a)A = 1 + 1/2 . (1 + 2) + 1/3 . (1 + 2 + 3) + 1/4 . (1 + 2 + 3 + 4) + ... + 1/200 . (1 + 2 + ... + 200)
b)B = \(\dfrac{1.5.6+2.10.12+4.20.24+9.45.54}{1.3.5+2.6.10+4.12.20+9.27.45}\)
c)C = \(5\dfrac{2}{5}\) + (-5,35) + \(\dfrac{1}{3}\) + (-2,46) + (-1,19) + \(\dfrac{4}{15}\)
d)D = \(\dfrac{25^{13}.9^6.24^7}{50^{12}.81^5.40^3}\)
e)E = (5/12 . 17) + (3/34 . 10) + (7/60 . 9) + (8/27 + 35)
f)F = \(\dfrac{5}{12.17}\) + \(\dfrac{3}{34.10}\) + \(\dfrac{7}{60.9}\) + \(\dfrac{8}{27.35}\)
a) Ta có: 1 + 2 + ... + n = `(n(n+1))/2`
\(A=1+\dfrac{1}{2}\left(1+2\right)+\dfrac{1}{3}\left(1+2+3\right)+...+\dfrac{1}{200}\left(1+2+3+...+200\right)\\ =1+\dfrac{1}{2}\cdot\dfrac{2\cdot\left(2+1\right)}{2}+\dfrac{1}{3}\cdot\dfrac{3\cdot\left(3+1\right)}{2}+...+\dfrac{1}{200}\cdot\dfrac{200\cdot\left(200+1\right)}{2}\\ =1+\dfrac{2+1}{2}+\dfrac{3+1}{2}+...+\dfrac{200+1}{2}\\ =1+\dfrac{3}{2}+\dfrac{4}{2}+...+\dfrac{201}{2}\\ =\dfrac{2+3+4+...+201}{2}\\ =\dfrac{\dfrac{201\cdot\left(201+1\right)}{2}-1}{2}\\ =\dfrac{\dfrac{201^2+201-2}{2}}{2}\\ =\dfrac{201^2+199}{4}\)
b)
\(B=\dfrac{1\cdot5\cdot6+2\cdot10\cdot12+4\cdot20\cdot24+9\cdot45\cdot54}{1\cdot3\cdot5+2\cdot6\cdot10+4\cdot12\cdot20+9\cdot27\cdot45}\\ =\dfrac{2\cdot\left(1\cdot3\cdot5+2\cdot6\cdot10+4\cdot12\cdot20+9\cdot27\cdot45\right)}{1\cdot3\cdot5+2\cdot6\cdot10+4\cdot12\cdot20+9\cdot27\cdot45}\\ =2\)
c)
\(C=5\dfrac{2}{5}+\left(-5,35\right)+\dfrac{1}{3}+\left(-2.46\right)+\left(-1,19\right)+\dfrac{4}{15}\\ =5+\dfrac{2}{5}+\left(-5,35+-2,46+-1,19\right)+\dfrac{4}{15}+\dfrac{1}{3}\\ =\left(\dfrac{2}{5}+\dfrac{4}{15}+\dfrac{1}{3}\right)+5\\ =\left(-9\right)+1+5=-3\)
d)
\(D=\dfrac{25^{13}\cdot9^6\cdot24^7}{50^{12}\cdot81^5\cdot40^3}\\ =\dfrac{\left(5^2\right)^{13}\cdot\left(3^2\right)^6\cdot3^7\cdot8^7}{25^{12}\cdot2^{12}\cdot\left(3^4\right)^5\cdot8^3\cdot5^3}\\ =\dfrac{5^{26}\cdot3^{12}\cdot3^7\cdot\left(2^3\right)^7}{\left(5^2\right)^{12}\cdot2^{12}\cdot3^{20}\cdot\left(2^3\right)^3\cdot5^3}\\ =\dfrac{5^{26}\cdot3^{19}\cdot2^{21}}{5^{27}\cdot2^{12}\cdot3^{20}\cdot2^9}\\ =\dfrac{5^{26}\cdot3^{19}\cdot2^{21}}{5^{27}\cdot2^{21}\cdot3^{20}}\\ =\dfrac{1}{5\cdot3}=\dfrac{1}{15}\)
e) Hình như e và f giống nhau
\(E=\dfrac{5}{12\cdot17}+\dfrac{3}{34\cdot10}+\dfrac{7}{60\cdot9}+\dfrac{8}{27\cdot35}\\ =\dfrac{5}{12\cdot17}+\dfrac{3}{17\cdot20}+\dfrac{7}{20\cdot27}+\dfrac{8}{27\cdot35}\\ =\dfrac{1}{12}-\dfrac{1}{17}+\dfrac{1}{17}-\dfrac{1}{20}+\dfrac{1}{20}-\dfrac{1}{27}+\dfrac{1}{27}-\dfrac{1}{35}\\ =\dfrac{1}{12}-\dfrac{1}{35}\\ =\dfrac{23}{420}\)
