Hướng dẫn trả lời:
Câu 15:
a) `3sqrt2 + sqrt50 - sqrt8`
`= 3sqrt2 + sqrt{5^2cdot2} - sqrt{2^2cdot2}`
`= 3sqrt2 + 5sqrt{2} - 2sqrt{2}`
`= 6sqrt2`
b) `{x + sqrtx}/{sqrtx} + {x - 4}/{sqrtx + 2}` `(x > 0)`
`= {sqrtxcdot(sqrtx + 1)}/{sqrtx} + {(sqrtx)^2 - 2^2}/{sqrtx + 2}`
`= (sqrtx + 1) + {(sqrtx + 2)cdot(sqrtx - 2)}/{sqrtx + 2}`
`= (sqrtx + 1) + (sqrtx - 2)`
`= sqrtx + 1 + sqrtx - 2`
`= 2sqrtx - 1`
Câu 16:
a) `A = (sqrt27 + 3sqrt12 - 2sqrt3) ÷ sqrt3`
`= (sqrt{3^2cdot3} + 3sqrt{2^2cdot3} - 2sqrt3) ÷ sqrt3`
`= (3sqrt{3} + 3cdot2sqrt{3} - 2sqrt3) ÷ sqrt3`
`= (3sqrt{3} + 6sqrt{3} - 2sqrt3) ÷ sqrt3`
`= 7sqrt3 ÷ sqrt3`
`= 7`
b) `B = ({1}/{sqrtx + 3} + {5}/{sqrtx - 3} + {6}/{x - 9}) ÷ {2}/{sqrtx - 3}` `(x ≥ 0; x ne 9)`
`= ({1cdot(sqrtx - 3)}/{x - 9} + {5cdot(sqrtx + 3)}/{x - 9} + {6}/{x - 9}) ÷ {2}/{sqrtx - 3}`
`= {1cdot(sqrtx - 3) + 5cdot(sqrtx + 3) + 6}/{x - 9} ÷ {2}/{sqrtx - 3}`
`= {sqrtx - 3 + 5sqrtx + 15 + 6}/{x - 9} ÷ {2}/{sqrtx - 3}`
`= {(sqrtx + 5sqrtx) + (- 3 + 15 + 6)}/{x - 9} ÷ {2}/{sqrtx - 3}`
`= {6sqrtx + 18}/{(sqrtx)^2 - 3^2} ÷ {2}/{sqrtx - 3}`
`= {6cdot(sqrtx + 3)}/{(sqrtx + 3)cdot(sqrtx - 3)} ÷ {2}/{sqrtx - 3}`
`= {6}/{sqrtx - 3} ÷ {2}/{sqrtx - 3}`
`= {6}/{sqrtx - 3} cdot {sqrtx - 3}/{2}`
`= 3`
Câu 17:
a) `A = sqrt{(sqrt7 - 3)^2} - sqrt{16 + 6sqrt7}`
`= |sqrt7 - 3| - sqrt{7 + 6sqrt7 + 9}`
`= (3 - sqrt7) - sqrt{3^2 + 2cdot3cdot sqrt7 + (sqrt7)^2}` (Vì `3 = sqrt9 > sqrt7 → 3 - sqrt7 > 0`)
`= (3 - sqrt7) - sqrt{(3 + sqrt7)^2}`
`= (3 - sqrt7) - |3 + sqrt7|`
`= (3 - sqrt7) - (3 + sqrt7)`
`= - 2sqrt7`
b) `B = {x + sqrtx}/{1 - x} + {(sqrtx - 2)^2 - sqrtx - x}/{1 - sqrtx}` `(x ≥ 0; x ne 1)`
`= {sqrtxcdot(sqrtx + 1)}/{1^2 - (sqrtx)^2} + {(sqrtx)^2 - 2cdot sqrtxcdot2 + 2^2 - sqrtx - x}/{1 - sqrtx}`
`= {sqrtxcdot(sqrtx + 1)}/{(1 - sqrtx)cdot(1 + sqrtx)} + {x - 4sqrtx + 4 - sqrtx - x}/{1 - sqrtx}`
`= {sqrtxcdot(sqrtx + 1)}/{(1 - sqrtx)cdot(sqrtx + 1)} + {x - 4sqrtx + 4 - sqrtx - x}/{1 - sqrtx}`
`= {sqrtx}/{1 - sqrtx} + {(x - x) + (- 4sqrtx - sqrtx) + 4}/{1 - sqrtx}`
`= {sqrtx}/{1 - sqrtx} + {4 - 5sqrtx}/{1 - sqrtx}`
`= {sqrtx + (4 - 5sqrtx)}/{1 - sqrtx}`
`= {sqrtx + 4 - 5sqrtx}/{1 - sqrtx}`
`= {4 - 4sqrtx}/{1 - sqrtx}`
`= {4cdot(1 - sqrtx)}/{1 - sqrtx}`
`= 4`
