Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\frac{\left(\frac{2}{5}\right)^7.5^7+\left(\frac{9}{4}\right)^3:\left(\frac{3}{16}\right)^3}{2^7.5^2.2^9}\)
=\(\frac{\left(\frac{2}{5}.5\right)^7+\left(\frac{9}{4}:\frac{3}{16}\right)^3}{2^7.5^2.2^9}\)
=\(\frac{2^7+\left(\frac{9}{4}.\frac{16}{3}\right)^3}{2^7.5^2.2^9}\)
=\(\frac{2^7+3^3.4^3}{2^7.5^2.2^9}\)
=\(\frac{2^7+3^3.2^6}{2^7.5^2.2^9}\)
=\(\frac{2^6.\left(27+2\right)}{2^6.5^2.2^{10}}\)
=\(\frac{29}{25600}\)
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!!!
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=(2.23.25.27.211)(52.54.58.516)
A=21+3+5+7+11.52+4+8+16
A=227.530
A=227.527.53
A=(227.527).125
A=(2.5)27.125
A=1027.125
Vậy A có 27 chữ số 0 tận cùng
\(=\frac{2^3.3^3.5^3.2^3.7}{3.2^4.5^3.2.7}=\frac{2^6.5^3.3^3}{3.2^5.5^3}=2.3^2=2.9=18\)
\(A=\frac{3^5\cdot1^7+3^9\cdot5}{3^7\cdot2^5}=\frac{3^5\left(1+405\right)}{3^7\cdot32}=\frac{406}{288}=\frac{203}{144}\)
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= \(\frac{203}{144}\).