H=\(\frac{1\cdot2\cdot3+2\cdot4\cdot6+3\cdot6\cdot9+5\cdot10\cdot15}{1\cdot3\cdot6+2\cdot6\cdot12+3\cdot9\cdot18+5\cdot15\cdot30}=\frac{1.2.3+2^3.\left(1.2.3\right)+3^3.\left(1.2.3\right)+5^3.\left(1.2.3\right)}{1.3.6+2^3.\left(1.3.6\right)+3^3.\left(1.3.6\right)+5^3.\left(1.3.6\right)}=\frac{1.2.3.\left(1+2^3+3^3+5^3\right)}{1.3.6.\left(1+2^3+3^3+5^3\right)}=\frac{2}{6}=\frac{1}{3}\)

\(\frac{4}{3}\)

\(\frac{3^{17}\cdot81^{11}}{27^{10}\cdot9^{15}}\)

\(=\frac{3^{17}\cdot\left(3^4\right)^{11}}{\left(3^3\right)^{10}\cdot\left(3^2\right)^{15}}\)

\(=\frac{3^{17}\cdot3^{44}}{3^{30}\cdot3^{30}}\)

\(=\frac{3^{61}}{3^{60}}\)

\(=3\)

\(\frac{9^2\cdot2^{11}}{16^2\cdot6^3}\)

\(=\frac{\left(3^2\right)^2\cdot2^{11}}{\left(2^4\right)^2\cdot\left(2\cdot3\right)^3}\)

\(=\frac{3^4\cdot2^{11}}{2^8\cdot2^3\cdot3^3}\)

\(=\frac{3^4\cdot2^{11}}{2^{11}\cdot3^3}\)

\(=\frac{3^4}{3^3}\)

\(=3\)

\(\frac{2^{10}\cdot3^{31}+2^{40}\cdot3^6}{2^{11}\cdot3^{31}+2^{41}\cdot3^6}\)

\(=\frac{2^{10}\cdot3^{31}+2^{40}\cdot3^6}{2\cdot\left(2^{10}\cdot3^{31}+2^{40}\cdot3^6\right)}\)

\(=\frac{1}{2}\)

=\(\frac{6\left(1+8+27+64\right)}{12\left(1+16+54+128\right)}\)

=\(\frac{6.100}{12.199}\)

=\(\frac{50}{199}\)

Tk mình với nha mọi người!!!!!

\(\frac{1x2x3+2x4x6+3x6x9+4x8x12}{1x3x4+4x6x8+6x9x12+8x12x16}\)

\(\frac{6x\left(1+8+27+64\right)}{12x\left(1+16+54+128\right)}=\frac{6x100}{12x199}=\frac{50}{199}\)

Ta có : \(S=\frac{3}{2\cdot3}+\frac{3}{3\cdot6}+\frac{3}{4\cdot9}+...+\frac{3}{6039\cdot2014}\)

\(S=3\cdot\left(\frac{3}{6\cdot3}+\frac{3}{9\cdot6}+\frac{3}{12\cdot9}+...+\frac{3}{6039\cdot6042}\right)\)

\(S=3\cdot\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+...+\frac{1}{6039}-\frac{1}{6042}\right)\)

\(S=3\cdot\left(\frac{1}{3}-\frac{1}{6042}\right)\)

\(S=3\cdot\frac{671}{2014}\)

\(S=\frac{2013}{2014}\)

a, A= \(5\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\right)\)

\(A=5\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)

\(A=5\left(1-\dfrac{1}{100}\right)\)

\(A=5.\dfrac{99}{100}=\dfrac{99}{20}.\)

b, \(C=1.2.3+2.3.4+...+8.9.10\)

\(4C=1.2.3.4+2.3.4.\left(5-1\right)+...+8.9.10.\left(11-7\right)\)\(4C=1.2.3.4+2.3.4.5-1.2.3.4+...+8.9.10.11-7.8.9.10\)\(4C=8.9.10.11\)

\(C=\dfrac{8.9.10.11}{4}=1980.\)

c, https://hoc24.vn/hoi-dap/question/384591.html

Câu này bạn vào đây mình đã giải câu tương tự nhé.

\(1)A=\dfrac{5}{1.2}+\dfrac{5}{2.3}+...+\dfrac{5}{99.100}\)

\(\Leftrightarrow A=5\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)

\(\Leftrightarrow A=5\left(1-\dfrac{1}{100}\right)\)

\(\Leftrightarrow A=5\cdot\dfrac{99}{100}\)

\(\Leftrightarrow A=\dfrac{99}{20}\)

bài này không khó. Nhưng đánh máy để giải cho bạn thì thực sự khó

\(S=\frac{3}{2.3}+\frac{3}{3.6}+...+\frac{3}{2014.6039}\)

\(=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2013.2014}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2013}-\frac{1}{2014}\)

\(=1-\frac{1}{2014}=\frac{2013}{2014}\)

a.\(\frac{3\cdot4\cdot7}{12\cdot8\cdot9}\)= \(\frac{3\cdot4\cdot7}{3\cdot4\cdot8\cdot9}\)= \(\frac{7}{72}\) 

b. \(\frac{4\cdot5\cdot6}{12\cdot10\cdot8}\)= \(\frac{4\cdot5\cdot2\cdot3}{3\cdot4\cdot5\cdot2\cdot8}\)= \(\frac{1}{8}\) 

c.\(\frac{5\cdot6\cdot7}{12\cdot14\cdot15}\)= \(\frac{5\cdot6\cdot7}{2\cdot6\cdot2\cdot7\cdot3\cdot5}\)= \(\frac{1}{12}\)

=\(\frac{30\left(1+8+27+64+125\right)}{5\left(3+24+81+64+375\right)}\)

= \(\frac{30.225}{5.574}\)

=\(\frac{6750}{2870}\)

=\(\frac{675}{287}\)

K mình với!!!!