mọi người giúp mình với ạ

`a)` Giả sử `\sqrt{a + \sqrt{b} } = \sqrt{\frac{a+\sqrt{a^2-b}}{2}}+\sqrt{\frac{a-\sqrt{a^2-b}}{2}}`

`<=> ( \sqrt{a + \sqrt{b}} )^2 = (\sqrt{\frac{a+\sqrt{a^2-b}}{2}}+\sqrt{\frac{a-\sqrt{a^2-b}}{2}})^2`

`<=> a + \sqrt{b} = ((\sqrt{a + \sqrt{a^2 - b} }+\sqrt{a - \sqrt{a^2 - b} })^2)/2`

`<=> a + \sqrt{b} = ( 2a  +2\sqrt{b} )/2`

`<=> a + \sqrt{b} = a + \sqrt{b}`

`=>` Điều G/S là đúng `=> ĐPCM`

`b)` `\sqrt{4\sqrt{2} + 2\sqrt{6} }`

`= \sqrt{ \sqrt{2} ( 4 + 2\sqrt{3} ) }`

`= \sqrt{ \sqrt{2} ( ( \sqrt{3} )^2 + 2 . \sqrt{3} . 1 + 1^2 ) }`

`= \sqrt{ \sqrt{2} ( \sqrt{3} + 1 )^2 }`

`= \sqrt{2} . | \sqrt{3}  +1 |`

`= \sqrt{2} . ( \sqrt{3}  +1 )`

`=>` ĐPCM

`c)` `\sqrt{17 - 4\sqrt{9 + 4\sqrt{5} } }`

 `= \sqrt{17 - 4 \sqrt{ ( \sqrt{5} )^2 + 2 . \sqrt{5} . 2 + 2^2 ) }`

`= \sqrt{17 - 4 \sqrt{ ( \sqrt{5} + 2 )^2 }`

`= \sqrt{17 - 4 ( sqrt{5} + 2 ) }`

`= \sqrt{17 - 4\sqrt{5} - 8}`

`= \sqrt{9 - 4\sqrt{5}`

`= \sqrt{ 2^2 - 2 . 2. \sqrt{5} + ( \sqrt{5} )^2 }`

`= \sqrt{ ( 2 - \sqrt{5} )^2 }`

`= | 2 - \sqrt{5} |`

`= \sqrt{5} - 2` ( Vì `\sqrt{5} > 2`

`=> ĐPCM` 

a. Xét: 

$A = \sqrt{\dfrac{a + \sqrt{a^2 - b}}{2} + \sqrt{\dfrac{a - \sqrt{a^2 - b}}{2} > 0$ 

$A^2 = \dfrac{a + \sqrt{a^2 - 2}}{2} + \dfrac{a - \sqrt{a^2 - b}}{2} + 2\sqrt{\dfrac{a + \sqrt{a^2 - b}}{2}.\dfrac{a - \sqrt{a^2 - b}}{2}}$ 

$A^2 = \dfrac{2a}{2} - 2\sqrt{\dfrac{(a + \sqrt{a^2 - b})(a - \sqrt{a^2 - b})}{4}}$ 

$A^2 = a - 2\sqrt{a^2 - (a^2 - b)} = a - \dfrac{2\sqrt{b}}{2} = a - \sqrt{b}$ 

Suy ra: $A = \sqrt{A^2} = \sqrt{a - \sqrt{b}} = VT$ (dpcm) 

b. Ta có: 

$B = \sqrt{\sqrt{2}}(\sqrt{3} + 1) > 0$ 

$B^2 = [\sqrt{\sqrt{2}}(\sqrt{3} + 1)]^2 = \sqrt{2} (\sqrt{3} + 1)^2 = \sqrt{2}(3 + 2\sqrt{3} + 1) = \sqrt{2}(4 + 2\sqrt{3}) = 4 \sqrt{2} + 2\sqrt{6}$ 

Suy ra: $B = \sqrt{B^2} = \sqrt{4\sqrt{2} + 2\sqrt{6}} = VT$ (đpcm) 

c. Ta có: 

$\sqrt{17 - 4\sqrt{9 + 4\sqrt{5}}} = \sqrt{17 - 4\sqrt{(\sqrt{5} + 2)^2}} $$= \sqrt{17 - 4(\sqrt{5} + 2)} = \sqrt{17 - 4\sqrt{5} - 8} =$$ \sqrt{9 - 4\sqrt{5}} = \sqrt{(\sqrt{5} - 2)^2} = \sqrt{5} - 2 = VP$. (đpcm)