Hướng dẫn trả lời:
Bài I:
1)
a) `A = sqrt{9a} - sqrt{16a} - sqrt{49a}` (`a ≥ 0`)
`= sqrt{3^2a} - sqrt{4^2a} - sqrt{7^2a}`
`= 3sqrt{a} - 4sqrt{a} - 7sqrt{a}`
`= - 8sqrt{a}`
b) `B = {1}/{2 - sqrt{3}} - {sqrt{6} - sqrt{2}}/{sqrt{3} - 1} - sqrt{5 - 2sqrt{6}}`
`= {2 + sqrt{3}}/{(2 + sqrt{3})cdot(2 - sqrt{3})} - {sqrt{2}cdot\sqrt{3} + sqrt{2}cdot(-1)}/{sqrt{3} - 1} - sqrt{3 - 2sqrt{6} + 2}`
`= {2 + sqrt{3}}/{2^2 - (sqrt{3})^2} - {sqrt{2}cdot(\sqrt{3} - 1)}/{sqrt{3} - 1} - sqrt{(sqrt{3})^2 - 2cdot\sqrt{3}cdot\sqrt{2} + (sqrt{2})^2}`
`= {2 + sqrt{3}}/{4 - 3} - sqrt{2} - sqrt{(sqrt{3} - sqrt{2})^2}`
`= {2 + sqrt{3}}/{1} - sqrt{2} - |sqrt{3} - sqrt{2}|`
`= (2 + sqrt{3}) - sqrt{2} - (sqrt{3} - sqrt{2})` (Vì `sqrt{3} > sqrt{2}`)
`= 2 + sqrt{3} - sqrt{2} - sqrt{3} + sqrt{2}`
`= 2 + (sqrt{3} - sqrt{3}) + (- sqrt{2} + sqrt{2})`
`= 2`
2) `4sqrt{2x + 3} - sqrt{8x + 12} + 1/3sqrt{18x + 27} = 15` (ĐK: `x ≥ - 3/2`)
`↔ 4sqrt{2x + 3} - sqrt{4cdot(2x + 3)} + 1/3sqrt{9cdot(2x + 3)} = 15`
`↔ 4sqrt{2x + 3} - sqrt{2^2cdot(2x + 3)} + 1/3sqrt{3^2cdot(2x + 3)} = 15`
`↔ 4sqrt{2x + 3} - 2sqrt{2x + 3} + 1/3cdot3sqrt{2x + 3} = 15`
`↔ 4sqrt{2x + 3} - 2sqrt{2x + 3} + sqrt{2x + 3} = 15`
`↔ 3sqrt{2x + 3} = 15`
`↔ sqrt{2x + 3} = 5`
`↔ [(5 > 0),((sqrt{2x + 3})^2 = 5^2):}`
`↔ 2x + 3 = 25`
`↔ 2x = 22`
`↔ x = 11` (TMĐK)
Vậy phương trình có tập nghiệm là `S = {11}`
Bài II:
1) `C = ({1}/{sqrt{a} - 1} - {1}/{sqrt{a}}) ÷ ({sqrt{a} + 1}/{sqrt{a} - 2} - {sqrt{a} + 2}/{sqrt{a} - 1})`
với `a > 0; a ne 1; a ne 4`
`= ({sqrt{a}}/{sqrt{a}cdot(sqrt{a} - 1)} - {sqrt{a} - 1}/{sqrt{a}cdot(sqrt{a} - 1)}) ÷ ({(sqrt{a} + 1)cdot(sqrt{a} - 1)}/{(sqrt{a} - 2)cdot(sqrt{a} - 1)} - {(sqrt{a} + 2)cdot(sqrt{a} - 2)}/{(sqrt{a} - 2)cdot(sqrt{a} - 1)})`
`= {sqrt{a} - (sqrt{a} - 1)}/{sqrt{a}cdot(sqrt{a} - 1)} ÷ {(sqrt{a} + 1)cdot(sqrt{a} - 1) - (sqrt{a} + 2)cdot(sqrt{a} - 2)}/{(sqrt{a} - 2)cdot(sqrt{a} - 1)}`
`= {sqrt{a} - sqrt{a} + 1}/{sqrt{a}cdot(sqrt{a} - 1)} ÷ {[(sqrt{a})^2 - 1^2] - [(sqrt{a})^2 - 2^2]}/{(sqrt{a} - 2)cdot(sqrt{a} - 1)}`
`= {1}/{sqrt{a}cdot(sqrt{a} - 1)} ÷ {(a - 1) - (a - 4)}/{(sqrt{a} - 2)cdot(sqrt{a} - 1)}`
`= {1}/{sqrt{a}cdot(sqrt{a} - 1)} ÷ {a - 1 - a + 4}/{(sqrt{a} - 2)cdot(sqrt{a} - 1)}`
`= {1}/{sqrt{a}cdot(sqrt{a} - 1)} ÷ {3}/{(sqrt{a} - 2)cdot(sqrt{a} - 1)}`
`= {1}/{sqrt{a}cdot(sqrt{a} - 1)}cdot{(sqrt{a} - 2)cdot(sqrt{a} - 1)}/{3}`
`= {sqrt{a} - 2}/{3sqrt{a}}`
2) Ta có:
`a = 4 + 2sqrt{3} = 3 + 2sqrt{3} + 1 = (sqrt{3})^2 + 2cdot\sqrt{3}cdot1 + 1^2 = (sqrt{3} + 1)^2`
Với `a = (sqrt{3} + 1)^2`, ta có: `C = {sqrt{(sqrt{3} + 1)^2} - 2}/{3sqrt{(sqrt{3} + 1)^2}}`
`= {|sqrt{3} + 1| - 2}/{3|sqrt{3} + 1|}`
`= {sqrt{3} + 1 - 2}/{3cdot(sqrt{3} + 1)}`
`= {sqrt{3} - 1}/{3cdot(sqrt{3} + 1)}`
`= {(sqrt{3} - 1)^2}/{3cdot(sqrt{3} + 1)cdot(sqrt{3} - 1)}`
`= {(sqrt{3})^2 - 2cdot\sqrt{3}cdot1 + 1^2}/{3cdot[(sqrt{3})^2 - 1^2]}`
`= {3 - 2sqrt{3} + 1}/{3cdot(3 - 1)}`
`= {4 - 2sqrt{3}}/{3cdot2}`
`= {2cdot(2 - sqrt{3})}/{3cdot2}`
`= {2 - sqrt{3}}/{3}`
3) Ta có: `a > 0` (gt) `→ sqrt{a} > 0` `→ 3sqrt{a} > 0`
`C ≤ 0`
`↔ {sqrt{a} - 2}/{3sqrt{a}} ≤ 0`
Mà `3sqrt{a} > 0` (cmt)
`→ sqrt{a} - 2 ≤ 0`
`↔ sqrt{a} ≤ 2`
`↔ (sqrt{a})^2 ≤ 2^2`
`↔ a ≤ 4`
Kết hợp với điều kiện đã cho của đề bài (`a > 0; a ne 1; a ne 4`)
`→ a ne 1; 0 < a < 4`
