a: =(căn x-1)(căn x+1)/căn x-1
=căn x+1
b: =(căn x+2)/(căn x+2)^2
=1/căn x+2
c: =(căn x-3)(căn x+3)/(căn x+3)^2
=(căn x-3)/(căn x+3)
d: \(=\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\dfrac{1}{x+\sqrt{x}+1}\)
d) ĐK: \(x\ge0\)
\(\dfrac{\sqrt{x}-1}{x\sqrt{x}-1}=\dfrac{\sqrt{x}-1}{\left(\sqrt{x}\right)^3-1^3}=\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\dfrac{1}{x+\sqrt{x}+1}\)
`d,`Viowis `x>=0;x\ne1` ta có:
`...=(sqrtx-1)/((sqrtx-1)(x+sqrtx+1))=1/(x+sqrtx+1)`
Hằng đẳng thức bậc 3: `A^3-B^3=(A-B)(A^2+AB+B^2)` với `A=x\sqrtx=(sqrtx)^3;B=1`
