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Question Number 197609 by dimentri last updated on 24/Sep/23

((x+3)/(2022)) + ((x+2)/(2023)) + ((x+1)/(2024)) + (x/(2025)) = −4

x+32022+x+22023+x+12024+x2025=−4

Answered by Rasheed.Sindhi last updated on 24/Sep/23

((x+3)/(2022)) + ((x+2)/(2023)) + ((x+1)/(2024)) + (x/(2025)) = −4 ((x+3)/(2022))+1 + ((x+2)/(2023))+1 + ((x+1)/(2024))+1 + (x/(2025))+1 = −4+4 ((x+2025)/(2022)) + ((x+2025)/(2023)) + ((x+2025)/(2024)) + ((x+2025)/(2025)) = 0 (x+2025)( (1/(2022)) + (1/(2023)) + (1/(2024)) + (1/(2025)))=0 x+2025=0 x=−2025

x+32022+x+22023+x+12024+x2025=−4x+32022+1+x+22023+1+x+12024+1+x2025+1=−4+4x+20252022+x+20252023+x+20252024+x+20252025=0(x+2025)(12022+12023+12024+12025)=0x+2025=0x=−2025

Answered by som(math1967) last updated on 24/Sep/23

(((x+3)/(2022))+1)+(((x+2)/(2023))+1)+(((x+1)/(2024))+1) + ((x/(2025))+1)=0 ⇒(x+2025)((1/(2022))+(1/(2023))+(1/(2024))+(1/(2025)))=0 ∴(x+2025)=0 [((1/(2022))+(1/(2023))+(1/(2024))+(1/(2025)))≠0] x=−2025

(x+32022+1)+(x+22023+1)+(x+12024+1)+(x2025+1)=0⇒(x+2025)(12022+12023+12024+12025)=0∴(x+2025)=0[(12022+12023+12024+12025)≠0]x=−2025

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