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x-3-2022-x-2-2023-x-1-2024-x-2025-4-

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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
$$\:\:\:\frac{{x}+\mathrm{3}}{\mathrm{2022}}\:+\:\frac{{x}+\mathrm{2}}{\mathrm{2023}}\:+\:\frac{{x}+\mathrm{1}}{\mathrm{2024}}\:+\:\frac{{x}}{\mathrm{2025}}\:=\:−\mathrm{4} \\ $$
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
$$\:\:\:\frac{{x}+\mathrm{3}}{\mathrm{2022}}\:+\:\frac{{x}+\mathrm{2}}{\mathrm{2023}}\:+\:\frac{{x}+\mathrm{1}}{\mathrm{2024}}\:+\:\frac{{x}}{\mathrm{2025}}\:=\:−\mathrm{4} \\ $$$$\:\:\:\frac{{x}+\mathrm{3}}{\mathrm{2022}}+\mathrm{1}\:+\:\frac{{x}+\mathrm{2}}{\mathrm{2023}}+\mathrm{1}\:+\:\frac{{x}+\mathrm{1}}{\mathrm{2024}}+\mathrm{1}\:+\:\frac{{x}}{\mathrm{2025}}+\mathrm{1}\:=\:−\mathrm{4}+\mathrm{4} \\ $$$$\:\:\:\frac{{x}+\mathrm{2025}}{\mathrm{2022}}\:+\:\frac{{x}+\mathrm{2025}}{\mathrm{2023}}\:+\:\frac{{x}+\mathrm{2025}}{\mathrm{2024}}\:+\:\frac{{x}+\mathrm{2025}}{\mathrm{2025}}\:=\:\mathrm{0} \\ $$$$\left({x}+\mathrm{2025}\right)\left(\:\frac{\mathrm{1}}{\mathrm{2022}}\:+\:\frac{\mathrm{1}}{\mathrm{2023}}\:+\:\frac{\mathrm{1}}{\mathrm{2024}}\:+\:\frac{\mathrm{1}}{\mathrm{2025}}\right)=\mathrm{0} \\ $$$${x}+\mathrm{2025}=\mathrm{0} \\ $$$${x}=−\mathrm{2025} \\ $$
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
$$\:\left(\frac{{x}+\mathrm{3}}{\mathrm{2022}}+\mathrm{1}\right)+\left(\frac{{x}+\mathrm{2}}{\mathrm{2023}}+\mathrm{1}\right)+\left(\frac{{x}+\mathrm{1}}{\mathrm{2024}}+\mathrm{1}\right) \\ $$$$\:\:\:+\:\:\:\left(\frac{{x}}{\mathrm{2025}}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow\left({x}+\mathrm{2025}\right)\left(\frac{\mathrm{1}}{\mathrm{2022}}+\frac{\mathrm{1}}{\mathrm{2023}}+\frac{\mathrm{1}}{\mathrm{2024}}+\frac{\mathrm{1}}{\mathrm{2025}}\right)=\mathrm{0} \\ $$$$\therefore\left({x}+\mathrm{2025}\right)=\mathrm{0} \\ $$$$\left[\left(\frac{\mathrm{1}}{\mathrm{2022}}+\frac{\mathrm{1}}{\mathrm{2023}}+\frac{\mathrm{1}}{\mathrm{2024}}+\frac{\mathrm{1}}{\mathrm{2025}}\right)\neq\mathrm{0}\right] \\ $$$$\:{x}=−\mathrm{2025} \\ $$$$ \\ $$

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