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\(a,A=\left(3\dfrac{5}{6}-1\dfrac{1}{3}\right)\left(3\dfrac{4}{15}-2\dfrac{3}{5}\right)\)
\(\Leftrightarrow A=\left(3+\dfrac{5}{6}-1+\dfrac{1}{3}\right)\left(3+\dfrac{4}{15}-2+\dfrac{3}{5}\right)\)
\(\Leftrightarrow A=\left[\left(3-1\right)+\left(\dfrac{5}{6}+\dfrac{1}{3}\right)\right]+\left[\left(3-2\right)+\left(\dfrac{4}{15}+\dfrac{3}{5}\right)\right]\)
\(\Leftrightarrow A=\left[2+\left(\dfrac{5}{6}+\dfrac{2}{6}\right)\right]+\left[1+\left(\dfrac{4}{15}+\dfrac{9}{15}\right)\right]\)
\(\Leftrightarrow A=\left(2+\dfrac{7}{6}\right)+\left(1+\dfrac{13}{15}\right)\)
\(\Leftrightarrow A=\left(2+1+\dfrac{1}{6}\right)+\left(1+\dfrac{13}{15}\right)\)
\(\Leftrightarrow A=3\dfrac{1}{6}+1\dfrac{13}{15}\)
Vậy...
\(b,B=\dfrac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}\)
\(\Leftrightarrow B=\dfrac{\left(2^2\right)^6.\left(3^2\right)^5+\left(2.3\right)^9.\left(2^3.3.5\right)}{\left(2^3\right)^4.3^{12}-\left(2.3\right)^{11}}\)
\(\Leftrightarrow B=\dfrac{2^{12}.3^{10}+2^9.3^9.2^3.3.5}{2^{12}.3^{12}-2^{11}.3^{11}}\)
\(\Leftrightarrow B=\dfrac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{12}.3^{12}-2^{11}.3^{11}}\)
\(\Leftrightarrow B=\dfrac{\left(2^{10}.3^{10}\right)\left(1+5\right)}{\left(2^{11}.3^{11}\right)\left(2.3-1\right)}\)
\(\Leftrightarrow B=\dfrac{6}{\left(2.3\right).5}\)
\(\Leftrightarrow B=\dfrac{6}{6.5}\)
\(\Leftrightarrow B=\dfrac{1}{5}\)
Vậy....
Ta có: \(M=\dfrac{\dfrac{1}{99}+\dfrac{2}{98}+\dfrac{3}{97}+\dfrac{4}{96}+...+\dfrac{97}{3}+\dfrac{98}{2}+\dfrac{99}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{100}}\)
\(=\dfrac{\left(1+\dfrac{1}{99}\right)+\left(1+\dfrac{2}{98}\right)+\left(1+\dfrac{3}{97}\right)+\left(1+\dfrac{4}{96}\right)+...+\left(1+\dfrac{98}{2}\right)+1}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{100}}\)
\(=\dfrac{\dfrac{100}{99}+\dfrac{100}{98}+\dfrac{100}{97}+...+\dfrac{100}{1}+\dfrac{100}{2}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{100}}\)
=100
Ta có: \(N=\dfrac{92-\dfrac{1}{9}-\dfrac{2}{10}-\dfrac{3}{11}-...-\dfrac{90}{98}-\dfrac{91}{99}-\dfrac{92}{100}}{\dfrac{1}{45}+\dfrac{1}{50}+\dfrac{1}{55}+...+\dfrac{1}{495}+\dfrac{1}{500}}\)
\(=\dfrac{\left(1-\dfrac{1}{9}\right)+\left(1-\dfrac{2}{10}\right)+\left(1-\dfrac{3}{11}\right)+...+\left(1-\dfrac{90}{98}\right)+\left(1-\dfrac{91}{99}\right)+\left(1-\dfrac{92}{100}\right)}{\dfrac{1}{5}\left(\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)}\)
\(=\dfrac{\dfrac{8}{9}+\dfrac{8}{10}+\dfrac{8}{11}+...+\dfrac{8}{99}+\dfrac{8}{100}}{\dfrac{1}{5}\left(\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)}\)
\(=\dfrac{8}{\dfrac{1}{5}}=40\)
\(\Leftrightarrow\dfrac{M}{N}=\dfrac{100}{40}=\dfrac{5}{2}\)
a) = (\(-\dfrac{141}{20}\)- \(\dfrac{1}{4}\)) : (-5) + \(\dfrac{1}{15}\) - \(\dfrac{1}{15}\)
= \(-\dfrac{73}{10}\) : - 5
= \(\dfrac{73}{50}\)
b) = \(\left(\dfrac{3}{25}-\dfrac{28}{25}\right)\). \(\dfrac{7}{3}\) : \(\left(\dfrac{7}{2}-\dfrac{11}{3}.14\right)\)
= \(-\dfrac{7}{3}\) . \(-\dfrac{6}{287}\)
= \(\dfrac{2}{41}\)
Nhận xét nè: ở mẫu số tại các phân số có tử số + mẫu số = 2012. Vì vậy mục tiêu là tạo ra con 2012 ở các phân số của mẫu số. E xử con tử số ở phân số mẫu số sao cho ra con 2012 là được =))
\(A=\left(3-\dfrac{1}{4}+\dfrac{3}{2}\right)-\left(5+\dfrac{1}{3}-\dfrac{5}{6}\right)-\left(6-\dfrac{7}{4}+\dfrac{2}{3}\right)\\ \Rightarrow A=3-\dfrac{1}{4}+\dfrac{3}{2}-5-\dfrac{1}{3}+\dfrac{5}{6}-6+\dfrac{7}{4}-\dfrac{2}{3}\\ \Rightarrow A=\left(3-5-6\right)-\left(\dfrac{1}{4}+\dfrac{7}{4}\right)+\left(\dfrac{3}{2}+\dfrac{5}{6}-\dfrac{2}{3}\right)\\ \Rightarrow A=-8-\dfrac{3}{2}+\dfrac{5}{3}\\ =-\dfrac{47}{6}.\\ B=0,5+\dfrac{1}{3}+0,4+\dfrac{5}{7}+\dfrac{1}{6}-\dfrac{4}{35}+\dfrac{1}{41}\)
\(\Rightarrow B=\left(0,5+0,4\right)+\left(\dfrac{1}{3}+\dfrac{1}{6}\right)+\left(\dfrac{5}{7}-\dfrac{4}{35}\right)+\dfrac{1}{41}\\ \Rightarrow B=\dfrac{9}{10}+\dfrac{1}{2}+\dfrac{3}{5}+\dfrac{1}{41}\\ \Rightarrow B=2+\dfrac{1}{41}\\ \Rightarrow B=\dfrac{83}{41}.\)
\(A=-\dfrac{1}{3}+\dfrac{1}{3^2}-...-\dfrac{1}{3^{99}}+\dfrac{1}{3^{100}}\)
\(=\dfrac{1}{3}\left(-1+\dfrac{1}{3}\right)+\dfrac{1}{3^3}\left(-1+\dfrac{1}{3}\right)+...+\dfrac{1}{3^{99}}\left(-1+\dfrac{1}{3}\right)\)
\(=\dfrac{-2}{3}\left(\dfrac{1}{3}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\right)\)
Ta có:
\(B=\dfrac{1}{3}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\)
\(9B=3+\dfrac{1}{3}+...+\dfrac{1}{3^{97}}\)
\(9B-B=3-\dfrac{1}{3^{99}}\)
\(B=\dfrac{3-\dfrac{1}{3^{99}}}{8}\)
\(A=-\dfrac{2}{3}B=\dfrac{-2}{3}.\dfrac{3-\dfrac{1}{99}}{8}=\dfrac{\dfrac{1}{3^{100}}-1}{4}\)
\(\Leftrightarrow A=\frac{1}{2}+\frac{1}{2\cdot7}+\frac{1}{7.5}+\frac{1}{5.13}+\frac{1}{13.8}+\frac{1}{8.19}\)
\(\Rightarrow A=\frac{1}{2}+\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{5}+\frac{1}{5}-\frac{1}{13}+\frac{1}{13}-\frac{1}{8}+\frac{1}{8}-\frac{1}{19}\)
\(\Rightarrow A=-\frac{1}{19}\)
\(A=-\dfrac{1}{5}+\dfrac{1}{5^2}-\dfrac{1}{5^3}+\dfrac{1}{5^4}-...-\dfrac{1}{5^{99}}+\dfrac{1}{5^{100}}\)
\(=-\dfrac{1}{5}\left(1-\dfrac{1}{5}\right)-\dfrac{1}{5^3}\left(1-\dfrac{1}{5}\right)-...-\dfrac{1}{5^{99}}\left(1-\dfrac{1}{5}\right)\)
\(=\left(1-\dfrac{1}{5}\right)\left(-\dfrac{1}{5}-\dfrac{1}{5^3}-...-\dfrac{1}{5^{99}}\right)\)
\(=\left(\dfrac{1}{5}-1\right)\left(\dfrac{1}{5}+\dfrac{1}{5^3}+...+\dfrac{1}{5^{99}}\right)\)
Mặt khác:
\(F=\dfrac{1}{5}+\dfrac{1}{5^3}+...+\dfrac{1}{5^{99}}\)
\(25F=5+\dfrac{1}{5}+...+\dfrac{1}{5^{97}}\)
\(25F-F=5-\dfrac{1}{5^{99}}\)
\(F=\dfrac{5-\dfrac{1}{5^{99}}}{24}\)
\(\Rightarrow A=\left(\dfrac{1}{5}-1\right).F\)
\(=\dfrac{-4}{5}.\dfrac{5-\dfrac{1}{5^{99}}}{24}=\dfrac{\dfrac{1}{5^{99}}-5}{5.6}=\dfrac{\dfrac{1}{5^{100}}-1}{6}\)