Cho a,b,ca,b,c là các số với |a|,|b|,|c|≤1|a|,|b|,|c|≤1chứng minh rằng, nếu a,b,ca,b,c thỏa mãn: ...
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Cho a,b,c" role="presentation" style="margin: 0px; padding: 0px; border: 0px; font-size: 13.696px; vertical-align: baseline; display: inline; line-height: normal; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; color: rgb(0, 0, 0); font-family: Arial, 'Liberation Sans', 'DejaVu Sans', sans-serif; position: relative; background: transparent;">a,b,c là các số với |a|,|b|,|c|≤1" role="presentation" style="margin: 0px; padding: 0px; border: 0px; font-size: 13.696px; vertical-align: baseline; display: inline; line-height: normal; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; color: rgb(0, 0, 0); font-family: Arial, 'Liberation Sans', 'DejaVu Sans', sans-serif; position: relative; background: transparent;">|a|,|b|,|c|≤1
chứng minh rằng, nếu a,b,c" role="presentation" style="margin: 0px; padding: 0px; border: 0px; font-size: 13.696px; vertical-align: baseline; display: inline-table; line-height: normal; word-spacing: normal; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; position: relative; background: transparent;">a,b,c thỏa mãn:
a2+b2+c2=1−2abc" role="presentation" style="margin: 0px; padding: 0px; border: 0px; font-size: 13.696px; vertical-align: baseline; display: inline; line-height: normal; word-spacing: normal; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; position: relative; background: transparent;">a2+b2+c2=1−2abc
thì
a+b+c=2(1−a)(1−b)(1−c)2+1" role="presentation" style="margin: 0px; padding: 0px; border: 0px; font-size: 13.696px; vertical-align: baseline; display: inline; line-height: normal; word-spacing: normal; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; position: relative; background: transparent;">a+b+c=2(1−a)(1−b)(1−c)2−−−−−−−−−−−−−−−−−√+1
Lời giải:
\(p_n=(1-\frac{1}{1+2})(1-\frac{1}{1+2+3})....(1-\frac{1}{1+2+3+...+n})\\ =(1-\frac{1}{\frac{2.3}{2}})(1-\frac{1}{\frac{3.4}{2}})....(1-\frac{1}{\frac{n(n+1)}{2}})\\ =(1-\frac{2}{2.3})(1-\frac{2}{3.4})....(1-\frac{2}{n(n+1)})\)
Xét thừa số tổng quát:
$1-\frac{2}{n(n+1)}=\frac{n(n+1)-2}{n(n+1)}=\frac{(n-1)(n+2)}{n(n+1)}$
Áp dụng vào tất cả các thừa số của $p_n$ suy ra:
$p_n=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}....\frac{(n-1)(n+2)}{n(n+1)}$
$=\frac{[1.2.3...(n-1)][(4.5.6....(n+2)]}{[2.3.4...n][3.4.5...(n+1)]}$
$=\frac{1.2.3...(n-1)}{2.3.4...n}.\frac{4.5.6...(n+2)}{3.4.5....(n+1)}$
$=\frac{1}{n}.\frac{n+2}{3}$
$=\frac{n+2}{3n}$
$\frac{1}{p_n}=\frac{3n}{n+2}$
Với $n$ nguyên dương, để $\frac{1}{p_n}$ là 1 số nguyên thì:
$3n\vdots n+2$
$\Rightarrow 3(n+2)-6\vdots n+2$
$\Rightarrow 6\vdots n+2$
$\Rightarrow n+2=6$ (do $n$ là số nguyên dương $\geq 2$)
$\Rightarrow n=4$