Let a,b,c,d,e are integers satisfying the following expression:
\(\frac{501}{2015}\)=\(\frac{1}{a+\frac{1}{b+\frac{1}{c+\frac{1}{d+\frac{1}{e}}}}}\)
Find the value of a+b+c+d+e
Toán tiếng anh,giải giúp mik
Let a , b and c be positive real numbers such that a + b + c = 3 . Find the minimum value of the expression .
\(P=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{a^2+b^2+c^2}\)
Cho:
:\(\frac{501}{2015}=\frac{1}{a+\frac{1}{b+\frac{1}{c+\frac{1}{d+\frac{1}{e}}}}}\)
Tính \(a+b+c+d+e\).
Cho: a+\(\frac{1}{b+\frac{1}{c+\frac{1}{d+\frac{1}{e+\frac{1}{f+\frac{1}{g}}}}}}\)=\(\frac{20102011}{2012}\)
Tìm các số tự nhiên a, b, c, d, e, f, g?
(Giải cụ thể giúp mình nha)
\(\frac{20102011}{2012}=9991+\frac{119}{2012}=9991+\frac{1}{\frac{2012}{119}}=9991+\frac{1}{16+\frac{108}{119}}=9991+\frac{1}{16+\frac{1}{\frac{119}{108}}}\)
\(=9991+\frac{1}{16+\frac{1}{1+\frac{11}{108}}}=9991+\frac{1}{16+\frac{1}{1+\frac{1}{\frac{108}{11}}}}=9991+\frac{1}{16+\frac{1}{1+\frac{1}{9+\frac{9}{11}}}}\)
=\(=9991+\frac{1}{16+\frac{1}{1+\frac{1}{9+\frac{1}{\frac{11}{9}}}}}=9991+\frac{1}{16+\frac{1}{1+\frac{1}{9+\frac{1}{1+\frac{2}{9}}}}}=9991+\frac{1}{16+\frac{1}{1+\frac{1}{9+\frac{1}{1+\frac{1}{4+\frac{1}{2}}}}}}\)
Nguyễn Thị Linh Chi có thể hướng dẫn cho mình cụ thể chút nữa được không.
Làm sao để \(\frac{20102011}{2012}\)=9991+\(\frac{119}{2012}\)vậy bạn?
(giúp mik nhé, mik cảm ơn nha!)
@Hồ Thị Điệp Lan @!!! Đặt tính rồi tính 20102011 chia 2012 bằng 9991 dư119
Hay 20102011=9991. 2012+119
=> \(\frac{20102011}{2012}=\frac{2012.9991+119}{2012}=9991+\frac{119}{2012}\)
Cho 5 số dương a b c d e
CMR
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{1}{e}>\frac{25}{a+b+c+d+e}\)
Áp dụng bất đẳng thức Cauchy- Schwartz ta có:
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{1}{e}\ge\frac{\left(1+1+1+1+1\right)^2}{a+b+c+d+e}=\frac{25}{a+b+c+d+e}\)
Dấu "=" xảy ra khi a = b = c = d = e
Biết a,b,c,d,e là các số nguyên dương thoả mãn \(a+\frac{1}{b+\frac{1}{c+\frac{1}{d+\frac{1}{e}}}}=\frac{2013}{1990}\) . TÍnh a+b+c+d+e
\(a+\frac{1}{b+\frac{1}{c+\frac{1}{d+\frac{1}{e}}}}=\frac{2013}{1990}\)
\(\Leftrightarrow a+\frac{1}{b+\frac{1}{c+\frac{1}{d+\frac{1}{e}}}}=1+\frac{23}{1990}\)
\(\Leftrightarrow a+\frac{1}{b+\frac{1}{c+\frac{1}{d+\frac{1}{e}}}}=1+\frac{1}{\frac{1990}{23}}\)
\(\Leftrightarrow a+\frac{1}{b+\frac{1}{c+\frac{1}{d+\frac{1}{e}}}}=1+\frac{1}{86+\frac{12}{23}}\)
\(\Leftrightarrow a+\frac{1}{b+\frac{1}{c+\frac{1}{d+\frac{1}{e}}}}=1+\frac{1}{86+\frac{1}{\frac{23}{12}}}\)
\(\Leftrightarrow a+\frac{1}{b+\frac{1}{c+\frac{1}{d+\frac{1}{e}}}}=1+\frac{1}{86+\frac{1}{1+\frac{11}{12}}}\)
\(\Leftrightarrow a+\frac{1}{b+\frac{1}{c+\frac{1}{d+\frac{1}{e}}}}=1+\frac{1}{86+\frac{1}{1+\frac{1}{\frac{12}{11}}}}\)
\(\Leftrightarrow a+\frac{1}{b+\frac{1}{c+\frac{1}{d+\frac{1}{e}}}}=1+\frac{1}{86+\frac{1}{1+\frac{1}{1+\frac{1}{11}}}}\)
Vậy a = 1; b = 86; c = 1; d = 1; e = 11
Vậy a + b + c + d + e = 1 + 86 + 1 + 1 + 11 = 100
\(\frac{1}{a+\frac{1}{b+\frac{1}{c+\frac{1}{d+\frac{1}{e}}}}}=2\)
\(a;b;c;d;e=?\)
\(\frac{25112012}{11}=a+\frac{1}{b+\frac{1}{c+\frac{1}{d+\frac{1}{e+\frac{1}{f+\frac{1}{g}}}}}}\)
Tìm a, b c d e f g
Tính a,b,c,d,e,g,h,i,k
\(\frac{654}{12254}=1+\frac{a}{1+\frac{b}{1+\frac{c}{1+\frac{d}{1+\frac{e}{1+\frac{g}{1+\frac{h}{1+\frac{i}{1+k}}}}}}}}\)
\(\frac{654}{12254}=\frac{12254-11600}{12254}=1+\frac{-11600}{12254}=1+\frac{1}{\frac{12254}{-11600}}=1+\frac{1}{1+\frac{23854}{-11600}}=1+\frac{1}{1+\frac{1}{-\frac{11600}{23854}}}=\)sức gõ công thức có hạn, cứ theo đó mà làm tiếp, đảm bảo sẽ ra ngay kết quả
đúng nha bạn
1) CMR: nếu \(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=\frac{d}{e}\) thì a\(\frac{a}{e}=\left(\frac{a-b+c-d}{b-c+d-e}\right)^4\)