Đáp án:
`a) 1/(3+√2)+1/(3-√2)`
`= (3-√2)/[(3+√2)(3-√2)]+(3+√2)/[(3-√2)(3+√2)]`
`= [(3-√2)+(3+√2)]/[(3-√2)(3+√2)]`
`= (3-√2+3+√2)/[(3-√2)(3+√2)]`
`= 9/(9-3)`
`= 9/6`
`= 3/2`
`b) 2/(3√2-4)-2/(3√2+4)`
`= [2(3√2+4)]/[(3√2-4)(3√2+4)]-[2/(3√2-4)]/[(3√2-4)(3√2+4)]`
`= [2(3√2+4)-2/(3√2-4)]/[(3√2-4)(3√2+4)]`
`= (6√2+8-6√2+8)/[(3√2-4)(3√2+4)]`
`= 16/(18-16)`
`= 16/2`
`= 8`
`c) (√5-√3)/(√5+√3)+(√5+√3)/(√5-√3)`
`= (√5-√3)^2/[(√5-√3)(√5+√3)]+(√5+√3)^2/[(√5-√3)(√5+√3)]`
`= [(√5-√3)^2+(√5+√3)^2]/[(√5-√3)(√5+√3)]`
`= (5-2√15+3+5+2√15+3)/(5-3)`
`= 16/2`
`= 8`
`d) 3/(2√2-3√3)-3/(2√2+3√3)`
`= [3(2√2+3√3)]/[(2√2-3√3)(2√2+3√3)]-[3(2√2-3√3)]/[(2√2-3√3)(2√2+3√3)]`
`= [3(2√2+3√3)-3(2√2-3√3)]/[(2√2-3√3)(2√2+3√3)]`
`= (6√2+9√3-6√2+9√3)/(8-27)`
`= (18√3)/(-19)`