\(\dfrac{1.2.4+2.4.8+4.8.16+8.16.32}{1.3.4+2.6.8+4.12.16+8.24.32}\)

\(=\dfrac{8.\left(1+8+4.16+16.32\right)}{12.\left(1+8+4.16+16.32\right)}\)

\(=\dfrac{8}{12}=\dfrac{2}{3}\)

\(=\dfrac{8+8\cdot8+8\cdot64+8\cdot512}{12+12\cdot8+12\cdot64+12\cdot512}=\dfrac{8}{12}=\dfrac{2}{3}\)

 \(A=\frac{8056}{2012.16-1982}\)= \(\frac{2014.4}{2012.15+2012-1982}\)=\(\frac{2014.4}{2012.15+30}\)=\(\frac{2014.4}{2012.15+2.15}\)=\(\frac{2014.4}{15.\left(2012+2\right)}=\frac{2014.4}{15.2014}=\frac{4}{15}\)

B = \(\frac{1.2.6+2.4.12+4.8.24+7.14.42}{1.6.9+2.12.18+4.24.36+7.42.63}\)

= \(\frac{1.2.3.2+2.2.2.12+4.4.2.24+7.7.2.42}{1.2.3.9+2.12.2.9+4.24.4.9+7.42.7.9}\)

= \(\frac{2\left(1.2.3+2.2.12+4.4.24+7.7.42\right)}{9\left(1.2.3+2.2.12+4.4.24+7.7.42\right)}\)

= \(\frac{2}{9}\)

Ta có: \(\frac{4}{15}=\frac{4.3}{15.3}=\frac{12}{45};\frac{2}{9}=\frac{2.5}{9.5}=\frac{10}{45}\)

Vì \(\frac{12}{45}>\frac{10}{45}\Rightarrow\frac{4}{15}>\frac{2}{9}\Rightarrow A>B\)

Vậy A > B

\(A=\dfrac{8056}{2012.16-1982}\)

\(A=\dfrac{8056}{32192-1982}\)

\(A=\dfrac{8056}{30210}=\dfrac{12}{45}\)

\(B=\dfrac{1.2.6+2.4.12+4.8.24+7.14.42}{1.6.9+2.12.18+4.24.36+7.42.63}\)

\(B=\dfrac{12+96+768+4116}{54+432+3456+18522}\)

\(B=\dfrac{4992}{22464}=\dfrac{10}{45}\)
Vậy: \(\dfrac{12}{45}>\dfrac{10}{45}\Rightarrow A>B\)

1100444-88888=

a)\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\)

\(=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right)\)

\(=\frac{1}{2}.\left(\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+\frac{9-7}{7.9}+\frac{11-9}{9.11}\right)\)

\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)\)

\(=\frac{1}{2}.\left(1-\frac{1}{11}\right)\)

\(\frac{10}{22}\)

\(\dfrac{1\cdot2\cdot3+2\cdot4\cdot6+4\cdot8\cdot12}{1\cdot3\cdot5+2\cdot6\cdot10+4\cdot12\cdot20}\\ =\dfrac{1\cdot2\cdot3+2\cdot1\cdot2\cdot2\cdot2\cdot3+4\cdot1\cdot4\cdot2\cdot4\cdot3}{1\cdot3\cdot5+2\cdot1\cdot2\cdot3\cdot2\cdot5+4\cdot1\cdot4\cdot3\cdot4\cdot5}\\ =\dfrac{1\cdot2\cdot3\cdot\left(1+2^3+4^3\right)}{1\cdot3\cdot5\cdot\left(1+2^3+4^3\right)}\\ =\dfrac{1\cdot2\cdot3}{1\cdot3\cdot5}\\ =\dfrac{6}{15}\)

cảm ơn bạn nhiều lắm

Ta có: \(5^2\cdot6^{11}\cdot16^2+6^2\cdot12^6\cdot15^2\)

\(=5^2\cdot2^{11}\cdot3^{11}\cdot2^8+2^2\cdot3^2\cdot2^{12}\cdot3^6\cdot5^2\cdot3^2\)

\(=2^{19}\cdot3^{11}\cdot5^2+2^{14}\cdot3^{10}\cdot5^2\)

\(=2^{14}\cdot3^{10}\cdot5^2\left(2^5\cdot3+1\right)=2^{14}\cdot3^{10}\cdot5^2\cdot97\)

Ta có: \(2\cdot6^{12}\cdot10^4-81^2\cdot960^3\)

\(=2\cdot2^{12}\cdot3^{12}\cdot2^4\cdot5^4-3^8\cdot\left(2^6\cdot3\cdot5\right)^3\)

\(=2^{17}\cdot3^{12}\cdot5^4-2^{18}\cdot3^{11}\cdot5^3\)

\(=2^{17}\cdot3^{11}\cdot5^3\left(3\cdot5-2\right)=2^{17}\cdot3^{11}\cdot5^3\cdot13\)

Ta có: \(\frac{5^2\cdot6^{11}\cdot16^2+6^2\cdot12^6\cdot15^2}{2\cdot6^{12}\cdot10^4-81^2\cdot960^3}\)

\(=\frac{2^{14}\cdot3^{10}\cdot5^2\cdot97}{2^{17}\cdot3^{11}\cdot5^3\cdot13}=\frac{97}{13\cdot2^3\cdot3\cdot5}=\frac{97}{1560}\)

Bài 1: \(A=3\cdot\frac{1}{1\cdot2}-5\cdot\frac{1}{2\cdot3}+7\cdot\frac{1}{3\cdot4}-\cdots+15\cdot\frac{1}{7\cdot8}-17\cdot\frac{1}{8\cdot9}\)

\(=\frac{3}{1\cdot2}-\frac{5}{2\cdot3}+\frac{7}{3\cdot4}-\cdots+\frac{15}{7\cdot8}-\frac{17}{8\cdot9}\)

\(=1+\frac12-\frac12-\frac13+\frac13+\frac14-\cdots+\frac17+\frac18-\frac18-\frac19\)

\(=1-\frac19=\frac89\)

Bài 2:

\(A=\frac{1}{1\cdot300}+\frac{1}{2\cdot301}+\cdots+\frac{1}{101\cdot400}\)

\(=\frac{1}{299}\left(\frac{299}{1\cdot300}+\frac{299}{2\cdot301}+\cdots+\frac{299}{101\cdot400}\right)\)

\(=\frac{1}{299}\left(1-\frac{1}{300}+\frac12-\frac{1}{301}+\cdots+\frac{1}{101}-\frac{1}{400}\right)\)

\(=\frac{1}{299}\left(1+\frac12+\cdots+\frac{1}{101}-\frac{1}{300}-\frac{1}{301}-...-\frac{1}{400}\right)\)

\(B=\frac{1}{1\cdot102}+\frac{1}{2\cdot103}+\cdots+\frac{1}{299\cdot400}\)

\(=\frac{1}{101}\left(\frac{101}{1\cdot102}+\frac{101}{2\cdot103}+\cdots+\frac{101}{299\cdot400}\right)\)

\(=\frac{1}{101}\left(1-\frac{1}{102}+\frac12-\frac{1}{103}+\cdots+\frac{1}{299}-\frac{1}{400}\right)\)

\(=\frac{1}{101}\left(1+\frac12+\cdots+\frac{1}{101}-\frac{1}{300}-\frac{1}{301}-\cdots-\frac{1}{400}\right)\)

DO đó: \(\frac{A}{B}=\frac{1}{299}:\frac{1}{101}=\frac{101}{299}\)

\(=\dfrac{1}{2x1x3x2}+\dfrac{1}{2x2x3x3}+\dfrac{1}{2x3x3x4}+...+\dfrac{1}{2x18x3x19}+\dfrac{1}{2x19x3x20}=\)

\(=\dfrac{1}{2x3}x\left(\dfrac{1}{1x2}+\dfrac{1}{2x3}+\dfrac{1}{3x4}+...+\dfrac{1}{18x19}+\dfrac{1}{19x20}\right)=\)

\(=\dfrac{1}{6}x\left(\dfrac{2-1}{1x2}+\dfrac{3-2}{2x3}+\dfrac{4-3}{3x4}+...+\dfrac{20-19}{19x20}\right)=\)

\(=\dfrac{1}{6}x\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{19}-\dfrac{1}{20}\right)=\)

\(=\dfrac{1}{6}x\left(1-\dfrac{1}{20}\right)=\dfrac{1}{6}x\dfrac{19}{20}=\dfrac{19}{120}\)