Toán $(2020 – x)^3 + (2021 – x)^3 + (2x – 4041)^3$ = 0 27/09/2021 By Skylar $(2020 – x)^3 + (2021 – x)^3 + (2x – 4041)^3$ = 0
`(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
`(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
`(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}`
`(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}`