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a) \(\dfrac{11\cdot8-11\cdot3}{17-6}\)
\(=\dfrac{11\cdot\left(8-3\right)}{11}=5\)
b) \(\dfrac{24-12\cdot13}{12+4\cdot9}\)
\(=\dfrac{12\cdot\left(2-13\right)}{12\left(1+3\right)}=\dfrac{-11}{4}\)
d)
\(\dfrac{3^9.3^{20}.2^8}{3^{24}.243.2^6}\\ =\dfrac{3^{29}.2^6.2^2}{3^{24}.3^5.2^6}\\ =\dfrac{3^{29}.2^6.4}{3^{29}.2^6}\\ =4\)
e)
\(\dfrac{2^{15}.5^3.2^6.3^4}{8.2^{18}.81.5}\\ =\dfrac{2^{21}.5^3.3^4}{2^3.2^{18}3^4.5}\\ =\dfrac{2^{21}.5.5^2.3^4}{2^{21}.3^4.5}\\ =5^2\\ =25\)
f)
\(=\dfrac{24\left(315+561+124\right)}{\dfrac{\left(1+99\right).50}{2}-500}\\ =\dfrac{24.1000}{2500-500}\\ =12\)
\(a,\dfrac{-14.15}{21.\left(-10\right)}=\dfrac{-7.2.3.5}{7.3.\left(-2\right).5}=1\)
\(b,\dfrac{5.7-7.9}{7.2+6.7}=\dfrac{7\left(5-9\right)}{7\left(2+6\right)}=\dfrac{-4}{8}=-\dfrac{1}{2}\)
\(c,\dfrac{\left(-7\right).3+2.\left(-14\right)}{\left(-5\right).7-2.7}=\dfrac{-7.\left(3+4\right)}{7\left(-5-2\right)}\)
\(=\dfrac{\left(-7\right).7}{7.\left(-7\right)}=1\)
\(d,\dfrac{3^9.3^{20}.2^8}{3^{24}.243.2^6}=\dfrac{3^{29}.2^8}{3^{24}.3^5.2^6}=\dfrac{3^{29}.2^8}{3^{29}.2^6}=2^2=4\)
\(e,\dfrac{2^{15}.5^3.2^6.3^4}{8.2^{18}.81.5}=\dfrac{2^{21}.3^4.5^3}{2^{18}.2^3.3^4.5}=\dfrac{2^{21}.3^4.5^3}{2^{21}.3^4.5}=5^2=25\)
\(f,\dfrac{24.315+3.561.8+4.124.6}{1+3+5+...+97+99-500}\)
\(=\dfrac{24.315+24.561+24.124}{1+3+5+...+97+99-500}\)
\(=\dfrac{24\left(315+561+124\right)}{1+3+5+...+97+99-500}\)
\(=\dfrac{24.1000}{1+3+5+...+97+99-500}\) (1)
Đặt A = 1 + 3 + 5 + ... + 97 + 99
Số số hạng trong A là: (99 - 1) : 2 + 1 = 50 (số)
Tổng A bằng: (99 + 1) . 50 : 2 = 2500
Thay A = 2500 vào biểu thức (1), ta được:
\(\dfrac{24.1000}{2500-500}=\dfrac{24.1000}{2.1000}=12\)
\(\dfrac{4\cdot5\cdot36}{35\cdot9\cdot2}=\dfrac{2^2\cdot5\cdot2^2\cdot3^2}{5\cdot7\cdot3^2\cdot2}=\dfrac{2^3}{7}=\dfrac{8}{7}\left(C\right)\)
a. \(\dfrac{8,5-8,2}{16}=\dfrac{0,3}{16}=\dfrac{3}{160}\)
b. \(\dfrac{2\cdot14}{7\cdot8}=\dfrac{1\cdot2}{1\cdot4}=\dfrac{2}{4}=\dfrac{1}{2}\)
c. \(\dfrac{11\cdot4-11}{2-13}=\dfrac{11\left(4-1\right)}{-11}=\dfrac{1\cdot3}{-1}=-3\)
d. \(\dfrac{49+7\cdot49}{49}=\dfrac{49\cdot\left(1+7\right)}{49}=\dfrac{8}{1}=8\)
a: \(=\dfrac{13\cdot\left(9-2\right)}{13}=7\)
b: \(=\dfrac{14\left(3-8\right)}{7\left(1+3\cdot3\right)}=2\cdot\dfrac{-5}{10}=-1\)
c: \(=\dfrac{54-72}{36}=\dfrac{-18}{36}=-\dfrac{1}{2}\)
d: \(=\dfrac{5^3}{10^2\cdot5}=\dfrac{5^2}{100}=\dfrac{1}{4}\)
b, B = \(\dfrac{1}{2}\) - \(\dfrac{1}{2^2}\) + \(\dfrac{1}{2^3}\) - \(\dfrac{1}{2^4}\)+.....+ \(\dfrac{1}{2^{99}}\) - \(\dfrac{1}{2^{100}}\)
2 \(\times\) B = 1 + \(\dfrac{1}{2}\) + \(\dfrac{1}{2^2}\) - \(\dfrac{1}{2^3}\) + \(\dfrac{1}{2^4}\)-.......-\(\dfrac{1}{2^{99}}\)
2 \(\times\) B + B = 1 - \(\dfrac{1}{2^{100}}\)
3B = ( 1 - \(\dfrac{1}{2^{100}}\))
B = ( 1 - \(\dfrac{1}{2^{100}}\)) : 3
A = 1 + \(\dfrac{1}{3}\) + \(\dfrac{1}{3^2}\)+ \(\dfrac{1}{3^3}\)+......+ \(\dfrac{1}{3^{n-1}}\) + \(\dfrac{1}{3^n}\)
A\(\times\) 3 = 3 + 1 + \(\dfrac{1}{3}\) + \(\dfrac{1}{3^2}\) + \(\dfrac{1}{3^2}\)+....+ \(\dfrac{1}{3^{n-1}}\)
A \(\times\) 3 - A = 3 - \(\dfrac{1}{3^n}\)
2A = 3 - \(\dfrac{1}{3^n}\)
A = ( 3 - \(\dfrac{1}{3^n}\)) : 2
b, \(\dfrac{56\cdot2+56\cdot3}{15\cdot56}=\dfrac{56\cdot\left(2+3\right)}{15\cdot56}\)
=\(\dfrac{56\cdot5}{15\cdot56}=\dfrac{1}{3}\)
c, \(\dfrac{20082008}{20112011}=\dfrac{2008\cdot10001}{2011\cdot10001}=\dfrac{2008}{2011}\)
d, \(\dfrac{2+2}{217\cdot32+217+33}=\dfrac{4}{217\cdot\left(32+33\right)}\)
=\(\dfrac{4}{217\cdot65}=\dfrac{4}{14105}\)