tìm tất cả các số nguyên x thoả mãn |x-3| +|x-26|+|x-101|+x+990|+|x+1000| = 2000 26/09/2021 Bởi Reagan tìm tất cả các số nguyên x thoả mãn |x-3| +|x-26|+|x-101|+x+990|+|x+1000| = 2000
|x-3| +|x-26|+|x-101|+x+990|+|x+1000| = 2000 ⇔|x-3| +|x-26|+|x-101|+|x+990|+|x+1000| = 2000 ⇔|x|-|√6x|+3+|x|-|√52x|+26+|x|-|√202x|+101+|x|+|√1980x|+990+|x|+|√2000x|+1000=2000 ⇔5|x|-|√6x|-|√52x|-|√202x|+|√1980x|+|√2000x|=2000-3-26-101-990-1000=-120 (Vô lý) ⇔Không tìm được nghiệm S=∅ Bình luận
|x-3| +|x-26|+|x-101|+x+990|+|x+1000| = 2000 ⇔|x-3| +|x-26|+|x-101|+|x+990|+|x+1000| = 2000 ⇔ $\sqrt{x-3}^{2}$ +$\sqrt{x-26}^{2}$ + $\sqrt{x-101}^{2}$ + $\sqrt{x+990}^{2}$+ $\sqrt{x+1000}^{2}$ =2000 ⇔$\sqrt{x}^{2}$-$\sqrt{6x}$+3+$\sqrt{x}^{2}$-$\sqrt{52x}$+26+$\sqrt{x}^{2}$-$\sqrt{202x}$+101+$\sqrt{x}^{2}$+$\sqrt{1980x}$+990+$\sqrt{x}^{2}$+$\sqrt{2000x}$+1000 =2000 ⇔5$\sqrt{x}^{2}$-$\sqrt{6x}$-$\sqrt{52x}$-$\sqrt{202x}$+$\sqrt{1980x}$+$\sqrt{2000x}$+2120=2000 ⇔5$\sqrt{x}^{2}$-$\sqrt{6x}$-$\sqrt{52x}$-$\sqrt{202x}$+$\sqrt{1980x}$+$\sqrt{2000x}$=2000-2120=-120 ⇔5|x|=-120+$\sqrt{6x}$+$\sqrt{52x}$+$\sqrt{202x}$-$\sqrt{1980x}$-$\sqrt{2000x}$ ⇔|x|=$\frac{-120+$\sqrt{6x}$+$\sqrt{52x}$+$\sqrt{202x}$-$\sqrt{1980x}$-$\sqrt{2000x}$}{5}$ ⇔$\left \{ {{\frac{-120+$\sqrt{6x}$+$\sqrt{52x}$+$\sqrt{202x}$-$\sqrt{1980x}$-$\sqrt{2000x}$}{5}\geq0} \atop {\left\{{{x=\frac{-120+$\sqrt{6x}$+$\sqrt{52x}$+$\sqrt{202x}$-$\sqrt{1980x}$-$\sqrt{2000x}}{5}} \atop {x=-\frac{-120+$\sqrt{6x}$+$\sqrt{52x}$+$\sqrt{202x}$-$\sqrt{1980x}$-$\sqrt{2000x}}{5}}} \right.}} \right.$ (Đã kết thúc bài toán) Chúc bạn thi tốt Bình luận
|x-3| +|x-26|+|x-101|+x+990|+|x+1000| = 2000
⇔|x-3| +|x-26|+|x-101|+|x+990|+|x+1000| = 2000
⇔|x|-|√6x|+3+|x|-|√52x|+26+|x|-|√202x|+101+|x|+|√1980x|+990+|x|+|√2000x|+1000=2000
⇔5|x|-|√6x|-|√52x|-|√202x|+|√1980x|+|√2000x|=2000-3-26-101-990-1000=-120 (Vô lý)
⇔Không tìm được nghiệm
S=∅
|x-3| +|x-26|+|x-101|+x+990|+|x+1000| = 2000
⇔|x-3| +|x-26|+|x-101|+|x+990|+|x+1000| = 2000
⇔ $\sqrt{x-3}^{2}$ +$\sqrt{x-26}^{2}$ + $\sqrt{x-101}^{2}$ + $\sqrt{x+990}^{2}$+ $\sqrt{x+1000}^{2}$ =2000
⇔$\sqrt{x}^{2}$-$\sqrt{6x}$+3+$\sqrt{x}^{2}$-$\sqrt{52x}$+26+$\sqrt{x}^{2}$-$\sqrt{202x}$+101+$\sqrt{x}^{2}$+$\sqrt{1980x}$+990+$\sqrt{x}^{2}$+$\sqrt{2000x}$+1000 =2000
⇔5$\sqrt{x}^{2}$-$\sqrt{6x}$-$\sqrt{52x}$-$\sqrt{202x}$+$\sqrt{1980x}$+$\sqrt{2000x}$+2120=2000
⇔5$\sqrt{x}^{2}$-$\sqrt{6x}$-$\sqrt{52x}$-$\sqrt{202x}$+$\sqrt{1980x}$+$\sqrt{2000x}$=2000-2120=-120
⇔5|x|=-120+$\sqrt{6x}$+$\sqrt{52x}$+$\sqrt{202x}$-$\sqrt{1980x}$-$\sqrt{2000x}$
⇔|x|=$\frac{-120+$\sqrt{6x}$+$\sqrt{52x}$+$\sqrt{202x}$-$\sqrt{1980x}$-$\sqrt{2000x}$}{5}$
⇔$\left \{ {{\frac{-120+$\sqrt{6x}$+$\sqrt{52x}$+$\sqrt{202x}$-$\sqrt{1980x}$-$\sqrt{2000x}$}{5}\geq0} \atop {\left\{{{x=\frac{-120+$\sqrt{6x}$+$\sqrt{52x}$+$\sqrt{202x}$-$\sqrt{1980x}$-$\sqrt{2000x}}{5}} \atop {x=-\frac{-120+$\sqrt{6x}$+$\sqrt{52x}$+$\sqrt{202x}$-$\sqrt{1980x}$-$\sqrt{2000x}}{5}}} \right.}} \right.$
(Đã kết thúc bài toán)
Chúc bạn thi tốt