`b)B=1/(sqrt3-sqrt2)-(sqrt{15}-sqrt{12})/(sqrt5-2)`
`B=(sqrt3+sqrt2)/(3-2)-(sqrt3(sqrt5-2))/(sqrt5-2)`
`B=sqrt3+sqrt2-sqrt3`
`B=sqrt2`
`c)C=((sqrta-2)/(sqrta+2)-(sqrta+2)/(sqrta-2))(sqrta-4/sqrta)(a>0,a ne 4)`
`C=(((sqrta-2)^2-(sqrta+2)^2)/((sqrta-2)(sqrta+2)))*(a-4)/sqrta`
`C=(a-4sqrta+4-a-4sqrta-4)/(a-4)*(a-4)/sqrta`
`C=(-8sqrta)/sqrta`
`C=-8`
`2a)4sqrt{x-1}-sqrt{4x-4}+sqrt{9x-9}=15(x>=1)`
`<=>4\sqrt{x-1}-2sqrt{x-1}+3sqrt{x-1}=15`
`<=>5\sqrt{x-1}=15`
`<=>\sqrt{x-1}=3`
`<=>x-1=9`
`<=>x=10(tmđk)`
Vậy `S={10}`
`b)sqrt{x^2-x+1/4}+x=3/2`
`<=>sqrt{(x-1/2)^2}=3/2-x`
`<=>|x-1/2|=3/2-x`
Vì `|x-1/2|>=0`
`=>3/2-x>=0<=>x<=3/2`
`=>` \(\left[ \begin{array}{l}x-\dfrac12=x-\dfrac32\\x-\dfrac12=\dfrac32-x\end{array} \right.\)
`=>` \(\left[ \begin{array}{l}-\dfrac12=-\dfrac32(\text{vô lý})\\2x=2\end{array} \right.\)
`<=>x=1(tmđk)`
Vậy `S={1}.`
