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Question Number 217186 by MrGaster last updated on 04/Mar/25
((x−4051)/(2024))+((x−4050)/(2025))+((x−4049)/(2026))=3
$$\frac{{x}−\mathrm{4051}}{\mathrm{2024}}+\frac{{x}−\mathrm{4050}}{\mathrm{2025}}+\frac{{x}−\mathrm{4049}}{\mathrm{2026}}=\mathrm{3} \\ $$
Answered by som(math1967) last updated on 04/Mar/25
((x−4051)/(2024)) −1+((x−4050)/(2025))−1+((x−4049)/(2026))−1=0 ⇒((x−6075)/(2024))+((x−6075)/(2025)) +((x−6075)/(2026))=0 ⇒(x−6075) =0 ⇒(x−6075)=0 [ ((1/(2024))+(1/(2025))+(1/(2026)))≠0] ∴ x=6075
$$\:\frac{{x}−\mathrm{4051}}{\mathrm{2024}}\:−\mathrm{1}+\frac{{x}−\mathrm{4050}}{\mathrm{2025}}−\mathrm{1}+\frac{{x}−\mathrm{4049}}{\mathrm{2026}}−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\frac{{x}−\mathrm{6075}}{\mathrm{2024}}+\frac{{x}−\mathrm{6075}}{\mathrm{2025}}\:+\frac{{x}−\mathrm{6075}}{\mathrm{2026}}=\mathrm{0} \\ $$$$\Rightarrow\left({x}−\mathrm{6075}\right)\:=\mathrm{0} \\ $$$$\Rightarrow\left({x}−\mathrm{6075}\right)=\mathrm{0} \\ $$$$\left[\:\left(\frac{\mathrm{1}}{\mathrm{2024}}+\frac{\mathrm{1}}{\mathrm{2025}}+\frac{\mathrm{1}}{\mathrm{2026}}\right)\neq\mathrm{0}\right] \\ $$$$\therefore\:{x}=\mathrm{6075} \\ $$
Answered by Rasheed.Sindhi last updated on 05/Mar/25
AnOther Approach ((x−4051)/(2024))+((x−4050)/(2025))+((x−4049)/(2026))=3 let 2025=k ((x−2k−1)/(k−1))+((x−2k)/k)+((x−2k+1)/(k+1))=3 ((x−2(k−1)−3)/(k−1))+((x−2k)/k)+((x−2(k+1)+3)/(k+1))=3 ((x−3)/(k−1))−2+(x/k)−2+((x+3)/(k+1))−2=3 ((x−3)/(k−1))+(x/k)+((x+3)/(k+1))=9 This leads to: x=3k=3(2025)=6075
$$\mathrm{AnOther}\:\mathrm{Approach} \\ $$$$\frac{{x}−\mathrm{4051}}{\mathrm{2024}}+\frac{{x}−\mathrm{4050}}{\mathrm{2025}}+\frac{{x}−\mathrm{4049}}{\mathrm{2026}}=\mathrm{3} \\ $$$${let}\:\mathrm{2025}={k} \\ $$$$\frac{{x}−\mathrm{2}{k}−\mathrm{1}}{{k}−\mathrm{1}}+\frac{{x}−\mathrm{2}{k}}{{k}}+\frac{{x}−\mathrm{2}{k}+\mathrm{1}}{{k}+\mathrm{1}}=\mathrm{3} \\ $$$$\frac{{x}−\mathrm{2}\left({k}−\mathrm{1}\right)−\mathrm{3}}{{k}−\mathrm{1}}+\frac{{x}−\mathrm{2}{k}}{{k}}+\frac{{x}−\mathrm{2}\left({k}+\mathrm{1}\right)+\mathrm{3}}{{k}+\mathrm{1}}=\mathrm{3} \\ $$$$\frac{{x}−\mathrm{3}}{{k}−\mathrm{1}}−\mathrm{2}+\frac{{x}}{{k}}−\mathrm{2}+\frac{{x}+\mathrm{3}}{{k}+\mathrm{1}}−\mathrm{2}=\mathrm{3} \\ $$$$\frac{{x}−\mathrm{3}}{{k}−\mathrm{1}}+\frac{{x}}{{k}}+\frac{{x}+\mathrm{3}}{{k}+\mathrm{1}}=\mathrm{9} \\ $$$${This}\:{leads}\:{to}: \\ $$$${x}=\mathrm{3}{k}=\mathrm{3}\left(\mathrm{2025}\right)=\mathrm{6075} \\ $$
Answered by MrGaster last updated on 12/Mar/25
Σ_(k=1) ^3 ((x−a_k )/b_k )=3,{a_k }={4051,4050,4069},{b_k }={2024,2025,2026} xΣ_(k=1) ^3 (1/b_k )−Σ_(k=1) ^3 (a_k /b_k )=3 x=((3+Σ_(k=1) ^3 (a_k /b_k ))/(Σ_(k=1) ^3 (1/b_k ))) Σ_(k=1) ^3 (a_k /b_k )=((4051)/(2024))+((4050)/(2025))+((4049)/(2026))=6+(3/(2024))−(3/(2026)) Σ_(k=1) ^3 (1/b_k )=(1/(2024))+(1/(2025))+(1/(2026)) x=((3+6+(3/(2024))−(3/(2026)))/((1/(2024))+(1/(2025))+(1/(2026))))=(9/((1/(2024))+(1/(2025))+(1/(2026))))=6075
$$\underset{{k}=\mathrm{1}} {\overset{\mathrm{3}} {\sum}}\frac{{x}−{a}_{{k}} }{{b}_{{k}} }=\mathrm{3},\left\{{a}_{{k}} \right\}=\left\{\mathrm{4051},\mathrm{4050},\mathrm{4069}\right\},\left\{{b}_{{k}} \right\}=\left\{\mathrm{2024},\mathrm{2025},\mathrm{2026}\right\} \\ $$$${x}\underset{{k}=\mathrm{1}} {\overset{\mathrm{3}} {\sum}}\frac{\mathrm{1}}{{b}_{{k}} }−\underset{{k}=\mathrm{1}} {\overset{\mathrm{3}} {\sum}}\frac{{a}_{{k}} }{{b}_{{k}} }=\mathrm{3} \\ $$$${x}=\frac{\mathrm{3}+\underset{{k}=\mathrm{1}} {\overset{\mathrm{3}} {\sum}}\frac{{a}_{{k}} }{{b}_{{k}} }}{\underset{{k}=\mathrm{1}} {\overset{\mathrm{3}} {\sum}}\frac{\mathrm{1}}{{b}_{{k}} }} \\ $$$$\underset{{k}=\mathrm{1}} {\overset{\mathrm{3}} {\sum}}\frac{{a}_{{k}} }{{b}_{{k}} }=\frac{\mathrm{4051}}{\mathrm{2024}}+\frac{\mathrm{4050}}{\mathrm{2025}}+\frac{\mathrm{4049}}{\mathrm{2026}}=\mathrm{6}+\frac{\mathrm{3}}{\mathrm{2024}}−\frac{\mathrm{3}}{\mathrm{2026}} \\ $$$$\underset{{k}=\mathrm{1}} {\overset{\mathrm{3}} {\sum}}\frac{\mathrm{1}}{{b}_{{k}} }=\frac{\mathrm{1}}{\mathrm{2024}}+\frac{\mathrm{1}}{\mathrm{2025}}+\frac{\mathrm{1}}{\mathrm{2026}} \\ $$$${x}=\frac{\mathrm{3}+\mathrm{6}+\frac{\mathrm{3}}{\mathrm{2024}}−\frac{\mathrm{3}}{\mathrm{2026}}}{\frac{\mathrm{1}}{\mathrm{2024}}+\frac{\mathrm{1}}{\mathrm{2025}}+\frac{\mathrm{1}}{\mathrm{2026}}}=\frac{\mathrm{9}}{\frac{\mathrm{1}}{\mathrm{2024}}+\frac{\mathrm{1}}{\mathrm{2025}}+\frac{\mathrm{1}}{\mathrm{2026}}}=\mathrm{6075} \\ $$

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