Rút gọn
\(A=\frac{3^5\cdot1^7+3^9\cdot5}{3^7\cdot2^5}\)
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Ta có : \(\frac{1}{10.9}-\frac{1}{9.8}-.....-\frac{1}{2.1}\)
\(=\frac{1}{90}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{9.8}\right)\)
\(=\frac{1}{90}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{8}-\frac{1}{9}\right)\)
\(=\frac{1}{90}-\left(1-\frac{1}{9}\right)\)
\(=\frac{1}{90}-\frac{8}{9}=\frac{-79}{90}\)
\(A=\frac{7^{48}.5^{30}.2^8-5^{30}.7^{49}.2^{10}}{5^{29}.2^8.7^{48}}=\frac{7^{48}.5^{30}.2^8.\left(1-7.2^2\right)}{5^{29}.2^8.7^{48}}=5.\left(-27\right)=-135\)
Vậy \(A=-135\)
\(\frac{\left(\frac{2}{5}\right)^7.5^7+\left(\frac{9}{4}\right)^3:\left(\frac{3}{16}\right)^3}{2^7.5^2.2^9}\)
=\(\frac{\left(\frac{2}{5}.5\right)^7+\left(\frac{9}{4}:\frac{3}{16}\right)^3}{2^7.5^2.2^9}\)
=\(\frac{2^7+\left(\frac{9}{4}.\frac{16}{3}\right)^3}{2^7.5^2.2^9}\)
=\(\frac{2^7+3^3.4^3}{2^7.5^2.2^9}\)
=\(\frac{2^7+3^3.2^6}{2^7.5^2.2^9}\)
=\(\frac{2^6.\left(27+2\right)}{2^6.5^2.2^{10}}\)
=\(\frac{29}{25600}\)
Đoạn \(\frac{2^6\cdot\left(27+2\right)}{2^6\cdot5^2\cdot2^{10}}\)là sai rồi bn ơi!!!
Bn phải lm như mục trên là:
\(\frac{2^6\left(2+3^3\right)}{2^7\left(5^2+2^2\right)}\)\(=\frac{2^6\cdot29}{2^7\cdot29}=\frac{1}{2}\)
Nhưng dù sao cx c.ơn bn vì đã giúp mk,mk sẽ cho bn 1 ths nka!!!
a) \(C=\frac{5}{2.1}+\frac{4}{1.11}+\frac{3}{11.2}+\frac{1}{2.15}+\frac{13}{15.4}\)
\(=7\left(\frac{5}{2.7}+\frac{4}{7.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.28}\right)\)
\(=7\left(\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{28}\right)\)
\(=7\left(\frac{1}{2}-\frac{1}{28}\right)\)
\(=7.\frac{13}{28}=\frac{7.13}{28}=\frac{13}{4}\)
b) \(B=\frac{6}{3.5}+\frac{6}{5.7}+\frac{6}{7.9}+...+\frac{6}{97.99}\)
\(=3\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\right)\)
\(=3\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(=3\left(\frac{1}{3}-\frac{1}{99}\right)\)
\(=3.\frac{32}{99}=\frac{3.32}{99}=\frac{32}{33}\)
két bn vớ mk . mk bày cho chớ làm vào đây tốn thời gian lắm
\(A=\frac{\left(-2\right)^3\cdot3^3\cdot5^3\cdot7\cdot8}{3\cdot5^3\cdot2^4\cdot42}\)
\(=\frac{\left(-2\right)^3\cdot3^3\cdot6^3\cdot5^3\cdot7\cdot2^3}{3\cdot5^3\cdot2^4\cdot2\cdot3\cdot7}\)
\(=\frac{\left(-2\right)^3\cdot3^8\cdot5^3\cdot2^3\cdot7}{3^2\cdot5^3\cdot2^5\cdot7}=-2\cdot3^6\)
câu b để nghĩ...
\(\frac{2^{12}.13+2^{12}.65}{2^{10}.104}+\frac{3^{10}.11+3^{10}.5}{3^9.2^4}\)
\(\Rightarrow\frac{2^{12}.\left(13+65\right)}{2^{10}.104}+\frac{3^{10}.\left(11+5\right)}{3^9.2^4}\)
\(\Rightarrow\frac{2^{12}.78}{2^{10}.104}+\frac{3^{10}.2^4}{3^9.2^4}\)
\(=\frac{2^2.3}{4}+3\)
\(=3+3=6\)
\(A=\frac{3^5\cdot1^7+3^9\cdot5}{3^7\cdot2^5}=\frac{3^5\left(1+405\right)}{3^7\cdot32}=\frac{406}{288}=\frac{203}{144}\)
A= \(\frac{3^5.1^7+3^9.5}{3^7.2^5}\)= \(\frac{3^5+3^5.3^4.5}{3^7.2^5}\)= \(\frac{3^5\left(1+81.5\right)}{3^7.2^5}\)= \(\frac{3^5\left(1+405\right)}{3^7.2^5}\)= \(\frac{406}{3^2.2^5}\)= \(\frac{406}{9.32}\)= \(\frac{203}{9.16}\)= \(\frac{203}{144}\).
Vậy A= \(\frac{203}{144}\).