`a)`

        `x^2-16=0`

`⇔x^2=16`

`⇔x^2=(±4)^2`

`⇔x=±4`

Vậy `x=±4`

`b)`

        `x^2=13`

`⇔x^2=(±\sqrt{13})^2`

`⇔x=±\sqrt{13}`

Vậy `x=±\sqrt{13}`

`c)`

        `x^2+9=0`

`⇔x^2=-9` `(` Vô lí `)`

Vậy không có giá trị nào của `x` thõa mãn

`d)`

        `\sqrt{x}=5`  `(` `ĐK: x≥0` `)`

`⇔(\sqrt{x})^2=5^2`

`⇔x=25` `(tm)`

Vậy `x=25`

`e)`  

        `-(\sqrt{x})/3+2=0`  `(` `ĐK: x≥0` `)`

`⇔(\sqrt{x})/3=2`

`⇔\sqrt{x}=6`

`⇔(\sqrt{x})^2=6^2`

`⇔x=36` `(tm)`

Vậy `x=36`

`f)`

        `\sqrt{x^2-2x+1}=4`

`⇔\sqrt{(x-1)^2}=4`

`⇔|x-1|=4`

`⇔[(x-1=4),(x-1=-4):}`

`⇔[(x=5),(x=-3):}`

Vậy `x∈{5;-3}`

Hoặc:

        `\sqrt{x^2}-2x+1=4`

`⇔|x|=2x+3`  `(` `ĐK:`  `x≥-3/2` `)`

`⇔ [(x=2x+3),(x=-2x-3):}`

`⇔ [(\text{x=-3 (ktm) } ),(\text{x=-1 (tm) } ):}`

Vậy `x=-1`

Hoặc:

        `\sqrt{x^2-2x}+1=4`  `(` `ĐK: x≥2` hoặc `x≤0` `)`

`⇔\sqrt{x^2-2x}=3`

`⇔(\sqrt{x^2-2x})^2=3^2`

`⇔x^2-2x=9`

`⇔x^2-2x+1=10`

`⇔(x-1)^2=(±\sqrt{10})^2`

`⇔[(x-1=\sqrt{10}),(x-1=-\sqrt{10}):}`

`⇔ [( x= 1+\sqrt{10} \text{  (tm) } ),(x=1-\sqrt{10} \text{ (tm) } ):}`

Vậy `x=1±\sqrt{10}`

`#` `ccUh`