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Đáp án + Giải thích các bước giải:

 `a)`

`x^3-25x=0`

`⇔ x(x^2-25)=0`

`⇔ x(x-5)(x+5)=0`

`⇔` $\left[\begin{matrix}x=0\\ x-5=0\\x+5=0\end{matrix}\right.$

`⇔` $\left[\begin{matrix}x=0\\ x=5\\x=-5\end{matrix}\right.$

Vậy ` x in {-5;0;5}`

`b)`

`x^4+4=5x^2`

`⇔ x^4 -5x^2 +4=0`

`⇔ x^4 -4x^2-x^2+4=0`

`⇔ x^2(x^2-4)-(x^2-4)=0`

`⇔ (x^2-4)(x^2-1)=0`

`⇔ (x-2)(x+2)(x-1)(x+1)=0`

`⇔` $\left[\begin{matrix} x-2=0\\ x+2=0\\x-1=0\\x+1=0 \end{matrix}\right.$

`⇔` $\left[\begin{matrix} x=2\\ x=-2\\x=1\\x=-1 \end{matrix}\right.$

Vậy ` x in {-2;-1;1;2}`

`c)`

`x^3+27 +(x+3)(x-9)=0`

`⇔ (x+3)(x^2 -6x+9)+(x+3)(x-9)=0`

`⇔ (x+3)(x^2-3x+9+x-9)=0`

`⇔ (x+3)(x^2 - 2x)=0`

`⇔ x(x+3)(x-2)=0`

`⇔` $\left[\begin{matrix} x=0\\ x+3=0\\ x-2=0\end{matrix}\right.$

`⇔` $\left[\begin{matrix} x=0\\ x=-3\\ x=2\end{matrix}\right.$

Vậy ` x in {-3;0;2}`

`d)`

`4(x-2)^2=25(1-2x)^2`

`⇔ 2^2(x-2)^2 = 5^2(1-2x)^2`

`⇔ [2(x-2)]^2=[5(1-2x)]^2`

`⇔ (2x-4)^2 = (5-10x)^2`

`⇔ (2x-4)^2-(5-10x)^2=0`

`⇔ (2x-4-5+10x)(2x-4+5-10x)=0`

`⇔ (12x-9)(-8x+1)=0`

`⇔` $\left[\begin{matrix} 12x-9=0\\-8x+1=0\end{matrix}\right.$

`⇔` $\left[\begin{matrix} 12x=9\\-8x=-1\end{matrix}\right.$

`⇔` $\left[\begin{matrix} x=\dfrac{3}{4}\\x=\dfrac{1}{8}\end{matrix}\right.$

Vậy ` x in {1/8;3/4}`.