a) (3x+5)²-4(x-3)²=0
⇔ (3x+5)²-(2x-6)²=0
⇔ (3x+5+2x-6)(3x+5-2x+6)=0
⇔ (x+11)(5x-1)=0
⇔\(\left[ \begin{array}{l}5x-1=0\\x+11=0\end{array} \right.\)\(\left[ \begin{array}{l}x=0,2\\x=-11\end{array} \right.\)
Vậy phương trình có tập nghiệm x∈{0,2 ;-11}
c)
(x-1)·(x-2)·(x-4)·(x-5)=40
⇔ [(x-1)(x-5)]·[(x-2)(x-4)]=40
⇔ (x²-6x+5)(x²-6x+8)=40
⇔(x²-6x+5)(x²-6x+5+3)=40
⇔ (x²-6x+5)²+3(x²-6x+5)-40=0
⇔ (x²-6x+5)²-5(x²-6x+5)+8(x²-6x+5)-40=0
⇔ (x²-6x)(x²-6x+13)=0
TH1: x²-6x=0
⇔ x(x-6)=0
x=0 hoặc x=6
TH2: x²-6x+13=0
⇔x²-6x+13=0
⇔x²-6x+9+4=0
⇔(x-3)²=-4
⇒ vô nghiệm
Vậy S={0;6}
d)
(x-2020) ³ + (x-2021) ³ = (2x-4041) ³
⇒(x-2020)³+ (x-2021)³=(2x-4041)³
⇔ (x-2020)³+(x-2021)³=(x-2020)³+3(x-2020)(x-2021)²+3(x-2020)²(x-2021)+ (x-2021)³
⇒ 3(x-2020)b²+3( x-2020)²(x-2021)=0
⇔ 3( x-2020)(x-2021)((2x-4041))=0
⇒ ( x-2020)=0 hoặc (x-2021)=0 hoặc (2x-4041)=0
⇒ x=2020 hoặc x=2021 hoặc x=2020,5
Vậy S={2020 ; 2021 ; 2020,5}
e)
x³+x²-4x-4=0
⇔ x²(x+1)-4(x+1)=0
⇔ (x+1)(x²-4)=0
⇔ (x+1)(x-2)(x+2)=0
⇒ x=-1 hoặc x=2 hoặc x=-2
Vậy S={-1;2;-2}
