Hướng dẫn trả lời:
Câu 12:
a) `A = sqrt28 + sqrt63 - 2sqrt7`
`= sqrt{2^2cdot7} + sqrt{3^2cdot7} - 2sqrt7`
`= 2sqrt{7} + 3sqrt{7} - 2sqrt7`
`= (2sqrt{7} - 2sqrt7) + 3sqrt{7}`
`= 3sqrt7`
b) Ta có: `{xsqrty + ysqrtx}/{sqrt{xy}} ÷ 1/{sqrtx - sqrty` `(x > 0, y > 0, x ne y)`
`= {sqrt{xy}cdot(sqrtx + sqrty)}/{sqrt{xy}} ÷ 1/{sqrtx - sqrty`
`= (sqrtx + sqrty)cdot{sqrtx - sqrty}/1`
`= (sqrtx + sqrty)cdot(sqrtx - sqrty)`
`= (sqrtx)^2 - (sqrty)^2`
`= x - y`
`→ đpcm`
Câu 13:
a) `A = 3sqrt2 - sqrt32 + sqrt50`
`= 3sqrt2 - sqrt{4^2cdot2} + sqrt{5^2cdot2}`
`= 3sqrt2 - 4sqrt{2} + 5sqrt{2}`
`= 4sqrt2`
b) `B = (1/{sqrtx - 2} - {sqrtx}/{x - 4}) ÷ 1/{sqrtx + 2}` `(x ≥ 0; x ne 4)`
`= ({sqrtx + 2}/{x - 4} - {sqrtx}/{x - 4}) ÷ 1/{sqrtx + 2}`
`= {(sqrtx + 2) - sqrtx}/{x - 4} ÷ 1/{sqrtx + 2}`
`= {sqrtx + 2 - sqrtx}/{x - 4} ÷ 1/{sqrtx + 2}`
`= {(sqrtx - sqrtx) + 2}/{x - 4} ÷ 1/{sqrtx + 2}`
`= 2/{x - 4} ÷ 1/{sqrtx + 2}`
`= 2/{(sqrtx + 2)cdot(sqrtx - 2)}cdot{sqrtx + 2}/1`
`= 2/{sqrtx - 2}`
Câu 14:
a) `A = sqrt{9cdot32} - sqrt2`
`= sqrt{3^2cdot4^2cdot2} - sqrt2`
`= 3cdot4sqrt{2} - sqrt2`
`= 12sqrt{2} - sqrt2`
`= 11sqrt2`
b) `B = {x - 5}/{sqrtx + sqrt5}` `(x ≥ 0)`
`= {(sqrtx)^2 - (sqrt5)^2}/{sqrtx + sqrt5}`
`= {(sqrtx + sqrt5)cdot(sqrtx - sqrt5)}/{sqrtx + sqrt5}`
`= sqrtx - sqrt5`
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`
`= (sqrtx + sqrtx) + (1 - 2)`
`= 2sqrtx - 1`
