a) \(D=\left(2\dfrac{2}{15}\times\dfrac{9}{17}\times\dfrac{3}{32}\right)\div\left(-\dfrac{3}{17}\right)\)

\(D=\dfrac{32}{15}\times\dfrac{9}{17}\times\dfrac{3}{32}\times\dfrac{-17}{3}\)

\(D=\dfrac{-3}{5}\)

b) \(\dfrac{1}{2}-\dfrac{1}{3\times7}-\dfrac{1}{7\times11}-\dfrac{1}{11\times15}-\dfrac{1}{15\times19}-\dfrac{1}{19\times23}-\dfrac{1}{23\times27}\)

\(=\dfrac{1}{2}-\left(\dfrac{1}{3\times7}+\dfrac{1}{7\times11}+\dfrac{1}{11\times15}+\dfrac{1}{15\times19}+\dfrac{1}{19\times23}+\dfrac{1}{23\times25}\right)\)

\(=\dfrac{1}{2}-\left[\dfrac{1}{4}\left(\dfrac{4}{3\times7}+\dfrac{4}{7\times11}+\dfrac{4}{11\times15}+\dfrac{4}{15\times19}+\dfrac{4}{19\times23}+\dfrac{4}{23\times27}\right)\right]\)

\(=\dfrac{1}{2}-\left[\dfrac{1}{4}\left(\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{15}+\dfrac{1}{15}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{23}+\dfrac{1}{23}-\dfrac{1}{27}\right)\right]\)

\(=\dfrac{1}{2}-\left[\dfrac{1}{4}\left(\dfrac{1}{3}-\dfrac{1}{27}\right)\right]\)

\(=\dfrac{1}{2}-\left[\dfrac{1}{4}\left(\dfrac{9-1}{27}\right)\right]\)

\(=\dfrac{1}{2}-\dfrac{1}{4}\times\dfrac{8}{27}\)

\(=\dfrac{1}{2}-\dfrac{2}{27}\)

\(=.....\)

Đó đến đây bn tự lm nốt. Câu c bn lm tương tự.

Mình cho bn dạng này, nếu bn chưa biết (để lm câu c)

\(\dfrac{x}{y\left(y+x\right)}=\dfrac{x}{y}-\dfrac{x}{y+x}\)

Chúc bn học tốt

\(B=\frac{5}{5\cdot10}+\frac{5}{10\cdot15}+...+\frac{5}{95\cdot100}\)

\(B=\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+...+\frac{1}{95}-\frac{1}{100}\)

\(B=\frac{1}{5}-\frac{1}{100}\)

\(B=\frac{19}{100}\)

\(B=\frac{5}{5.10}+\frac{5}{10.15}+...+\frac{5}{95.100}\)

\(B=\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+...+\frac{1}{95}-\frac{1}{100}\)

\(B=\frac{1}{5}-\frac{1}{100}\)

\(B=\frac{19}{100}\)

\(\dfrac{1}{5.10}+\dfrac{1}{10.15}+...+\dfrac{1}{395.400}\\ =\dfrac{1}{5}\left(\dfrac{5}{5.10}+\dfrac{5}{10.15}+...+\dfrac{5}{395.400}\right)\\ =\dfrac{1}{5}\left(\dfrac{1}{5}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{15}+...+\dfrac{1}{395}-\dfrac{1}{400}\right)\\ =\dfrac{1}{5}\left(\dfrac{1}{5}-\dfrac{1}{400}\right)\\ =\dfrac{1}{5}.\dfrac{79}{400}\\ =\dfrac{79}{2000}\)

Cảm ơn

\(\dfrac{2}{5.10}+\dfrac{2}{10.15}+...+\dfrac{2}{995.1000}\\ =2\left(\dfrac{1}{5.10}+\dfrac{1}{10.15}+...+\dfrac{1}{995.1000}\right)\\ =\dfrac{2}{5}\left(\dfrac{5}{5.10}+\dfrac{5}{10.15}+...+\dfrac{5}{995.1000}\right)\\ =\dfrac{2}{5}\left(\dfrac{1}{5}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{15}+...+\dfrac{1}{995}-\dfrac{1}{1000}\right)\\ =\dfrac{2}{5}\left(\dfrac{1}{5}-\dfrac{1}{1000}\right)\)

\(=\dfrac{2}{5}.\dfrac{199}{1000}\\ =\dfrac{199}{2500}\)

5/5.10 + 5/10.15 + ... + 5/45.50

= 1/5 - 1/10 + 1/10 - 1/15 + ... + 1/45 - 1/50

= 1/5 - 1/50

= 9/50

\(\frac{5}{5\times10}+\frac{5}{10\times15}+...+\frac{5}{45\times50}\)

\(=\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+...+\frac{1}{45}-\frac{1}{50}\)

\(=\frac{1}{5}-\frac{1}{50}\)

\(=\frac{9}{50}\)

~Study well~

#Thạc_Trân

\(A=\dfrac{3}{5\cdot10}+\dfrac{3}{10\cdot15}+...+\dfrac{3}{95\cdot100}\)

\(=\dfrac{3}{5}\left(\dfrac{5}{5\cdot10}+\dfrac{5}{10\cdot15}+...+\dfrac{5}{95\cdot100}\right)\)

\(=\dfrac{3}{5}\left(\dfrac{1}{5}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{15}+...+\dfrac{1}{95}-\dfrac{1}{100}\right)\)

\(=\dfrac{3}{5}\left(\dfrac{1}{5}-\dfrac{1}{100}\right)\)\(=\dfrac{3}{5}\cdot\dfrac{19}{100}=\dfrac{57}{500}\)

\(A=\dfrac{3}{5.10}+\dfrac{3}{10.15}+.....+\dfrac{3}{95.100}\)

\(A=\dfrac{3}{5}\left(\dfrac{1}{5}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{15}+.....+\dfrac{1}{95}-\dfrac{1}{100}\right)\)

\(A=\dfrac{3}{5}\left(\dfrac{1}{5}-\dfrac{1}{100}\right)\)

\(=\dfrac{3}{5}.\dfrac{19}{100}=\dfrac{19}{500}\)

C=1/5.10+1/10.15+...+1/95.100

   = 5/5.10+5/10.15+...+5/95.100

   = 1/5-1/10+1/10-1/15+...+1/95-1/100

   = 1/5-1/100

   = 19/100

\(C=5\times\left(1+\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+..+\frac{1}{95}-\frac{1}{100}\right)\)

\(C=5\times\left(1-\frac{1}{100}\right)\)

\(C=5\times\frac{99}{100}\)

\(C=\frac{99}{20}\)

a: \(\frac{\left(-5\right)^{60}\cdot30^5}{15^5\cdot5^{61}}\)

\(=\frac{5^{60}}{5^{61}}\cdot\left(\frac{30}{15}\right)^5\)

\(=\frac{2^5}{5}=\frac{32}{5}\)

b: \(\frac{\left(-3\right)^{10}\cdot15^5}{25^3\cdot\left(-9\right)^7}=\frac{3^{10}\cdot3^5\cdot5^5}{5^6\cdot\left(-1\right)\cdot3^{14}}=\frac{-3^{15}}{3^{14}}\cdot\frac15=-\frac35\)

c: \(4^6\cdot9^5+6^9\cdot120\)

\(=2^{12}\cdot3^{10}+2^9\cdot3^9\cdot2^3\cdot3\cdot5\)

\(=2^{12}\cdot3^{10}+2^{12}\cdot3^{10}\cdot5=2^{12}\cdot3^{10}\cdot6=2^{13}\cdot3^{11}\)

\(8^4\cdot3^{12}-6^{11}\)

\(=2^{12}\cdot3^{12}-2^{11}\cdot3^{11}\)

\(=2^{11}\cdot3^{11}\left(2\cdot3-1\right)=2^{11}\cdot3^{11}\cdot5\)

Ta có: \(\frac{4^6\cdot9^5+6^9\cdot120}{8^4\cdot3^{12}-6^{11}}\)

\(=\frac{2^{13}\cdot3^{11}}{2^{11}\cdot3^{11}\cdot5}=\frac{2^2}{5}=\frac45\)

d: \(4^2\cdot25^2+32\cdot125\)

\(=2^4\cdot5^4+2^5\cdot5^3\)

\(=2^4\cdot5^3\left(5+2\right)=2^4\cdot5^3\cdot7\)

\(\frac{4^2\cdot25^2+32\cdot125}{2^3\cdot5^2}\)

\(=\frac{2^4\cdot5^3\cdot7}{2^3\cdot5^2}=2\cdot5\cdot7=10\cdot7=70\)