Đáp án+Giải thích các bước giải:
`a)Q=((x+3)/(x-9)+1/(sqrtx+3)):sqrtx/(sqrtx-3)`
Điều kiện:`{(x>=0),(x-9 ne 0),(sqrtx ne 0),(sqrtx-3 ne 0):}`
`<=>{(x>=0),(x ne 9),(x ne 0),(sqrtx ne 3):}`
`<=>x>0,x ne 9`
`Q=((x+3)/((sqrtx-3)(sqrtx+3))+(sqrtx-3)/((sqrtx-3)(sqrtx+3)))*(sqrtx-3)/sqrtx`
`Q=(x+3+sqrtx-3)/((sqrtx-3)(sqrtx+3))*(sqrtx-3)/sqrtx`
`Q=(x+sqrtx)/((sqrtx-3)(sqrtx+3))*(sqrtx-3)/sqrtx`
`Q=(sqrtx(sqrtx+1))/((sqrtx-3)(sqrtx+3))*(sqrtx-3)/sqrtx`
`C=(sqrtx+1)/(sqrtx+3)`
`b)x=sqrt{6+4sqrt2}-sqrt{3+2sqrt2}`
`x=sqrt{4+2.2sqrt2+2}-sqrt{2+2sqrt2+1}`
`x=sqrt{(2+sqrt2)^2}-sqrt{(sqrt2+1)^2}`
`x=2+sqrt2-sqrt2-1=1`
`=>sqrtx=1`
`=>C=(1+1)/(1+3)=2/4=1/2`
`c)1/Q=(sqrtx+3)/(sqrtx+1)`
`1/Q in ZZ`
`=>sqrtx+3 vdots sqrtx+1`
`=>2 vdots sqrtx+1`
`=>sqrtx+1 in Ư(2)={+-1,+-2}`
Mà `sqrtx+1>1`
`=>sqrtx+1=2`
`<=>sqrtx=1<=>x=1(tm)`
