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\(A=\dfrac{3}{4}\cdot\dfrac{8}{9}\cdot\dfrac{15}{16}\cdot...\cdot\dfrac{9999}{10000}\\ =\dfrac{1\cdot3}{2\cdot2}\cdot\dfrac{2\cdot4}{3\cdot3}\cdot\dfrac{3\cdot5}{4\cdot4}\cdot...\cdot\dfrac{99\cdot101}{100\cdot100}\\ =\dfrac{1\cdot3\cdot2\cdot4\cdot3\cdot5\cdot...\cdot99\cdot101}{2\cdot2\cdot3\cdot3\cdot4\cdot4\cdot...\cdot100\cdot100}\\ =\dfrac{\left(1\cdot2\cdot3\cdot...\cdot99\right)\cdot\left(3\cdot4\cdot5\cdot...\cdot101\right)}{\left(2\cdot3\cdot4\cdot...\cdot100\right)\cdot\left(2\cdot3\cdot4\cdot...\cdot100\right)}\\ =\dfrac{1\cdot101}{100\cdot2}\\ =\dfrac{101}{200}\)
\(C=\left(1+\dfrac{1}{1\cdot3}\right)\cdot\left(1+\dfrac{1}{2\cdot4}\right)\cdot\left(1+\dfrac{1}{3\cdot5}\right)\cdot...\left(1+\dfrac{1}{99\cdot101}\right)\\ =\left(\dfrac{1\cdot3}{1\cdot3}+\dfrac{1}{1\cdot3}\right)\cdot\left(\dfrac{2\cdot4}{2\cdot4}+\dfrac{1}{2\cdot4}\right)\cdot\left(\dfrac{3\cdot5}{3\cdot5}+\dfrac{1}{3\cdot5}\right)\cdot...\cdot\left(\dfrac{99\cdot101}{99\cdot101}+\dfrac{1}{99\cdot101}\right)\\ =\left(\dfrac{2^2-1}{1\cdot3}+\dfrac{1}{1\cdot3}\right)\cdot\left(\dfrac{3^2-1}{2\cdot4}+\dfrac{1}{2\cdot4}\right)\cdot\left(\dfrac{4^2-1}{3\cdot5}+\dfrac{1}{3\cdot5}\right)\cdot...\cdot\left(\dfrac{100^2-1}{99\cdot101}+\dfrac{1}{99\cdot101}\right)\\ =\dfrac{2^2}{1\cdot3}\cdot\dfrac{3^2}{2\cdot4}\cdot\dfrac{4^2}{3\cdot5}\cdot...\cdot\dfrac{100^2}{99\cdot101}\\ =\dfrac{2^2\cdot3^2\cdot4^2\cdot...\cdot100^2}{1\cdot3\cdot2\cdot4\cdot3\cdot5\cdot...\cdot99\cdot101}\\ =\dfrac{\left(2\cdot3\cdot4\cdot...\cdot100\right)\cdot\left(2\cdot3\cdot4\cdot...\cdot100\right)}{\left(1\cdot2\cdot3\cdot...\cdot99\right)\cdot\left(3\cdot4\cdot5\cdot...\cdot101\right)}\\ =\dfrac{100\cdot2}{1\cdot101}=\dfrac{200}{101}\)
A=\(\dfrac{3}{4}.\dfrac{8}{9}.....\dfrac{9999}{10000}\)
A=\(\dfrac{1.3}{2.2}.\dfrac{2.4}{3.3}.....\dfrac{99.101}{100.100}\)
A=\(\dfrac{1.2.3.....99}{2.3.4.....100}.\dfrac{3.4.....101}{2.3.4.....100}\)
A=\(\dfrac{1}{100}.\dfrac{101}{2}\)
A=\(\dfrac{101}{200}\)
\(A=\dfrac{1.3}{2.2}.\dfrac{2.4}{3.3}.\dfrac{3.5}{4.4}.....\dfrac{99.101}{100.100}\\ =\dfrac{1}{2}.\dfrac{101}{100}=\dfrac{101}{200}\)
\(B=\left(1-\dfrac{1}{4}\right)\left(1-\dfrac{1}{9}\right)...\left(1-\dfrac{1}{10000}\right)\\ =\dfrac{3}{4}.\dfrac{8}{9}...\dfrac{9999}{10000}\)
(làm như câu a)
a)
\(A=\dfrac{3}{4}.\dfrac{8}{9}...\dfrac{9999}{10000}\)
\(=\dfrac{1.3}{2.2}.\dfrac{2.4}{3.3}...\dfrac{99.101}{100.100}\)
\(=\dfrac{1.2...99}{2.3...100}.\dfrac{3.4...101}{2.3...100}\)
\(=\dfrac{1}{100}.\dfrac{101}{2}\)
\(=\dfrac{101}{200}\)
a)\(1-2+3-4+5-6+7-8+8-9+9-10\)
=\(\left(1-2\right)+\left(3-4\right)+\left(5-6\right)+\left(7-8\right)+\left(8-9\right)+\left(9-10\right)\)
\(=\left(-1\right)+\left(-1\right)+\left(-1\right)+\left(-1\right)+\left(-1\right)+\left(-1\right)\)
\(=\left(-1\right).6\)
\(=-6\)
b)\(1-2+3-4+...+99-100\)
\(=\left(1-2\right)+\left(3-4\right)+...+\left(99-100\right)\)}\(\left[\left(100-1\right):1+1\right]:2=50\)(cặp)
\(=\left(-1\right)+\left(-1\right)+\left(-1\right)+...+\left(-1\right)\)} 50 số (-1)
\(=\left(-1\right).50\)
\(=-50\)
c)\(1-3+5-7+9-11+13-15\)
\(=\left(1-3\right)+\left(5-7\right)+\left(9-11\right)+\left(13-15\right)\)
\(=\left(-2\right)+\left(-2\right)+\left(-2\right)+\left(-2\right)\)
\(=\left(-2\right).4\)
\(=-8\)
d)\(1-3+5-7+...-99+101\) (Đối với bài này, có vẻ đề sai, mình đã sửa lại rồi
\(=\left(1-3\right)+\left(5-7\right)+...+\left(97-99\right)+101\) } \(\left[\left(99-1\right):2+1\right]:2=25\)(cặp)
\(=\left(-2\right)+\left(-2\right)+\left(-2\right)+...+\left(-2\right)\) } 25 số (-2)
\(=\left(-2\right).25\)
\(=-50\)
e)\(-1-2-3-4-...-99-100\)
\(=\left(-1\right)+\left(-2\right)+\left(-3\right)+...+\left(-99\right)+\left(-100\right)\)
\(=\left[\left(-1\right)+\left(-100\right)\right]+\left[\left(-2\right)+\left(-99\right)\right]+...+\left[\left(-51\right)+\left(-50\right)\right]\) } \(\left[\left(100-1\right):1+1\right]:2=50\)(cặp) (phần này của đề bài, không thay được như (-100) hoặc (-1))
\(=\left(-100\right)+\left(-100\right)+\left(-100\right)+...+\left(-100\right)\)} 50 số (-100)
\(=\left(-100\right).50\)
\(=-5000\)
câu 1)
\(\dfrac{-12}{18}+\left(\dfrac{-21}{35}\right)=\dfrac{-19}{15}\)
câu 2)
\(-\dfrac{3}{21}+\dfrac{6}{42}=0\)
câu 3)
\(-\dfrac{18}{24}+\dfrac{15}{21}=-\dfrac{1}{28}\)
câu 4)
\(\dfrac{1}{6}+\dfrac{2}{5}=\dfrac{17}{30}\)
câu 5)
\(\dfrac{3}{5}+\left(-\dfrac{7}{4}\right)=-\dfrac{23}{20}\)
câu 6)
\(\left(-2\right)+\left(\dfrac{-5}{8}\right)=\dfrac{-21}{8}\)
câu 7)
\(\dfrac{1}{-8}+\left(-\dfrac{5}{9}\right)=-\dfrac{49}{72}\)
câu 8)
\(\dfrac{4}{13}+\dfrac{12}{39}=\dfrac{8}{13}\)
câu 9)
\(\dfrac{1}{21}+\dfrac{1}{28}=\dfrac{1}{12}\)
câu 10)
\(-\dfrac{3}{29}+\dfrac{16}{58}=\dfrac{5}{29}\)
câu 11)
\(\dfrac{8}{40}+\left(-\dfrac{36}{45}\right)=-\dfrac{3}{5}\)
câu 12)
\(-\dfrac{8}{18}+\left(-\dfrac{15}{27}\right)=-1\)
câu 13)
\(\dfrac{13}{30}+\left(-\dfrac{1}{5}\right)=\dfrac{7}{30}\)
câu 14)
\(\dfrac{2}{21}+\dfrac{1}{28}=\dfrac{11}{84}\)
câu 15)
\(5+\left(-\dfrac{3}{4}\right)=\dfrac{17}{4}\)
câu 16)
\(\dfrac{18}{24}+\dfrac{45}{-10}=-\dfrac{15}{4}\)
biết mỗi câu a
ukm thế cx dc