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Hai câu a) và b) bạn chỉ cần xem số mũ rồi trừ số mũ là xong
\(c)\) \(\frac{2^{10}.3^{10}-2^{10}.3^9}{2^9.3^{10}}=\frac{2^{10}.3^9\left(3-1\right)}{2^9.3^{10}}=\frac{2^{10}.3^9.2}{2^9.3^{10}}=\frac{2^{11}.3^9}{2^9.3^{10}}=\frac{2^2}{3}=\frac{4}{3}\)
\(d)\) \(\frac{5^{11}.7^{12}+5^{11}.7^{11}}{5^{12}.7^{12}+9.5^{11}.7^{11}}=\frac{5^{11}.7^{11}\left(7+1\right)}{5^{11}.7^{11}\left(5.7+9\right)}=\frac{8}{35+9}=\frac{8}{44}=\frac{2}{11}\)
Chúc bạn học tốt
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nhìu thế
đăng từng câu một mới có cơ có người trả lời đó bạn
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(-\right)\frac{1989\cdot1990+3970}{1992\cdot1991+3984}=\frac{1989\cdot\left(1990+2\right)}{1992\cdot\left(1991+2\right)}=\frac{1989}{1993}\)
\(\left(-\right)\frac{3^{10}\cdot\left(-5\right)^{21}}{\left(-5\right)^{20}\cdot3^{12}}=\frac{3^{10}\cdot\left(-5\right)^{20}\cdot\left(-5\right)}{3^{10}\cdot\left(-5\right)^{20}\cdot3^2}=-\frac{5}{9}\)
\(\left(-\right)\frac{\left(-11\right)^5\cdot13^7}{11^5\cdot13^8}=\frac{11^5\cdot13^7\cdot\left(-1\right)}{11^5\cdot13^7\cdot13}=-\frac{1}{13}\)
\(\left(-\right)\frac{2^{10}\cdot3^{10}-2^{10}\cdot3^9}{2^9\cdot3^{10}}=\frac{2^{10}\cdot3^9\left(3-1\right)}{2^9\cdot3^{10}}=\frac{2^{11}\cdot3^9}{2^9\cdot3^{10}}=\frac{4}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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:
\(\dfrac{24.1000}{2500-500}=\dfrac{24.1000}{2.1000}=12\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)\(\dfrac{-10}{11}.\dfrac{8}{9}+\dfrac{7}{18}.\dfrac{10}{11}\)
=\(\dfrac{10}{11}.\dfrac{-8}{9}+\dfrac{7}{18}.\dfrac{10}{11}\)
=\(\dfrac{10}{11}(\dfrac{-8}{9}+\dfrac{7}{18})\)
=\(\dfrac{10}{11}.\dfrac{-1}{2}\)
=\(\dfrac{-5}{11}\)
b;
B = \(\dfrac{3}{14}\) : \(\dfrac{1}{28}\) - \(\dfrac{13}{21}\): \(\dfrac{1}{28}\) + \(\dfrac{29}{42}\) : \(\dfrac{1}{28}\) - 8
B = (\(\dfrac{3}{14}\) - \(\dfrac{13}{21}\) + \(\dfrac{29}{42}\)) - 8
B = (\(\dfrac{9}{42}\) - \(\dfrac{26}{42}\) + \(\dfrac{29}{42}\)) - 8
B = (\(\dfrac{-17}{42}\) + \(\dfrac{29}{42}\)) - 8
B = \(\dfrac{2}{7}\) - 8
B = \(\dfrac{2}{7}-\dfrac{56}{7}\)
B = - \(\dfrac{54}{7}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{-11^5.13^7}{11^5.13^8}=\frac{\left(-1\right).11^5.13^7}{11^5.13^7.13}=\frac{-1}{13}\)
\(\frac{2^{10}.3^{10}-2^{10}.3^9}{2^9.3^{10}}=\frac{2^{10}.\left(3^{10}-3^9\right)}{2^9.3^{10}}=\frac{2^9.2.3^9.\left(3-1\right)}{2^9.3.3^9}=\frac{2.2}{3}=\frac{4}{3}\)
a, \(\frac{-11^5.13^7}{11^5.13^8}=\frac{-1}{13}\)
b, \(\frac{2^{10}.3^{10}-2^{10}.3^9}{2^9.3^{10}}=\frac{2^{10}.3^9.\left(1-3\right)}{2^9.3^{10}}=\frac{-4}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
mình gõ hơi khó nhìn nhưng bài giải này là chính xác 100%
a,3^10.(-5)^21/(-50)^20.3^12
=3^10.(-5)^20.(-5)/(-5)^20.3^10.3^2
=-5/3^2=-5/9
câu b, bạn tách ra làm tương tự
a)\(\dfrac{3^{10}.\left(-5\right)^{21}}{\left(-5\right)^{20}.3^{12}}=\dfrac{\left(-5\right)}{9}\)
b)\(\dfrac{5^{11}.7^{12}+5^{11}.7^{11}}{5^{12}.7^{12}+9.5^{11}.7^{11}}=\dfrac{7.35^{11}+35^{11}}{35^{12}+9.35^{11}}=\dfrac{8.35^{11}}{44.35^{11}}=\dfrac{2}{11}\)
a. \(\dfrac{3^{10}.\left(-5\right)^{21}}{\left(-5\right)^{20}.3^{12}}\) = \(\dfrac{3^{10}.\left(-5\right)^{20}.\left(-5\right)}{\left(-5\right)^{20}.3^{10}.3^2}\) = \(\dfrac{-5}{3^2}\)= \(\dfrac{-5}{9}\)
b. \(\dfrac{-11^5.13^7}{11^5.13^8}\) = \(\dfrac{-11^5.13^7}{11^5.13^7.13}\)= \(\dfrac{-1}{13}\)
c. \(\dfrac{2^{10}.3^{10}-2^{10}.3^9}{2^9.3^{10}}\)= \(\dfrac{2^{10}\left(3^{10}-3^9\right)}{2^9.3^{10}}\)= \(\dfrac{2^{10}.3}{2^9.3^{10}}\)= \(\dfrac{2^9.2.3}{2^9.3.3^9}\)= \(\dfrac{2}{3^9}\)=\(\dfrac{2}{19683}\)