`A=(1+(a+sqrta)/(sqrta+1))(1-(a-sqrta)/(sqrta-1))+(a-1)/(sqrta+1)`

a)Điều kiện:

`{(a>=0),(sqrta+1 ne 0(\text{luôn đúng})),(sqrta-1 ne 0):}`

`<=>{(a>=0),(sqrta ne 1):}`

`<=>{(a>=0),(a ne 1):}`

`b)A=(1+(sqrta(sqrta+1))/(sqrta+1))(1-(sqrta(sqrta-1))/(sqrta-1))+((sqrta-1)(sqrta+1))/(sqrta+1)`

`A=(1+sqrta)(1-sqrta)+sqrta-1`

`A=1-a+sqrta-1`

`A=-a+sqrta`

`c)a=4-2sqrt3`

`a=4-2sqrt3+1`

`a=(sqrt3-1)^2`

`=>sqrta=sqrt3-1`

`=>A=-(4-2sqrt3)+sqrt3-1`

`A=-4+2sqrt3+sqrt3-1`

`A=3sqrt3-5`

Vậy với `a=4-2sqrt3` thì `A=3sqrt3-5`

Đáp án:

Giải thích các bước giải:

 `A=(1+\frac{a+\sqrt{a}}{\sqrt{a}+1})(1-\frac{a-\sqrt{a}}{\sqrt{a}-1})+\frac{a-1}{\sqrt{a}+1}`

a) ĐK: \(\begin{cases} \sqrt{a}+1 \ne 0\\\sqrt{a}-1 \ne 0\\a \ge 0\end{cases}\)

`⇒` \(\begin{cases} a \ne 1\\a \ge 0\end{cases}\)

b) `A=[1+\frac{\sqrt{a}(\sqrt{a}+1)}{\sqrt{a}+1}][1-\frac{\sqrt{a}(\sqrt{a}-1)}{\sqrt{a}-1}]+\frac{(\sqrt{a}-1)(\sqrt{a}+1)}{\sqrt{a}+1}`

`A=(1+\sqrt{a})(1-\sqrt{a})+\sqrt{a}-1`

`A=1-a+\sqrt{a}-1`

`A=\sqrt{a}-a`

c) `a=4-2\sqrt{3}`

`a=3+1-2\sqrt{3}`

`a=(\sqrt{3}-1)^2`

`⇒ \sqrt{a}=\sqrt{(\sqrt{3}-1)^2}=|\sqrt{3}-1|=\sqrt{3}-1`

Thay vào BT A ta có:

`A=\sqrt{3}-1-(4-2\sqrt{3})`

`A=\sqrt{3}-1-4+2\sqrt{3}`

`A=3\sqrt{3}-5`

Vậy khi `a=4-2\sqrt{3}` thì `A=3\sqrt{3}-5`