ĐỀ : Tính
\(M=\dfrac{\left(1+\dfrac{2012}{1}\right)\left(1+\dfrac{2012}{1}\right)...\left(1+\dfrac{2012}{1000}\right)}{\left(1+\dfrac{1000}{1}\right)\left(1+\dfrac{1000}{2}\right)...\left(1+\dfrac{1000}{2012}\right)}\)
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Những câu hỏi liên quan
cho \(f\left(x\right)=\dfrac{x^3}{1-3x-3x^2}\). hãy tính giá trị biểu thức sau: \(A=f\left(\dfrac{1}{2012}\right)+f\left(\dfrac{2}{2012}\right)+...+f\left(\dfrac{2010}{2012}\right)+f\left(\dfrac{2011}{2012}\right)\)
Bạn kiểm tra lại đề, \(f\left(x\right)=\dfrac{x^3}{1-3x-3x^2}\) hay \(f\left(x\right)=\dfrac{x^3}{1-3x+3x^2}\)
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Tính
P=\(\left(1-\dfrac{28}{10}\right)\left(1-\dfrac{52}{22}\right)\left(1-\dfrac{80}{36}\right)...\left(1-\dfrac{21808}{10900}\right)\)
Q=\(\dfrac{1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2012}}{\dfrac{2013}{1}+\dfrac{2014}{2}+\dfrac{2015}{3}+...+\dfrac{4023}{2011}+\dfrac{4024}{2012}-2012}\)
(5+10+15+...+1000)\(\left[\dfrac{2}{5}:0.5+2\left(-0.4\right)\right]:\left(\dfrac{1}{5}+\dfrac{1}{10}+\dfrac{1}{15}+...\dfrac{1}{1000}\right)\)
cho \(f\left(x\right)=\dfrac{x^3}{1-3x+3x^2}\) hãy tính giá trị của biểu thức sau:
\(A=f\left(\dfrac{1}{2012}\right)+f\left(\dfrac{2}{2012}\right)+...+f\left(\dfrac{2010}{2012}\right)+f\left(\dfrac{2011}{2012}\right)\)
D=\(\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)...\left(1-\dfrac{1}{2005}\right)\)
E=\(\dfrac{1^2}{1.3}.\dfrac{2^2}{2.4}.\dfrac{3^2}{3.5}....\dfrac{999^2}{999.1000}.\dfrac{1000^2}{1000.1001}\)
Ta có: D\(=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)...\left(1-\dfrac{1}{2005}\right)\)
\(\Leftrightarrow D=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{2004}{2005}=\dfrac{1.2.3...2004}{2.3.4...2005}=\dfrac{1}{2005}\)
Ta có: \(E=\dfrac{1^2}{1.3}.\dfrac{2^2}{2.4}.\dfrac{3^2}{3.5}...\dfrac{999^2}{999.1000}.\dfrac{1000^2}{1000.1001}=\dfrac{\left(1.2.3.4...1000\right)\left(1.2.3.4...1000\right)}{\left(1.2.3....1000\right)\left(3.4.5....1001\right)}=\dfrac{2}{1001}\)
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Tính
a) \(\left(1-\dfrac{1}{2}\right).\left(1-\dfrac{1}{3}\right).\left(1-\dfrac{1}{4}\right).....\left(1-\dfrac{1}{2012}\right)\)
\(\left(1-\dfrac{1}{2}\right).\left(1-\dfrac{1}{3}\right).\left(1-\dfrac{1}{4}\right).....\left(1-\dfrac{1}{2012}\right)\)
\(=\dfrac{2}{3}.\dfrac{3}{4}.\dfrac{4}{5}.....\dfrac{2011}{2012}\)
\(=\dfrac{2.3.4...2011}{3.4.5...2012}\)
\(=\dfrac{2}{2012}=\dfrac{1}{1006}.\)
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Tính B=\(\left(\dfrac{1}{2013^2}-1\right)\left(\dfrac{1}{2012^2}-1\right)\left(\dfrac{1}{2011^2}-1\right)....\left(\dfrac{1}{100^2-1}\right)\)
Số số hạng của B là 1914(là 1 số chẵn)
\(\Rightarrow B=\left(1-\dfrac{1}{2013^2}\right)\left(1-\dfrac{1}{2012^2}\right)\left(1-\dfrac{1}{2011^2}\right)\cdot\cdot\cdot\cdot\cdot\left(1-\dfrac{1}{100^2}\right)\)
\(B=\dfrac{2013^2-1}{2013^2}\cdot\dfrac{2012^2-1}{2012^2}\cdot\dfrac{2011^2-1}{2011^2}\cdot\cdot\cdot\cdot\cdot\dfrac{100^2-1}{100^2}\)
\(B=\dfrac{2014\cdot2012\cdot2013\cdot2011\cdot2012\cdot2010\cdot...\cdot101\cdot99}{2013\cdot2013\cdot2012\cdot2012\cdot2011\cdot2011\cdot...\cdot100\cdot100}\)
\(B=\dfrac{2014\cdot99}{2013\cdot100}=\dfrac{3021}{3050}\)
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B=\(\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2013}\right):\left(2012+\dfrac{2011}{2}+...+\dfrac{2}{2011}+\dfrac{1}{2012}\right)\).Tính B
Đặt \(B=A\div C\)
\(C=2012+\dfrac{2011}{2}+...+\dfrac{1}{2012}=2012+\dfrac{2013-2}{2}+\dfrac{2013-3}{3}+...+\dfrac{2013-2012}{2012}\)
\(C=2012+\dfrac{2013}{2}+\dfrac{2013}{3}+...+\dfrac{2013}{2012}-1-1-...-1\)
\(C=2012+2013\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2012}\right)-2011\)
\(C=1+2013\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2012}\right)=\dfrac{2013}{2013}+2013\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2012}\right)\)
\(C=2013\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2013}\right)=2013.A\)
\(\Rightarrow B=\dfrac{A}{C}=\dfrac{1}{2013}\)
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Bai 5. Tinh nhanh
a, \(\dfrac{1}{5.8}+\dfrac{1}{8.7}+\dfrac{1}{11.14}+.......+\dfrac{1}{605.606}\)
b,\(\left(\dfrac{1}{10}-1\right)\left(\dfrac{1}{11}-1\right)\left(\dfrac{1}{12}-1\right)....\left(\dfrac{1}{2012}-1\right)\)
Mk ko biết nhưng mk chúc bn sớm tìm đc câu trả lời
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b)
\(\left(\dfrac{1}{10}-1\right)\left(\dfrac{1}{11}-1\right)\left(\dfrac{1}{12}-1\right)...\left(\dfrac{1}{2012}-1\right)\\ =\dfrac{-9}{10}\cdot\dfrac{-10}{11}\cdot\dfrac{-11}{12}\cdot...\cdot\dfrac{-2011}{2012}\\ =\left(-1\right)\cdot\dfrac{9}{10}\cdot\left(-1\right)\cdot\dfrac{10}{11}\cdot\left(-1\right)\cdot\dfrac{11}{12}\cdot...\cdot\left(-1\right)\cdot\dfrac{2011}{2012}\\ =\left[\left(-1\right)\cdot\left(-1\right)\cdot...\cdot\left(-1\right)\right]\cdot\left(\dfrac{9}{10}\cdot\dfrac{10}{11}\cdot\dfrac{11}{12}\cdot...\cdot\dfrac{2011}{2012}\right)\\ =\left[\left(-1\right)\cdot\left(-1\right)\cdot...\cdot\left(-1\right)\right]\cdot\dfrac{9}{2012}\)
(Có tất cả 2003 thừa số -1)
\(=\left(-1\right)\cdot\dfrac{9}{2012}=\dfrac{-9}{2012}\)
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