BT1: Giải phương trình b, (2x-3)($x^{2}$+2) = 0 c, ($x^{2}$-16)(x+1) = 0 d, 3x(x-1)-2(1-x) = 0 e, $x^{2}$-4-(x+5)(2-x) = 0

`a)`

`(2x-3)(x^2+2)=0` (*)

Vì `x^2>=0AAx`

`=>x^2+2>=2`

Khi đó, (*) xảy ra `<=>2x-3=0`

`<=>2x=3<=>x=3/2`

Vậy `S={3/2}`

`b)`

`(x^2-16)(x+1)=0`

`<=>(x-4)(x+4)(x+1)=0`

`<=>`\(\left[ \begin{array}{l}x-4=0\\x+4=0\\x+1=0\end{array} \right.\) 

`<=>`\(\left[ \begin{array}{l}x=4\\x=-4\\x=-1\end{array} \right.\) 

Vậy `S={4;-4;-1}`

`c)`

`3x(x-1)-2(1-x) = 0`

`<=>3x(x-1)+2(x-1)=0`

`<=>(3x+2)(x-1)=0`

`<=>`\(\left[ \begin{array}{l}3x+2=0\\x-1=0\end{array} \right.\) 

`<=>`\(\left[ \begin{array}{l}x=-\dfrac{2}{3}\\x=1\end{array} \right.\) 

Vậy `S={1;-2/3}`

`d)`

`x^2-4-(x+5)(2-x)=0`

`<=>(x-2)(x+2)+(x+5)(x-2)=0`

`<=>(x-2)(x+2+x+5)=0`

`<=>(x-2)(2x+7)=0`

`<=>`\(\left[ \begin{array}{l}x-2=0\\2x+7=0\end{array} \right.\) 

`<=>`\(\left[ \begin{array}{l}x=2\\x=-\dfrac{7}{2}\end{array} \right.\) 

Vậy `S={2;-7/2}`