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A = - 522 - { - 222 - [ - 122 - (100 - 522) + 2022] }
A = - 522 - { -222 - [- 122 - 100 + 522 ] + 2022}
A = - 522 - { -222 - { - 222 + 522 } + 2022}
A = - 522 - {- 222 + 222 - 522 + 2022}
A = -522 + 522 - 2022
A = - 2022
B = 1 + \(\dfrac{1}{2}\)(1 + 2) + \(\dfrac{1}{3}\).(1 + 2 + 3) + ... + \(\dfrac{1}{20}\).(1 + 2+ 3 + ... + 20)
B = 1+\(\dfrac{1}{2}\)\(\times\)(1+2)\(\times\)[(2-1):1+1]:2+ ... + \(\dfrac{1}{20}\)\(\times\) (20 + 1)\(\times\)[(20-1):1+1]:2
B = 1 + \(\dfrac{1}{2}\) \(\times\) 3 \(\times\) 2:2 + \(\dfrac{1}{3}\) \(\times\)4 \(\times\) 3 : 2+....+ \(\dfrac{1}{20}\) \(\times\)21 \(\times\) 20 : 2
B = 1 + \(\dfrac{3}{2}\) + \(\dfrac{4}{2}\) + ....+ \(\dfrac{21}{2}\)
B = \(\dfrac{2+3+4+...+21}{2}\)
B = \(\dfrac{\left(21+2\right)\left[\left(21-2\right):1+1\right]:2}{2}\)
B = \(\dfrac{23\times20:2}{2}\)
B = \(\dfrac{23\times10}{2}\)
B = 23
\(C=\dfrac{5\cdot2^{12}\cdot3^8-3^9\cdot2^{12}}{2^{15}\cdot3^8-2^{13}\cdot3^9}\)
\(=\dfrac{2^{12}\cdot3^8\left(5-3\right)}{2^{13}\cdot3^8\left(2^2-3\right)}=\dfrac{1}{2}\cdot\dfrac{2}{1}=1\)
\(C=\dfrac{5\times2^{12}\times3^8-3^9\times2^{12}}{2^2\times2^{13}\times3^8+2\times2^{12}\times\left(-3^9\right)}=\dfrac{3^8\times2^{12}\times\left(5-3\right)}{2^{15}\times3^8+2^{13}\times\left(-3\right)^9}\)
\(=\dfrac{3^8\times2^{12}\times2}{2^{13}\times3^8\times\left(4-3\right)}=\dfrac{1}{1}=1\)
\(#PaooNqoccc\)
a) Ta có: \(\dfrac{5}{8}+\dfrac{3}{17}+\dfrac{4}{18}+\dfrac{20}{-17}+\dfrac{-2}{9}+\dfrac{21}{56}\)
\(=\left(\dfrac{3}{17}-\dfrac{20}{17}\right)+\left(\dfrac{2}{9}-\dfrac{2}{9}\right)+\left(\dfrac{5}{8}+\dfrac{3}{8}\right)\)
\(=-1+1=0\)
b) Ta có: \(\left(\dfrac{9}{16}+\dfrac{8}{-27}\right)+\left(1+\dfrac{7}{16}+\dfrac{-19}{27}\right)\)
\(=\left(\dfrac{9}{16}+\dfrac{7}{16}\right)+\left(\dfrac{-8}{27}-\dfrac{19}{27}\right)+1\)
=1-1+1=1
\(=\dfrac{2^{12}\cdot3^5-2^{12}\cdot3^4}{2^{18}\cdot3^6+2^{12}\cdot3^5}-\dfrac{5^{10}\cdot7^3-5^{10}\cdot7^4}{5^9\cdot7^3+5^9\cdot2^3\cdot7^3}\)
\(=\dfrac{2^{12}\cdot3^4\cdot2}{2^{12}\cdot3^5\left(2^6\cdot3+1\right)}-\dfrac{5^{10}\cdot7^3\cdot\left(-6\right)}{5^9\cdot7^3\cdot9}\)
\(=\dfrac{2}{193}+\dfrac{5\cdot2}{3}=\dfrac{1936}{579}\)
\(a,15\dfrac{3}{13}-\left(3\dfrac{4}{7}+8\dfrac{3}{13}\right)=15\dfrac{3}{13}-3\dfrac{4}{7}-8\dfrac{3}{13}=\left(15\dfrac{3}{13}-8\dfrac{3}{13}\right)-\dfrac{25}{7}=7-\dfrac{25}{7}=\dfrac{49}{7}-\dfrac{25}{7}=\dfrac{24}{7}\)
\(b,\left(7\dfrac{4}{9}+4\dfrac{7}{11}\right)-3\dfrac{4}{9}=\left(7\dfrac{4}{9}-3\dfrac{4}{9}\right)+4\dfrac{4}{9}=4+\dfrac{40}{9}=\dfrac{36}{9}+\dfrac{40}{9}=\dfrac{76}{9}\)
\(c,\dfrac{-7}{9}.\dfrac{4}{11}+\dfrac{-7}{9}.\dfrac{7}{11}+5\dfrac{7}{9}=\dfrac{-7}{9}\left(\dfrac{4}{11}+\dfrac{7}{11}\right)+\dfrac{52}{9}=\dfrac{-7}{9}.1+\dfrac{52}{9}=\dfrac{-7}{9}+\dfrac{52}{9}=\dfrac{45}{9}=5\)
\(d,50\%.1\dfrac{1}{3}.10.\dfrac{7}{35}.0,75=\dfrac{1}{2}.\dfrac{4}{3}.10.\dfrac{1}{5}.\dfrac{3}{4}=\left(\dfrac{1}{2}.\dfrac{1}{5}.10\right).\left(\dfrac{4}{3}.\dfrac{3}{4}\right)=1.1=1\)
\(e,\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{40.43}=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{40}-\dfrac{1}{43}=1-\dfrac{1}{43}=\dfrac{42}{43}\)
\(A=\dfrac{5\cdot4^6\cdot9^4-3^9\cdot\left(-8\right)^4}{4\cdot2^{13}\cdot3^8+2\cdot8^4\cdot\left(-27\right)^3}\)
\(=\dfrac{5\cdot2^{12}\cdot3^8-3^9\cdot2^{12}}{2^{15}\cdot3^8-2^{13}\cdot3^9}\)
\(=\dfrac{2^{12}\cdot3^8\left(5-3\right)}{2^{13}\cdot3^8\left(2^2-3\right)}\)
\(=\dfrac{2^{13}}{2^{13}}=1\)
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