\(\frac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.7^3.8}\)
helppppppppppp meeeeeeeeeeeeeee
K
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15 tháng 8 2017
\(\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6.2^{12}.3^5}\)- \(\frac{5^{10}.7^3-5^{10}.7^4}{5^9.7^3+5^9.7^3.2^3}\)= \(\frac{2^{12}.3^4.\left(3-1\right)}{2^{24}.3^{11}}\)-\(\frac{5^{10}.7^3.\left(1-7\right)}{5^9.7^3.\left(1+8\right)}\)=\(\frac{1}{2^{11}.3^7}\)-\(\frac{-10}{3}\)
3 tháng 10 2019
TA có\(\frac{2^{12}.3^5.4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}-\frac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3}\)
=\(\frac{2^{12}.3^5.2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}-\frac{5^{10}.7^3-5^{10}.7^4}{5^9.7^3}\)
=\(\frac{2^{24}.3^9}{2^{12}.3^5.\left(3+1\right)}-\frac{5^{10}.7^3.\left(1-7\right)}{5^9.7^3}\)
=\(2^{20}.3^4-30\)
K
4 tháng 7 2017
\(B=\frac{2^{13}.3^5-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}-\frac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}\)
\(=\frac{2^{13}.3^5-\left(2^2\right)^6.\left(3^2\right)^2}{2^{12}.3^6+\left(2^3\right)^4.3^5}-\frac{5^{10}.7^3-\left(5^2\right)^5.\left(7^2\right)^2}{\left(5^3.7\right)^3+5^9.\left(2.7\right)^3}\)
\(=\frac{2^{13}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}-\frac{5^{10}.7^3-5^{10}.7^4}{5^9.7^3+5^9.2^3.7^3}\)
\(=\frac{2^{12}.3^4\left(2.3-1\right)}{2^{12}.3^5\left(3+1\right)}-\frac{5^{10}.7^3\left(1-7\right)}{5^9.7^3.\left(1+2^3\right)}\)
\(=\frac{2^{12}.3^4.5}{2^{12}.3^5.4}-\frac{5^{10}.7^3.\left(-6\right)}{5^9.7^3.9}\)
\(=\frac{5}{12}-\frac{-10}{3}=\frac{5}{12}+\frac{40}{12}=\frac{45}{12}=\frac{15}{4}=3\frac{3}{4}\)
14 tháng 11 2018
\(B=\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}-\frac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}\)
\(=\frac{2^{12}.3^5-2^{12}.3^6}{2^{12}.3^6+2^{12}.3^5}-\frac{5^{10}.7^3-5^{10}.7^4}{5^9.7^3+5^9.2^3.7^3}\)
\(=\frac{2^{12}.3^5.\left(1-3\right)}{2^{12}.3^6.\left(3+1\right)}-\frac{5^9.7^3.\left(1-7\right)}{5^9.7^3\left(1+2^3.1\right)}\)
\(=\frac{-1}{2}-\frac{-2}{3}=\frac{-7}{6}\)
20 tháng 11 2018
Gợi ý : Phân tích hết ra thành tích các thừa số nguyên tố rồi đặt cái chung ra ngoài
-> rút gọn
-> kết quả
P/S : bài này cx ko dài lắm nhưg lười ^^
\(\frac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.7^3.8}=\frac{5^{10}.7^3-\left(5^2\right)^5.\left(7^2\right)^2}{\left(5^3.7\right)^3+5^9.7^3.8}\)
\(=\frac{5^{10}.7^3-5^{10}.7^4}{\left(5^3\right)^3.7^3+5^9.7^3.8}=\frac{5^{10}.7^3\left(1-7\right)}{5^9.7^3+5^9.7^3.8}\)
\(=\frac{5^{10}.7^3.\left(-6\right)}{5^9.7^3.\left(1+8\right)}=\frac{5.\left(-6\right)}{9}=\frac{-10}{3}\)
TL:
-10/3
-HT-
sorry mik không thể viết phân số vì mik dùng máy tính