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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
a: \(=\left(\dfrac{5}{15}-\dfrac{12}{9}\right)+\left(\dfrac{14}{15}+\dfrac{11}{25}\right)+\dfrac{2}{7}\)
\(=\left(\dfrac{1}{3}-\dfrac{4}{3}\right)+\dfrac{70+33}{75}+\dfrac{2}{7}\)
\(=-1+\dfrac{2}{7}+\dfrac{103}{75}=\dfrac{-5}{7}+\dfrac{103}{75}=\dfrac{346}{525}\)
b: \(4\cdot\left(-\dfrac{1}{2}\right)^3+\dfrac{1}{2}\)
\(=4\cdot\dfrac{-1}{8}+\dfrac{1}{2}=\dfrac{-1}{2}+\dfrac{1}{2}=0\)
c: \(\dfrac{10^3+5\cdot10^2+5^3}{6^3+3\cdot6^2+3^3}=\dfrac{5^3\cdot8+5\cdot5^2\cdot2^2+5^3}{3^3\cdot2^3+3\cdot2^2\cdot3^2+3^3}\)
\(=\dfrac{5^3\left(8+4+1\right)}{3^3\left(8+4+1\right)}=\dfrac{125}{27}\)
e: \(\dfrac{2^8\cdot9^2}{6^4\cdot8^2}=\dfrac{2^8\cdot3^4}{3^4\cdot2^4\cdot2^6}=\dfrac{1}{4}\)
\(\dfrac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}=\dfrac{2^{10}.3^8-2.3^9.2^9}{2^{10}.3^8+2^8.3^8.2^2.5}=\dfrac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}\)
\(=\dfrac{2^{10}.\left(3^8-3^9\right)}{2^{10}.3^8.\left(1+5\right)}=\dfrac{3^8-3^9}{3^8.6}=\dfrac{3^8.\left(1-3\right)}{3^8.6}=\dfrac{-2}{6}=-\dfrac{1}{3}\)
~ Học tốt ~
Bài 1:
1) \(3^2.\dfrac{1}{243}.81^2.\dfrac{1}{3^3}\)
\(=3^2.\left(\dfrac{1}{3}\right)^5.\left(3^4\right)^2.\dfrac{1}{3^3}\)
\(=3^2.\dfrac{1}{3^5}.3^8.\dfrac{1}{3^3}\)
\(=3^2=9\)
2) \(\left(4.2^5\right):\left(2^3.\dfrac{1}{16}\right)\)
\(=\left(2^2.2^5\right):[2^3.\left(\dfrac{1}{2}\right)^4]\)
\(=2^7:2^3:\dfrac{1}{2^4}\)
\(=2^4.2^4=256\)
3)\(\left(2^{-1}+3^{-1}\right)+\left(2^{-1}.2^0\right):2^3\)
\(=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{2}.1:2^3\)
\(=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{2^4}\)
\(=\dfrac{43}{48}\)
4)\(\left(-\dfrac{1}{3}\right)^{-1}-\left(-\dfrac{6}{7}\right)^0+\left(\dfrac{1}{2}\right)^2:2\)
\(=-3-1+\dfrac{1}{4}.\dfrac{1}{2}\)
\(=-3-1+\dfrac{1}{8}\)
\(=-4+\dfrac{1}{8}\\ \)
\(=-\dfrac{31}{8}\)
5)\([\left(0,1\right)^2]^0+[\left(\dfrac{1}{7}\right)^{-1}]^2.\dfrac{1}{49}.[\left(2^2\right)^3:2^5]\\ =1+7^2.\dfrac{1}{7^2}.2^6:2^5\\ =1+1.2\\ =3\)
Chúc bạn học tốt 
a: x>-3/5 nên x+3/5>0
x<1/7 nên x-1/7<0
A=1/7-x-x-3/5+4/5=-2x+12/35
b: B=|x-1/7|+|x+3/5|-1/3
x>-3/5 nên x+3/5>0
x<1/7 nên x-1/7<0
B=1/7-x+3/5+x-1/3=43/105
\(\left(\dfrac{1}{2}-2+\dfrac{2}{3}\right)-\left(7+\dfrac{1}{2}-\dfrac{2}{3}\right)-\left(\dfrac{1}{2}-\dfrac{2}{3}-5\right)\)
\(=\dfrac{1}{2}-2+\dfrac{2}{3}-7-\dfrac{1}{2}+\dfrac{2}{3}-\dfrac{1}{2}+\dfrac{2}{3}+5\)
\(=\left(\dfrac{1}{2}-\dfrac{1}{2}-\dfrac{1}{2}\right)+\left(\dfrac{2}{3}+\dfrac{2}{3}+\dfrac{2}{3}\right)-\left(2+7-5\right)\)
\(=-\dfrac{1}{2}+2-4\)
\(=-\dfrac{1}{2}-2\)
\(=-\dfrac{1}{2}-\dfrac{4}{2}\)
\(=-\dfrac{5}{2}\)
a: \(=-\dfrac{1}{15}x^6y\)
b: \(=\dfrac{4}{5}ab^5\cdot2x^3y\cdot\left(-y\right)=-\dfrac{8}{5}ab^5\cdot x^3y^2\)
c: \(=-16\cdot\dfrac{3}{4}v^3\cdot\dfrac{-2}{5}uv=\dfrac{24}{5}v^4u\)
d: \(=8\cdot\left(-64\right)\cdot5\cdot u^2v^2\cdot\left(-27\right)v^3=69120u^2v^5\)
e: \(=-10y\cdot8y^3z^3\cdot25z^2=-2000y^4z^5\)
\(D=1-\dfrac{2}{5\cdot10}-\dfrac{2}{10\cdot15}-\dfrac{2}{15\cdot20}-...-\dfrac{2}{2020\cdot2025}\)
\(D=1-\left(\dfrac{2}{5\cdot10}+\dfrac{2}{10\cdot15}+\dfrac{2}{15\cdot20}+...+\dfrac{2}{2020\cdot2025}\right)\)
Đặt \(A=\dfrac{2}{5\cdot10}+\dfrac{2}{10\cdot15}+\dfrac{2}{15\cdot20}+...+\dfrac{2}{2020\cdot2025}\)
\(A=\dfrac{2}{5}\cdot\left(\dfrac{1}{5}-\dfrac{1}{10}\right)+\dfrac{2}{5}\cdot\left(\dfrac{1}{10}-\dfrac{1}{15}\right)+\dfrac{2}{5}\cdot\left(\dfrac{1}{15}-\dfrac{1}{20}\right)+...+\dfrac{2}{5}\cdot\left(\dfrac{1}{2020}-\dfrac{1}{2025}\right)\)
\(A=\dfrac{2}{5}\cdot\left(\dfrac{1}{5}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{15}+\dfrac{1}{15}-\dfrac{1}{20}+...+\dfrac{1}{2020}-\dfrac{1}{2025}\right)\)
\(A=\dfrac{2}{5}\cdot\left(\dfrac{1}{5}-\dfrac{1}{2025}\right)\)
\(A=\dfrac{2}{5}\cdot\dfrac{404}{2025}\)
\(A=\dfrac{808}{10125}\)
Thay vào D được:
\(D=1-\dfrac{808}{10125}\)
\(D=\dfrac{9317}{10125}\)
Vậy \(D=\dfrac{9317}{10125}\)