$(2020 – x)^3 + (2021 – x)^3 + (2x – 4041)^3$ = 0

0 bình luận về “$(2020 – x)^3 + (2021 – x)^3 + (2x – 4041)^3$ = 0”

1. `(2020-x)^3+(2021-x)^3+(2x-4041)^3=0`

   `⇔(2020-x)^3+(2021-x)^3-(2020-x+2021-x)^3=0`

   `⇔(2020-x)^3+(2021-x)^3-[(2020-x)+(2021-x)]^3=0`

   `⇔(2020-x)^3+(2021-x)^3-[(2020-x)^3+3(2020-x)^2(2021-x)+3(2020-x)(2021-x)^2+(2021-x)^3=0`

   `⇔(2020-x)^3+(2021-x)^3-(2020-x)^3-3(2020-x)^2(2021-x)-3(2020-x)(2021-x)^2-(2021-x)^3]=0`

   `⇔-3(2020-x)^2(2021-x)-3(2020-x)(2021-x)^2=0`

   `⇔-3(2020-x)(2021-x)(2020-x+2021-x)=0` 

   `⇔3(2020-x)(2021-x)(-2020+x-2021+x)=0`

   `⇔3(2020-x)(2021-x)(2x-4041)=0`

   `⇔` \(\left[ \begin{array}{l}2020-x=0\\2021-x=0\\2x-4041=0\end{array} \right.\) 

   `⇔` \(\left[ \begin{array}{l}x=2020\\x=2021\\x=\dfrac{4041}{2}\end{array} \right.\) 

   Vậy phương trình có tập nghiệm `S={2020;2021;{4041}/2}`

   Trả lời

2. `(2020-x)^3+(2021-x)^3+(2x-4041)^3=0` (1)

   Đặt `2020-x=a; 2021-x=b`

   `-> 2x-4041=-(2020-x+2021-x)=-a-b`

   Khi đó pt (1) trở thành

   `a^3+b^3+(-a-b)^3=0`

   `<=> a^3+b^3-(a^3+3a^2b+3ab^2+b^3)=0`

   `<=> a^3+b^3-a^3-3a^2b-3ab^2-b^3=0`

   `<=> -3a^2b-3ab^2=0`

   `<=> -3ab(a+b)=0`

   `<=>`\(\left[ \begin{array}{l}a=0\\b=0\\a+b=0\end{array} \right.\) 

   `->`\(\left[ \begin{array}{l}2020-x=0\\2021-x=0\\2x-4041=0\end{array} \right.\) 

   `<=>`\(\left[ \begin{array}{l}x=2020\\x=2021\\2x=4041\end{array} \right.\) 

   `<=>`\(\left[ \begin{array}{l}x=2020\\x=2021\\x=\dfrac{4041}{2}\end{array} \right.\) 

   Vậy pt có tập nghiệm `S={2020;2021; 4041/2}`

    

    

   Trả lời