`1)`
`(x>0;x\ne1)`
`R=((1)/(\sqrt{x}+1)-(\sqrt{x})/(1-\sqrt{x})+(2)/(x-1)):((\sqrt{x}+1)/(\sqrt{x}-1)-(\sqrt{x}-1)/(\sqrt{x}+1))`
`=((1)/(\sqrt{x}+1)+(\sqrt{x})/(\sqrt{x}-1)+(2)/((\sqrt{x}-1).(\sqrt{x}+1))):((\sqrt{x}+1)/(\sqrt{x}-1)-(\sqrt{x}-1)/(\sqrt{x}+1))`
`=((\sqrt{x}-1)/((\sqrt{x}+1).(\sqrt{x}-1))+(\sqrt{x}.(\sqrt{x}+1))/((\sqrt{x}-1).(\sqrt{x}+1))+(2)/((\sqrt{x}-1).(\sqrt{x}+1))):(((\sqrt{x}+1)^2)/((\sqrt{x}-1).(\sqrt{x}+1))-((\sqrt{x}-1)^2)/((\sqrt{x}+1).(\sqrt{x}-1)))`
`=(\sqrt{x}-1+x+\sqrt{x}+2)/((\sqrt{x}-1).(\sqrt{x}+1)):(x+2\sqrt{x}+1-x+2\sqrt{x}-1)/((\sqrt{x}+1).(\sqrt{x}-1))`
`=(x+2\sqrt{x}+1)/((\sqrt{x}-1).(\sqrt{x}+1)).((\sqrt{x}+1).(\sqrt{x}-1))/(4\sqrt{x})`
`=((\sqrt{x}+1)^2)/((\sqrt{x}-1).(\sqrt{x}+1)).((\sqrt{x}+1).(\sqrt{x}-1))/(4\sqrt{x})`
`=((\sqrt{x}+1)^2)/(4\sqrt{x})`
Vì `x>0<=>\sqrt{x}>0<=>4\sqrt{x}>0`
Mà `(\sqrt{x}+1)^2>0∀x>0`
`=>((\sqrt{x}+1)^2)/(4\sqrt{x})>0`
Vậy `R>0` với `x>0;x\ne1`
`2)`
`R=(9)/(8)(x>0;x\ne1)`
`<=>((\sqrt{x}+1)^2)/(4\sqrt{x})=9/8`
`<=>((\sqrt{x}+1)^2)/(4\sqrt{x})-9/8=0`
`<=>(8.(\sqrt{x}+1)^2)/(8.4.\sqrt{x})-(9.4.\sqrt{x})/(8.4.\sqrt{x})=0`
`<=>8.(\sqrt{x}+1)^2-36\sqrt{x}=0`
`<=>8x+16\sqrt{x}+8-36\sqrt{x}=0`
`<=>8x-20\sqrt{x}+8=0`
`<=>4.(2x-5\sqrt{x}+2)=0`
`<=>4.(\sqrt{x}-2).(2\sqrt{x}-1)=0`
`<=>`\(\left[ \begin{array}{l}\sqrt{x}=2\\\sqrt{x}=\dfrac{1}{2}\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=4\\x=\dfrac{1}{4}\end{array} (tm)\right.\)
Vậy `x=4` hoặc `x=1/4` thì `R=9/8`
