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Hai câu a) và b) bạn chỉ cần xem số mũ rồi trừ số mũ là xong
\(c)\) \(\frac{2^{10}.3^{10}-2^{10}.3^9}{2^9.3^{10}}=\frac{2^{10}.3^9\left(3-1\right)}{2^9.3^{10}}=\frac{2^{10}.3^9.2}{2^9.3^{10}}=\frac{2^{11}.3^9}{2^9.3^{10}}=\frac{2^2}{3}=\frac{4}{3}\)
\(d)\) \(\frac{5^{11}.7^{12}+5^{11}.7^{11}}{5^{12}.7^{12}+9.5^{11}.7^{11}}=\frac{5^{11}.7^{11}\left(7+1\right)}{5^{11}.7^{11}\left(5.7+9\right)}=\frac{8}{35+9}=\frac{8}{44}=\frac{2}{11}\)
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Rút gọn phân số:\(\frac{1989.1990+3978}{1992.1991-3984}\)
\(=\frac{1989.1990+1989.2}{1992.1991-1992.2}=\frac{1989.\left(1990+2\right)}{1992.\left(1991-2\right)}=\frac{1989.1992}{1992.1989}=1\)
học giỏi nha
\(\frac{1989\cdot1990+3978}{1992\cdot1991-3984}\)
\(=\frac{1989\cdot1990+1989\cdot2}{1992\cdot1991-1992\cdot2}\)
\(=\frac{1989\cdot\left(1990+2\right)}{1992\cdot\left(1991-2\right)}\)
\(=\frac{1989\cdot1992}{1992\cdot1989}=1\)
\(\frac{\left(-11\right)^5.13^7}{11^5.13^8}=\frac{\left(-11\right)^5.13^7}{11^5.13^7}.\frac{1}{13}=-1.\frac{1}{13}=-\frac{1}{13}\)
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. \(\dfrac{2^{10}.3^{10}-2^{10}.3^9}{2^9.3^{10}}\)= \(\dfrac{2^{10}\left(3^{10}-3^9\right)}{2^9.3^{10}}\)= \(\dfrac{2^{10}.3}{2^9.3^{10}}\)= \(\dfrac{2^9.2.3}{2^9.3.3^9}\)= \(\dfrac{2}{3^9}\)=\(\dfrac{2}{19683}\)
\(a.\frac{2\cdot\left(-13\right)\cdot9\cdot10}{\left(-3\right)\cdot4\cdot\left(-5\right)\cdot26}\)
\(=\frac{2\cdot\left(-13\right)\cdot3\cdot3\cdot2\cdot5}{\left(-3\right)\cdot2\cdot2\cdot\left(-5\right)\cdot13\cdot2}\)
\(=-\frac{3}{2}\)
b) \(\frac{2^3\cdot3^4}{2^2\cdot3^2\cdot5}=\frac{2\cdot3^2}{5}=\frac{2\cdot9}{5}=\frac{18}{5}\)
\(\frac{2^4\cdot5^2\cdot11^2\cdot7}{2^3\cdot5^3\cdot7^2\cdot11}=\frac{2\cdot1\cdot11\cdot1}{1\cdot5\cdot7\cdot1}=\frac{22}{35}\)
c) \(\frac{121\cdot75\cdot130\cdot169}{39\cdot60\cdot11\cdot198}=\frac{11\cdot11\cdot13\cdot10\cdot169}{13\cdot3\cdot6\cdot10\cdot11\cdot11\cdot6\cdot3}\)
\(=\frac{169}{3\cdot6\cdot6\cdot3}=\frac{169}{324}\)
d) \(\frac{1998\cdot1990+3978}{1992\cdot1991-3984}\)
1989.1990+3978 / 1992.1991-3984 = 1989.1990+1989x2 / 1992.1991-1992x2
= 1989.(1990=2) / 1992.(1991-2) = 1989.1992 / 1992.1989 = 1
\(\left(-\right)\frac{1989\cdot1990+3970}{1992\cdot1991+3984}=\frac{1989\cdot\left(1990+2\right)}{1992\cdot\left(1991+2\right)}=\frac{1989}{1993}\)
\(\left(-\right)\frac{3^{10}\cdot\left(-5\right)^{21}}{\left(-5\right)^{20}\cdot3^{12}}=\frac{3^{10}\cdot\left(-5\right)^{20}\cdot\left(-5\right)}{3^{10}\cdot\left(-5\right)^{20}\cdot3^2}=-\frac{5}{9}\)
\(\left(-\right)\frac{\left(-11\right)^5\cdot13^7}{11^5\cdot13^8}=\frac{11^5\cdot13^7\cdot\left(-1\right)}{11^5\cdot13^7\cdot13}=-\frac{1}{13}\)
\(\left(-\right)\frac{2^{10}\cdot3^{10}-2^{10}\cdot3^9}{2^9\cdot3^{10}}=\frac{2^{10}\cdot3^9\left(3-1\right)}{2^9\cdot3^{10}}=\frac{2^{11}\cdot3^9}{2^9\cdot3^{10}}=\frac{4}{3}\)