Viết chương trình tính tổng S=\(\frac{10}{11}\)+\(\frac{11}{12}\)+\(\frac{12}{13}\)...+\(\frac{100}{101}\)
Những câu hỏi liên quan
\(\frac{10}{11}x\frac{12}{13}:\frac{50}{51}-\frac{19}{20}x\frac{12}{13}:\frac{101}{102}+\frac{99}{100}\)
so sánh
\(\frac{100}{10^{11}}+\frac{100}{10^{12}}va\frac{99}{10^{11}}+\frac{101}{10^{12}}\)
\(\frac{10^{10}+1}{10^{11}+1}va\frac{10^{11}+1}{10^{12}+1}\)
s2 Lắc Lư s2 cko hỏi ôg lp mấy z?
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Bài 1: Chứng minh B = \(3^{21}+3^{22}+3^{23}+.........+3^{29}\) chia hết cho 13
Bài 2: So sánh \(\frac{100}{11^{11}}+\frac{100}{11^{12}}\)và \(\frac{99}{11^{11}}+\frac{101}{11^{12}}\)
3^21*(1+3+3^2)+3^24*(1+3+3^2)+3^27*(1+3+3^2)=13*3^21+13*3^24+13*3^27=13*(3^21+3^24+3^27)chia hết cho 13
Giải nghĩa ^:mũ
*:nhân
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Tính:
a, A =\(\frac{101+100+99+98+...+3+2+1}{101-100+99-98+...+3-2+1}\)
b, B = \(\frac{3737.43-4343.37}{2+4+6+...+100}\)
c, D = \(\frac{2^{12}.13+2^{12}.65}{2^{10}.104}+\frac{3^{10}.11+3^{10}.5}{3^9.2^4}\)
(Các bn giải chi tiết giúp mik nha)
b, \(3737.43-4343.37=\left(37.101\right).43-\left(43.101\right).37=0\)
suy ra B = 0
c, \(D=\frac{2^{12}\left(13+65\right)}{2^{10}.104}+\frac{3^{10}\left(11+5\right)}{3^9.2^4}=\frac{2^{12}.78}{2^{10}.104}+\frac{3^{10}.16}{3^9.2^4}\)
\(=\frac{2^{12}.2.39}{2^{10}.2^3.13}+\frac{3^{10}.2^4}{3^9.2^4}=\frac{39}{13}+3=6\)
cho \(S=\frac{3}{10}+\frac{3}{11}+\frac{3}{12}+\frac{3}{13}+\frac{3}{14}\) không tính tổng S, hãy chứng minh S không phải 1 số tự nhiên
cho \(A=\frac{1}{61}+\frac{1}{62}+\frac{1}{63}+...+\frac{1}{99}+\frac{1}{100}\) . Chứng minh \(A>\frac{9}{20}\)
a,Ta có: \(\frac{3}{10}=\frac{3}{10};\frac{3}{11}< \frac{3}{10};\frac{3}{12}< \frac{3}{10};\frac{3}{13}< \frac{3}{10};\frac{3}{14}< \frac{3}{10}\)
\(\Rightarrow S< \frac{3}{10}+\frac{3}{10}+\frac{3}{10}+\frac{3}{10}+\frac{3}{10}=\frac{15}{10}=\frac{3}{2}=1,5\left(1\right)\)
Lại có: \(\frac{3}{10}>\frac{3}{15};\frac{3}{11}>\frac{3}{15};\frac{3}{12}>\frac{3}{15};\frac{3}{13}>\frac{3}{15};\frac{3}{14}>\frac{3}{15}\)
\(\Rightarrow S>\frac{3}{15}+\frac{3}{15}+\frac{3}{15}+\frac{3}{15}+\frac{3}{15}=\frac{15}{15}=1\left(2\right)\)
Từ (1) và (2) => 1 < S < 1,5
Vậy...
b, \(A=\frac{1}{61}+\frac{1}{62}+...+\frac{1}{100}\)
\(=\left(\frac{1}{61}+\frac{1}{62}+...+\frac{1}{80}\right)+\left(\frac{1}{81}+\frac{1}{82}+...+\frac{1}{100}\right)\)
Ta có: \(\frac{1}{61}>\frac{1}{80};\frac{1}{62}>\frac{1}{80};...;\frac{1}{80}=\frac{1}{80}\)
\(\Rightarrow\frac{1}{61}+\frac{1}{62}+...+\frac{1}{80}>\frac{1}{80}+\frac{1}{80}+...+\frac{1}{80}=\frac{20}{80}=\frac{1}{4}\left(1\right)\)
Lại có: \(\frac{1}{81}>\frac{1}{100};\frac{1}{82}>\frac{1}{100};...;\frac{1}{100}=\frac{1}{100}\)
\(\Rightarrow\frac{1}{81}+\frac{1}{82}+...+\frac{1}{100}>\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}=\frac{20}{100}=\frac{1}{5}\left(2\right)\)
Từ (1) và (2) => \(A>\frac{1}{4}+\frac{1}{5}=\frac{9}{20}\)
Vậy...
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Tính giá trị của biểu thức:
\(A=\frac{1}{9}.\frac{1}{10}+\frac{1}{10}.\frac{1}{11}+\frac{1}{11}.\frac{1}{12}+\frac{1}{12}.\frac{1}{13}+\frac{1}{13}.\frac{1}{14}+\frac{1}{14}.\frac{1}{15}\)
nhờ các bn giúp mình nha
\(A\)\(=\)\(\frac{1}{9}\)\(-\)\(\frac{1}{10}\)\(+\)\(\frac{1}{10}\)\(-\)\(\frac{1}{11}\)\(+\)\(\frac{1}{11}\)\(-\)\(\frac{1}{12}\)\(+\)\(\frac{1}{12}\)\(-\)\(\frac{1}{13}\)\(+\)\(\frac{1}{13}\)\(-\)\(\frac{1}{14}\)\(+\)\(\frac{1}{14}\)\(-\)\(\frac{1}{15}\)
\(A\)\(=\)\(\frac{1}{9}\)\(-\)\(\frac{1}{15}\)
\(A\)\(=\)\(\frac{2}{45}\)
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\(A=\left(\frac{1}{9}.\frac{1}{10}+\frac{1}{10}.\frac{1}{11}\right)+\left(\frac{1}{11}.\frac{1}{12}+\frac{1}{12}.\frac{1}{13}\right)+\left(\frac{1}{13}.\frac{1}{14}+\frac{1}{14}.\frac{1}{15}\right)\)
Sau đó nhân phân phối ra là xong nhé bạn
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Tính nhanh : \(B=\frac{25.49-24}{25+49.24}:\frac{4+\frac{4}{7}-\frac{4}{11}+\frac{4}{2011}-\frac{4}{13}}{\frac{12}{2011}-\frac{12}{13}+\frac{12}{7}-\frac{12}{11}+12}\)
Chứng minh rằng:
\(\frac{10}{11!}+\frac{11}{12!}+\frac{12}{13!}+...+\frac{2014}{2015!}< \frac{1}{10!}\)
Đặt \(A=\frac{10}{11!}+\frac{11}{12!}+\frac{12}{13!}+...+\frac{2014}{2015!}\)
\(=\frac{11-1}{11!}+\frac{12-1}{12!}+\frac{13-1}{13!}+...+\frac{2015-1}{2015!}\)
\(=\frac{11}{11!}-\frac{1}{11!}+\frac{12}{12!}-\frac{1}{12!}+\frac{13}{13!}-\frac{1}{13!}+...+\frac{2015}{2015!}-\frac{1}{2015!}\)
\(=\frac{11}{10!.11}-\frac{1}{11!}+\frac{12}{11!.12}-\frac{1}{12!}+\frac{13}{12!.13}-\frac{1}{13!}+...+\frac{2015}{2014!.2015}-\frac{1}{2015!}\)
\(=\frac{1}{10!}-\frac{1}{11!}+\frac{1}{11!}-\frac{1}{12!}+\frac{1}{12!}-\frac{1}{13!}+...+\frac{1}{2014!}-\frac{1}{2015!}\)
\(=\frac{1}{10!}-\frac{1}{2015!}< \frac{1}{10!}\)
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Tìm x biết
\(\frac{x+2}{10^{10}}+\frac{x+2}{11^{11}}=\frac{x+2}{12^{12}}+\frac{x+2}{13^{13}}\)
\(\frac{x+2}{10^{10}}+\frac{x+2}{11^{11}}=\frac{x+2}{12^{12}}\frac{x+2}{13^{13}}\)
=> x + 2 = 0
=> x = 0 - 2
=> x = -2
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