Đáp án:

Giải thích các bước giải:

 b)

$\frac{2}{\sqrt{3 - 2\sqrt{2}}} - \frac{2}{\sqrt{3 + 2\sqrt{2}}}$

= $\frac{2}{\sqrt{2 - 2\sqrt{2} + 1}} - \frac{2}{\sqrt{2 + 2\sqrt{2} + 1}}$

= $\frac{2}{\sqrt{(\sqrt{2} - 1)²}} - \frac{2}{\sqrt{(\sqrt{2} + 1)²}}$

= $\frac{2}{\sqrt{2} - 1} - \frac{2}{\sqrt{2} + 1}$

= $\frac{2(\sqrt{2} +1 - \sqrt{2} + 1)}{(\sqrt{2} + 1)(\sqrt{2} - 1)}$ 

= $\frac{4}{2 - 1}$

= 4

c)

$(4 + \sqrt{15})(\sqrt{10} - \sqrt{6})\sqrt{4 - \sqrt{15}}$

= $(\sqrt{4 + \sqrt{15}})(\sqrt{10} - \sqrt{6})\sqrt{(4 - \sqrt{15})(4 + \sqrt{15})}$

= $(\sqrt{4 + \sqrt{15}})(\sqrt{10} - \sqrt{6})\sqrt{16 - 15}$

= $(\sqrt{4 + \sqrt{15}})(\sqrt{10} - \sqrt{6})$

= $\sqrt{2}(\sqrt{4 + \sqrt{15}})(\sqrt{5} - \sqrt{3})$

= $(\sqrt{8 + 2\sqrt{15}})(\sqrt{5} - \sqrt{3})$

= $(\sqrt{\sqrt{5}² + 2\sqrt{15} + \sqrt{3}²})(\sqrt{5} - \sqrt{3})$

= $\sqrt{(\sqrt{5} + \sqrt{3})²}(\sqrt{5} - \sqrt{3})$

= $(\sqrt{5} + \sqrt{3})(\sqrt{5} - \sqrt{3})$

= 5 - 3 = 2

d)

Đặt A = $\sqrt{4 + \sqrt{7}} - \sqrt{4 - \sqrt{7}} - \sqrt{2}$

⇔$\sqrt{2}A = \sqrt{8 + 2\sqrt{7}} - \sqrt{8 - 2\sqrt{7}} - 2$

= $\sqrt{(\sqrt{7} + 1)²} - \sqrt{(\sqrt{7} - 1)²} - 2$

= $\sqrt{7} + 1 - \sqrt{7} + 1 - 24$

= 0

⇔ A = 0

Hướng dẫn trả lời:

3)

a) `sqrt{2 - sqrt{3}} - sqrt{2 + sqrt{3}}`

Đặt `A = sqrt{2 - sqrt{3}} - sqrt{2 + sqrt{3}}`

`→ sqrt{2}A = sqrt{2}cdot(sqrt{2 - sqrt{3}} - sqrt{2 + sqrt{3}})`

`= sqrt{2}cdot sqrt{2 - sqrt{3}} - sqrt{2}cdot sqrt{2 + sqrt{3}}`

`= sqrt{2cdot(2 - sqrt{3})} - sqrt{2cdot(2 + sqrt{3})}`

`= sqrt{4 - 2sqrt{3}} - sqrt{4 + 2sqrt{3}}`

`= sqrt{3 - 2sqrt{3} + 1} - sqrt{3 + 2sqrt{3} + 1}`

`= sqrt{(sqrt{3})^2 - 2cdot sqrt{3}cdot1 + 1^2} - sqrt{(sqrt{3})^2 + 2cdot sqrt{3}cdot1 + 1^2}`

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

`= |sqrt{3} - 1| - |sqrt{3} + 1|`

`= (sqrt{3} - 1) - (sqrt{3} + 1)` (Vì `sqrt{3} > 1`)

`= sqrt{3} - 1 - sqrt{3} - 1`

`= (sqrt{3} - sqrt{3}) + (- 1 - 1)`

`= - 2`

`→ A = - 2 ÷ sqrt{2} = - sqrt{2}`

b) `{2}/{sqrt{3 - 2sqrt{2}}} - {2}/{sqrt{3 + 2sqrt{2}}}`

`= {2}/{sqrt{2 - 2sqrt{2} + 1}} - {2}/{sqrt{2 + 2sqrt{2} + 1}}`

`= {2}/{sqrt{(sqrt{2})^2 - 2cdot sqrt{2}cdot1 + 1^2}} - {2}/{sqrt{(sqrt{2})^2 + 2cdot sqrt{2}cdot1 + 1^2}}`

`= {2}/{sqrt{(sqrt{2} - 1)^2}} - {2}/{sqrt{(sqrt{2} + 1)^2}}`

`= {2}/{|sqrt{2} - 1|} - {2}/{|sqrt{2} + 1|}`

`= {2}/{sqrt{2} - 1} - {2}/{sqrt{2} + 1}` (Vì `sqrt{2} > 1`)

`= {2cdot(sqrt{2} + 1)}/{(sqrt{2} + 1)cdot(sqrt{2} - 1)} - {2cdot(sqrt{2} - 1)}/{(sqrt{2} + 1)cdot(sqrt{2} - 1)}`

`= {2cdot(sqrt{2} + 1) - 2cdot(sqrt{2} - 1)}/{(sqrt{2} + 1)cdot(sqrt{2} - 1)}`

`= {2sqrt{2} + 2 - 2sqrt{2} + 2}/{(sqrt{2})^2 - 1^2}`

`= {(2sqrt{2} - 2sqrt{2}) + (2 + 2)}/{2 - 1}`

`= {4}/{1}`

`= 4`

c) `(4 + sqrt{15})cdot(sqrt{10} - sqrt{6})cdot sqrt{4 - sqrt{15}}`

`= (4 + sqrt{15})cdot(sqrt{2}cdot sqrt{5} - sqrt{2}cdot sqrt{3})cdot sqrt{4 - sqrt{15}}`

`= (4 + sqrt{15})cdot sqrt{2}cdot(sqrt{5} - sqrt{3})cdot sqrt{4 - sqrt{15}}`

`= (4 + sqrt{15})cdot(sqrt{5} - sqrt{3})cdot[sqrt{2}cdot sqrt{4 - sqrt{15}}]`

`= (4 + sqrt{15})cdot(sqrt{5} - sqrt{3})cdot sqrt{2cdot(4 - sqrt{15})}`

`= (4 + sqrt{15})cdot(sqrt{5} - sqrt{3})cdot sqrt{8 - 2sqrt{15}}`

`= (4 + sqrt{15})cdot(sqrt{5} - sqrt{3})cdot sqrt{5 - 2sqrt{15} + 3}`

`= (4 + sqrt{15})cdot(sqrt{5} - sqrt{3})cdot sqrt{(sqrt{5})^2 - 2cdot sqrt{5}cdot sqrt{3} + (sqrt{3})^2}`

`= (4 + sqrt{15})cdot(sqrt{5} - sqrt{3})cdot sqrt{(sqrt{5} - sqrt{3})^2}`

`= (4 + sqrt{15})cdot(sqrt{5} - sqrt{3})cdot|sqrt{5} - sqrt{3}|`

`= (4 + sqrt{15})cdot(sqrt{5} - sqrt{3})cdot(sqrt{5} - sqrt{3})` (Vì `sqrt{5} > sqrt{3}`)

`= (4 + sqrt{15})cdot(sqrt{5} - sqrt{3})^2`

`= (4 + sqrt{15})cdot[(sqrt{5})^2 - 2cdot sqrt{5}cdot sqrt{3} + (sqrt{3})^2]`

`= (4 + sqrt{15})cdot(8 - 2sqrt{15})`

`= (4 + sqrt{15})cdot2cdot(4 - sqrt{15})`

`= 2cdot[(4 + sqrt{15})cdot(4 - sqrt{15})]`

`= 2cdot[4^2 - (sqrt{15})^2]`

`= 2cdot(16 - 15)`

`= 2cdot1`

`= 2`

d) `sqrt{4 + sqrt{7}} - sqrt{4 - sqrt{7}} - sqrt{2}`

Đặt `D = sqrt{4 + sqrt{7}} - sqrt{4 - sqrt{7}} - sqrt{2}`

`↔ sqrt{2}D = sqrt{2}cdot(sqrt{4 + sqrt{7}} - sqrt{4 - sqrt{7}}) -  (sqrt{2})^2`

`= sqrt{2}cdot sqrt{4 + sqrt{7}} - sqrt{2}cdot sqrt{4 - sqrt{7}} - 2`

`= sqrt{2cdot(4 + sqrt{7})} - sqrt{2cdot(4 - sqrt{7})} - 2`

`= sqrt{8 + 2sqrt{7}} - sqrt{8 - 2sqrt{7}} - 2`

`= sqrt{7 + 2sqrt{7} + 1} - sqrt{7 - 2sqrt{7} + 1} - 2`

`= sqrt{(sqrt{7})^2 + 2cdot sqrt{7}cdot1 + 1^2} - sqrt{(sqrt{7})^2 - 2cdot sqrt{7}cdot1 + 1^2} - 2`

`= sqrt{(sqrt{7} + 1)^2} - sqrt{(sqrt{7} - 1)^2} - 2`

`= |sqrt{7} + 1| - |sqrt{7} - 1| - 2`

`= (sqrt{7} + 1) - (sqrt{7} - 1) - 2` (Vì `sqrt{7} > 1`)

`= sqrt{7} + 1 - sqrt{7} + 1 - 2`

`= (sqrt{7} - sqrt{7}) + (1 + 1 - 2)`

`= 0`

`→ D = 0`