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\(\left(-\right)\frac{1989\cdot1990+3970}{1992\cdot1991+3984}=\frac{1989\cdot\left(1990+2\right)}{1992\cdot\left(1991+2\right)}=\frac{1989}{1993}\)
\(\left(-\right)\frac{3^{10}\cdot\left(-5\right)^{21}}{\left(-5\right)^{20}\cdot3^{12}}=\frac{3^{10}\cdot\left(-5\right)^{20}\cdot\left(-5\right)}{3^{10}\cdot\left(-5\right)^{20}\cdot3^2}=-\frac{5}{9}\)
\(\left(-\right)\frac{\left(-11\right)^5\cdot13^7}{11^5\cdot13^8}=\frac{11^5\cdot13^7\cdot\left(-1\right)}{11^5\cdot13^7\cdot13}=-\frac{1}{13}\)
\(\left(-\right)\frac{2^{10}\cdot3^{10}-2^{10}\cdot3^9}{2^9\cdot3^{10}}=\frac{2^{10}\cdot3^9\left(3-1\right)}{2^9\cdot3^{10}}=\frac{2^{11}\cdot3^9}{2^9\cdot3^{10}}=\frac{4}{3}\)
\(a)\frac{3^{10}.\left(-5\right)^{21}}{\left(-5\right)^{20}.3^{12}}=\frac{-5}{3^2}=\frac{-5}{9}\)
\(b)\frac{-11.13^7}{11^5.13^8}=\frac{-1}{11^4.13}\) (Bạn xem thử xem có sai đề không nhé)
\(c)\frac{2^{10}.3^{10}-2^{10}.3^9}{2^9.3^{10}}=\frac{2^{10}.3^9\left(3+1\right)}{2^9.3^{10}}=\frac{2.4}{3}=\frac{8}{3}\)
\(d)\frac{5^{11}.7^{12}+5^{11}.7^{11}}{5^{12}.7^{12}+9.5^{11}.7^{11}}=\frac{5^{11}.7^{11}\left(7+1\right)}{5^{11}.7^{11}\left(5.4+9\right)}=\frac{8}{20+9}=\frac{8}{29}\)
\(a)\frac{3^{10}\cdot\left(-5\right)^{21}}{\left(-5\right)^{20}\cdot3^{12}}=\frac{-5}{3^2}=\frac{-5}{9}\)
\(b)\frac{\left(-11\right)\cdot13^7}{11^5\cdot13^8}=\frac{-1}{11^4\cdot13}=\frac{-1}{14641\cdot13}=\frac{-1}{190333}\)
\(c)\frac{2^{10}\cdot3^{10}-2^{10}\cdot3^9}{2^9\cdot3^{10}}=\frac{2^{10}\left(3^{10}-3^9\right)}{2^9\cdot3^{10}}=\frac{2^{10}\cdot3^9\left(3-1\right)}{2^9\cdot3^{10}}=\frac{2^{10}\cdot3^9\cdot2}{2^9\cdot3^{10}}=\frac{2\cdot2}{3}=\frac{4}{3}\)
d)
đặt A = 1 + 2 + 22 + ... + 280
2A = 2 + 22 + 23 + ... + 281
2A - A = ( 2 + 22 + 23 + ... + 281 ) - ( 1 + 2 + 22 + ... + 280 )
A = 281 - 1 > 281 - 2
e)
đặt \(A=\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+...+\frac{899}{900}\)
\(A=\left(1-\frac{1}{4}\right)+\left(1-\frac{1}{9}\right)+\left(1-\frac{1}{16}\right)+...+\left(1-\frac{1}{900}\right)\)
\(A=\left(1+1+1+...+1\right)-\left(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{900}\right)\)
\(A=29-\left(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{900}\right)\)
đặt \(B=\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{900}\)
\(B=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{30^2}\)
\(B< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{29.30}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{29}-\frac{1}{30}\)
\(=1-\frac{1}{30}=\frac{29}{30}< 1\)
\(\Rightarrow A< 29\)
So sánh C và D biết
C=1+13+13^2+...+13^13/1+13+13^2+...+13^12
D=1+11+11^2+...+11^13/1+11+11^2+...+11^12
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
Bài 1 : \(\frac{x-12}{4}=\frac{1}{2}\)
\(\Rightarrow2\cdot(x-12)=1\cdot4\)
\(\Rightarrow2x-24=4\)
\(\Rightarrow2x=28\)
\(\Rightarrow x=14\)
Vậy x = 14
Bài 2 : Rút gọn phân số
\(a,\frac{-315}{540}=\frac{-7}{12}\)
\(b,\frac{25\cdot13}{26\cdot35}=\frac{5\cdot1}{2\cdot7}=\frac{5}{14}\)
\(c,\frac{6\cdot9-2\cdot17}{63\cdot3-119}=\frac{54-34}{70}=\frac{20}{70}=\frac{2}{7}\)
\(d,\frac{1989\cdot1990+3978}{1992\cdot1991-3984}=1\)
Bài 3 tự so sánh nhé :v