Đáp án:
`a,G={2x+1}/{4\sqrt{x}}`
`b,x=1/4`
`c,`$Min$`G={\sqrt{2}}/{2}` khi `x=1/2`
Giải thích các bước giải:
`ĐKXĐ:x>0,x\ne1`
`a,G=({\sqrt{x}+1}/{\sqrt{x}-1}+{\sqrt{x}}/{\sqrt{x}+1}+{\sqrt{x}}/{1-x}):({\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}}/{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}+1}/{\sqrt{x}-1}+{\sqrt{x}}/{\sqrt{x}+1}-{\sqrt{x}}/{(\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}+1)}/{(\sqrt{x}+1)(\sqrt{x}-1)}+{\sqrt{x}(\sqrt{x}-1)}/{(\sqrt{x}+1)(\sqrt{x}-1)}-{\sqrt{x}}/{(\sqrt{x}+1)(\sqrt{x}-1)}]:{(\sqrt{x}+1)^2-(\sqrt{x}-1)^2}/{(\sqrt{x}+1)(\sqrt{x}-1)}`
`={(\sqrt{x}+1)(\sqrt{x}+1)+\sqrt{x}(\sqrt{x}-1)-\sqrt{x}}/{(\sqrt{x}+1)(\sqrt{x}-1)}:{x+2\sqrt{x}+1-(x-2\sqrt{x}+1)}/{(\sqrt{x}+1)(\sqrt{x}-1)}`
`={(\sqrt{x}+1)^2+x-\sqrt{x}-\sqrt{x}}/{(\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+x-\sqrt{x}-\sqrt{x}}/{(\sqrt{x}+1)(\sqrt{x}-1)}:{4\sqrt{x}}/{(\sqrt{x}+1)(\sqrt{x}-1)}`
`={2x+1}/{(\sqrt{x}+1)(\sqrt{x}-1)}.{(\sqrt{x}+1)(\sqrt{x}-1)}/{4\sqrt{x}}`
`={2x+1}/{4\sqrt{x}}`
Vậy với `x>0,x\ne1` thì `G={2x+1}/{4\sqrt{x}}`
`b,G=3/4`
`⇔{2x+1}/{4\sqrt{x}}=3/4`
`⇔{2x+1}/{4\sqrt{x}}={3\sqrt{x}}/{4\sqrt{x}}`
`⇒2x+1=3\sqrt{x}`
`⇔2x-3\sqrt{x}+1=0`
`⇔2x-2\sqrt{x}-\sqrt{x}+1=0`
`⇔2\sqrt{x}(\sqrt{x}-1)-(\sqrt{x}-1)=0`
`⇔(\sqrt{x}-1)(2\sqrt{x}-1)=0`
⇔\(\left[ \begin{array}{l}\sqrt{x}-1=0\\2\sqrt{x}-1=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}\sqrt{x}=1\\2\sqrt{x}=1\end{array} \right.\)
⇔\(\left[ \begin{array}{l}\sqrt{x}=1\\\sqrt{x}=\frac{1}{2}\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=1(KTM)\\x=\frac{1}{4}(TM)\end{array} \right.\)
Vậy với `x=1/4` thì `G=3/4`
`c,G={2x+1}/{4\sqrt{x}}`
`={2x}/{4\sqrt{x}}+{1}/{4\sqrt{x}}`
`={\sqrt{x}}/{2}+{1}/{4\sqrt{x}}`
Áp dụng bất đẳng thức Côsi cho hai số thực dương `{\sqrt{x}}/{2}` và `{1}/{4\sqrt{x}` (vì `x>0`) có:
`{\sqrt{x}}/{2}+{1}/{4\sqrt{x}}\ge2.\sqrt{{\sqrt{x}}/{2}.{1}/{4\sqrt{x}}}=2.\sqrt{{1}/{8}}={\sqrt{2}}/{2}`
`⇒G\ge{\sqrt{2}}/{2}`
`⇒`$Min$`G={\sqrt{2}}/{2}`. Dấu "=" xảy ra
`⇔{\sqrt{x}}/{2}={1}/{4\sqrt{x}}`
`⇔{2x}/{4\sqrt{x}}={1}/{4\sqrt{x}}`
`⇒2x=1`
`⇔x=1/2`
Vậy $Min$`G={\sqrt{2}}/{2}` khi `x=1/2`
