sqrt{a^{2}-b^{2}}}-\dfrac{b^{2}}{b \cdot \sqrt{a^{2}-b^{2}}}=a2−b2​a​−b⋅a2−b2​b2​

=\dfrac{a}{\sqrt{a^{2}-b^{2}}}-\dfrac{b}{\sqrt{a^{2}-b^{2}}}=a2−b2​a​−a2−b2​b​

=\dfrac{a-b}{\sqrt{a^{2}-b^{2}}}=a2−b2​a−b​

=\dfrac{\sqrt{a-b} \cdot \sqrt{a-b}}{\sqrt{a-b} \cdot \sqrt{a+b}}=a−b​⋅a+b​a−b​⋅a−b​​ (do $a>b>0$)$

=\dfrac{\sqrt{a-b}}{\sqrt{a+b}}=a+b​a−b​​

Vậy Q=\dfrac{\sqrt{a-b}}{\sqrt{a+b}}Q=a+b​a−b​​.
b) Thay $a=3b$ vào Q=\dfrac{\sqrt{a-b}}{\sqrt{a+b}}Q=a+b​a−b​​, ta được:

Q=\dfrac{\sqrt{3 b-b}}{\sqrt{3 b+b}}=\dfrac{\sqrt{2 b}}{\sqrt{4 b}}Q=3b+b​3b−b​​=4b​2b​​

=\dfrac{\sqrt{2 b}}{\sqrt{2} \cdot \sqrt{2 b}}=\dfrac{1}{\sqrt{2}}=\dfrac{\sqrt{2}}{2}=2​⋅2b​2b​​=2​1​=22​​.

Answer 20

a) Q = \(\dfrac{\sqrt{a-b}}{\sqrt{a+b}}\)

b) \(\dfrac{\sqrt{2}}{2}\)

Answer 21

a,Q=\(\dfrac{\sqrt{a-b}}{\sqrt{a+b}}\)

b,thay a=3b =>Q=\(\dfrac{\sqrt{2}}{2}\)

Answer 22

aa2−b2−(1+aa2−b2):ba−a2−b2

aa2−b2−a+a2−b2a2−b2⋅a−a2−b2b

aa2−b2−a2−(a2−b2)2ba2−b2

aa2−b2−a2−(a2−b2)ba2−b2

aa2−b2−b2b⋅a2−b2

aa2−b2−ba2−b2

a−ba2−b2

a−b⋅a−ba−b⋅a+b (do $a>b>0$)$

a−ba+b

a−ba+b.
b) Thay , ta được:

3b−b3b+b=2b4b

2b2⋅2b=12=22.

Answer 23

 

 

 

Answer 24

 

 

Answer 25

 

Answer 26

 

 

Answer 27

Answer 28

 

Answer 29

Answer 30