Hướng dẫn trả lời:
Bài 1:
a) `B = {2 + sqrt{x}}/{sqrt{x} + 1} + {2 - sqrt{x}}/{sqrt{x} - 1}` (Đk: `x ≥ 0; x ne 1`)
`= {(2 + sqrt{x})cdot(sqrt{x} - 1)}/{(sqrt{x} + 1)cdot(sqrt{x} - 1)} + {(sqrt{x} + 1)cdot(2 - sqrt{x})}/{(sqrt{x} + 1)cdot(sqrt{x} - 1)}`
`= {(2 + sqrt{x})cdot(sqrt{x} - 1) + (sqrt{x} + 1)cdot(2 - sqrt{x})}/{(sqrt{x} + 1)cdot(sqrt{x} - 1)}`
`= {2cdot(sqrt{x} - 1) + sqrt{x}cdot(sqrt{x} - 1) + sqrt{x}cdot(2 - sqrt{x}) + 1cdot(2 - sqrt{x})}/{(sqrt{x})^2 - 1^2}`
`= {2sqrt{x} - 2 + x - sqrt{x} + 2sqrt{x} - x + 2 - sqrt{x}}/{x - 1}`
`= {(x - x) + (2sqrt{x} - sqrt{x} + 2sqrt{x} - sqrt{x}) + (- 2 + 2)}/{x - 1}`
`= {2sqrt{x}}/{x - 1}`
b) `x = sqrt{17 + 12sqrt{2}} = sqrt{9 + 12sqrt{2} + 8} = sqrt{3^2 + 2cdot3cdot2sqrt{2} + (2sqrt{2})^2`
`= sqrt{(3 + 2sqrt{2})^2} = |3 + 2sqrt{2}| = 3 + 2sqrt{2}`
Với `x = 3 + 2sqrt{2}`, ta có:
`B = {2sqrt{3 + 2sqrt{2}}}/{(3 + 2sqrt{2}) - 1}`
`= {2sqrt{2 + 2sqrt{2} + 1}}/{3 + 2sqrt{2} - 1}`
`= {2sqrt{(sqrt{2})^2 + 2cdot sqrt{2}cdot1 + 1^2}}/{2sqrt{2} + (3 - 1)}`
`= {2sqrt{(sqrt{2} + 1)^2}}/{2sqrt{2} + 2}`
`= {2|sqrt{2} + 1|}/{2cdot(sqrt{2} + 1)}`
`= {2cdot(sqrt{2} + 1)}/{2cdot(sqrt{2} + 1)}`
`= 1`
Vậy giá trị của biểu thức `B` khi `x = sqrt{17 + 12sqrt{2}}` là `1`
