Giải thích các bước giải:
`1, \ x^4+4x^3-16x-16=0`
`<=> (x^4-16)+(4x^3-16x)=0`
`<=> (x^2-4)(x^2+4)+4x(x^2-4)=0`
`<=> (x^2-4)(x^2+4x+4)=0`
`<=> (x-2)(x+2)(x+2)^2=0`
`<=> (x-2)(x+2)^3=0`
`<=> [(x-2=0),(x+2=0):} <=> [(x=2),(x=-2):}`
Vậy `x in {-2;2}`
`2, \ (2x-3)^2-(x+5)^2=0`
`<=> (2x-3-x-5)(2x-3+x+5)=0`
`<=> (x-8)(3x+2)=0`
`<=> [(x-8=0),(3x+2=0):} <=>` \(\left[ \begin{array}{l}x=8\\x=-\dfrac{2}{3}\end{array} \right.\)
Vậy `x in {8;-2/3}`
`3, \ x^2-2x+y^2+4y+5+(2z-3)^2=0`
`<=> (x^2-2x+1)+(y^2+4y+4)+(2z-3)^2=0`
`<=> (x-1)^2+(y-2)^2+(2z-3)^2=0`
Vì : `{((x-1)^2 ge0),((y-2)^2 ge0),((2z-3)^2 ge0):} to (x-1)^2+(y-2)^2+(2z-3)^2 ge0`
Dấu "=" xảy ra khi : `{(x-1=0),(y-2=0),(2z-3=0):} <=>` $\begin{cases}x=1\\y=2\\z=\dfrac{3}{2}\end{cases}$
Vậy `(x;y;z)=(1;2;3/2)`
