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\(1+\frac{1}{1+2}+\frac{1}{1+2+3}+....+\frac{1}{1+2+3+...+2015}\)
\(=\frac{2}{1.2}+\frac{1}{\frac{\left(1+2\right).2}{2}}+\frac{1}{\frac{\left(1+2+3\right).3}{2}}+.....+\frac{1}{\frac{\left(2015+1\right).2015}{2}}\)
\(=\frac{2}{1.2}+\frac{2}{2.3}+....+\frac{2}{2015.2016}\)
\(\frac{1}{3.8}+\frac{1}{8.13}+...+\frac{1}{2018.2023}\)
Ta có : \(\frac{1}{3.8}+\frac{1}{8.13}+...+\frac{1}{2018.2023}\)
\(=\frac{1}{5}.\left(\frac{5}{3.8}+\frac{5}{8.13}+...+\frac{5}{2018.2023}\right)\)
\(=\frac{1}{5}.\left(\frac{1}{3}-\frac{1}{8}+\frac{1}{8}-\frac{1}{13}+...+\frac{1}{2018}-\frac{1}{2023}\right)\)
\(=\frac{1}{5}.\left(\frac{1}{3}-\frac{1}{2023}\right)\)
\(=\frac{1}{5}.\frac{2020}{6069}=\frac{404}{6069}\)
\(a)\) \(427-98=329\)
\(b)\) \(2\cdot19\cdot15+3\cdot43\cdot10+62\cdot80\)
\(=\left(2\cdot15\right)\cdot19+\left(3\cdot10\right)\cdot43+62\cdot80\)
\(=30\cdot19+30\cdot43+62\cdot80\)
\(=30\cdot\left(19+43\right)+62\cdot80\)
\(=30\cdot62+62\cdot80\)
\(=62\cdot\left(30+80\right)\)
\(=62\cdot110=6820\)
\(c)\) Đặt \(M=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)
\(=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}+\frac{1}{3^6}\)
\(\Rightarrow3M=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}\)
\(\Rightarrow3M-M=\left(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}+\frac{1}{3^6}\right)\)
\(\Rightarrow2M=1-\frac{1}{3^6}\)
\(\Rightarrow M=\frac{728}{2\cdot729}=\frac{364}{729}\)
Vậy \(M=\frac{364}{729}\)
(1/12+3 1/6-30,75).x -8 = (3/5+0,415+1/200):0,01
(1/12+19/6-123/4).x-8=(3/5+83/200+1/200):1/100
-55/2.x-8=51/50:1/100
-55/2.x-8=102
-55/2.x=102+8=110
x=110:-55/2=-4
\(a,A=\frac{1}{25\cdot27}+\frac{1}{27\cdot29}+...+\frac{1}{73\cdot75}\)
\(A=\frac{1}{2}\left[\frac{2}{25\cdot27}+\frac{2}{27\cdot29}+...+\frac{2}{73\cdot75}\right]\)
\(A=\frac{1}{2}\left[\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+...+\frac{1}{73}-\frac{1}{75}\right]\)
\(A=\frac{1}{2}\left[\frac{1}{25}-\frac{1}{75}\right]=\frac{1}{2}\cdot\frac{2}{75}=\frac{1}{75}\)
\(b,B=\frac{1}{8\cdot11}+\frac{1}{11\cdot14}+\frac{1}{14\cdot17}+...+\frac{1}{197\cdot200}\)
\(3B=\frac{3}{8\cdot11}+\frac{3}{11\cdot14}+\frac{3}{14\cdot17}+...+\frac{3}{197\cdot200}\)
\(3B=\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{197}-\frac{1}{200}\)
\(3B=\frac{1}{8}-\frac{1}{200}\)
\(3B=\frac{3}{25}\)
\(B=\frac{3}{25}:3=\frac{1}{25}\)
#)Giải :
a, \(A=\frac{1}{25.27}+\frac{1}{27.29}+...+\frac{1}{73.75}\)
\(A=\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+...+\frac{1}{73}-\frac{1}{75}\)
\(A=\frac{1}{25}-\frac{1}{75}\)
\(A=\frac{2}{75}\)
b, \(B=\frac{1}{8.11}+\frac{1}{11.14}+\frac{1}{14.17}+...+\frac{1}{197.200}\)
\(B=\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+...+\frac{1}{197}-\frac{1}{200}\)
\(B=\frac{1}{8}-\frac{1}{200}\)
\(B=\frac{3}{25}\)
#~Will~be~Pens~#
E = 1/2*5 + 1/5*8 + 1/8*11 + ....... + 1/29*32
E = 1/2 - 1/5 +1/5 - 1/8 + 1/8 - 1/11 + ........+1/29 - 1/32
E = 1/2 - 1/32
E = 15/32
nếu đúng thì k mik với nha