Đáp án:
`f(a)=-2-4\sqrt{3}` với `a=\sqrt{3}/2`
Giải thích các bước giải:
Ta có: `a=\sqrt{3}/2`
`=>1+a=1+\sqrt{3}/2={2+\sqrt{3}}/2`
`={4+2\sqrt{3}}/4`
`={(\sqrt{3})^2+2.\sqrt{3}.1+1^2}/4`
`={(\sqrt{3}+1)^2}/{2^2}`
`=>\sqrt{1+a}=|{\sqrt{3}+1}/2|={\sqrt{3}+1}/2`
$\\$
`\qquad 1-a=1-\sqrt{3}/2={2-\sqrt{3}}/2`
`={4-2\sqrt{3}}/4`
`={(\sqrt{3})^2-2.\sqrt{3}.1+1^2}/4`
`={(\sqrt{3}-1)^2}/{2^2}`
`=>\sqrt{1-a}=|{\sqrt{3}-1}/2|={\sqrt{3}-1}/2`
$\\$
`\qquad f(x)={1+\sqrt{1+x}}/{x+1}+{1+\sqrt{1-x}}/{x-1}` `(x\ne ±1; -1<x<1)`
`a=\sqrt{3}/2` thỏa điều kiện
`=>f(a)={1+\sqrt{1+a}}/{a+1}+{1+\sqrt{1-a}}/{a-1}`
$=\dfrac{1+\dfrac{\sqrt{3}+1}{2}}{\dfrac{\sqrt{3}}{2}+1}+\dfrac{1+\dfrac{\sqrt{3}-1}{2}}{\dfrac{\sqrt{3}}{2}-1}$
$=\dfrac{\dfrac{2+\sqrt{3}+1}{2}}{\dfrac{\sqrt{3}+2}{2}}+\dfrac{\dfrac{2+\sqrt{3}-1}{2}}{\dfrac{\sqrt{3}-2}{2}}$
`={3+\sqrt{3}}/2 . 2/{\sqrt{3}+2}+{\sqrt{3}+1}/{2}. 2/{\sqrt{3}-2}`
`={\sqrt{3}.(\sqrt{3}+1)}/{\sqrt{3}+2} +{\sqrt{3}+1}/{\sqrt{3}-2}`
`=(\sqrt{3}+1). (\sqrt{3}/{\sqrt{3}+2}+1/{\sqrt{3}-2})`
`=(\sqrt{3}+1). {\sqrt{3}.(\sqrt{3}-2)+\sqrt{3}+2}/{(\sqrt{3}+2)(\sqrt{3}-2)}`
`=(\sqrt{3}+1).{3-2\sqrt{3}+\sqrt{3}+2}/{3-4}`
`=(\sqrt{3}+1).[-(5-\sqrt{3})]`
`=(\sqrt{3}+1).(\sqrt{3}-5)`
`=3-5\sqrt{3}+\sqrt{3}-5=-2-4\sqrt{3}`
Vậy `f(a)=-2-4\sqrt{3}` với `a=\sqrt{3}/2`
