Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\frac{1}{1}\)x 2 x 3 + \(\frac{1}{2}\)x 3 x 4 + \(\frac{1}{3}\)x 4 x 5 + \(\frac{1}{4}\)x 5 x 6
= 1 x 2 + \(\frac{1}{2}\)+ \(\frac{1}{3}\)+ \(\frac{1}{4}\)x 6
= 2 +\(\frac{1}{2}\)+ \(\frac{1}{3}\)+ 1, 5
=
a) \(\frac{75^3.3^7}{81^4.5^6}=\frac{5^3.3^3.5^3.3^7}{\left(3^4\right)^4.5^6}=\frac{5^6.3^3.3^7}{3^{16}.5^6}=\frac{3^{10}}{3^{16}}=\frac{1}{3^6}=\frac{1}{729}\)
b) \(\frac{6^6.4^2}{3^{12}.2^8}=\frac{2^6.3^6.\left(2^2\right)^2}{3^{12}.2^8}=\frac{2^6.3^6.2^4}{3^{12}.2^8}=\frac{2^{10}.3^6}{3^{12}.2^8}=\frac{2^2.1}{3^6}=\frac{4}{729}\)
c) \(\frac{34^5.2^5}{2^{14}.17^5}=\frac{2^5.17^5.2^5}{2^{14}.17^5}=\frac{2^{10}}{2^{14}}=\frac{1}{2^4}=\frac{1}{16}\)
Bạn nên viết đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo) để được hỗ trợ tốt hơn.
\(A=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{18.19.20}\)
\(2A=\dfrac{3-1}{1.2.3}+\dfrac{4-2}{2.3.4}+\dfrac{5-3}{3.4.5}+...+\dfrac{20-18}{18.19.20}=\)
\(=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{18.19}-\dfrac{1}{19.20}=\dfrac{1}{2}-\dfrac{1}{19.20}\)
\(\Rightarrow A=\left(\dfrac{1}{2}-\dfrac{1}{19.20}\right):2\)
\(a=4^5.9^4-2.\dfrac{6^9}{2^{10}}.3^8+6^8.20\)
Đề là như vầy đúng ko bn?
a) \(\frac{10^4.81-16.15^2}{\left(-8\right)^4.3^{12}+6^{11}}=\frac{10^4.3^4-4^2.15^2}{8^4.3^{12}+6^{11}}=\frac{30^4-60^2}{2^{12}.3^{12}+6^{11}}=\frac{\left(30^2\right)^2-60^2}{6^{12}+6^{11}}\)
\(=\frac{900^2-60^2}{6^{11}.\left(6+1\right)}=\frac{60^2.\left(15^2-1\right)}{6^{11}.7}=\frac{60^2.224}{6^{11}.7}=\frac{2^9.3^2.5^2.7}{2^{11}.3^{11}.7}=\frac{5^2}{2^2.3^9}=\frac{25}{78732}\)
(63 + 2. 62 + 23)/37
= \(\dfrac{6^2.(6+2)+2^3}{37}\)
= \(\dfrac{6^2.8+8}{37}\)
= \(\dfrac{8.(36+1)}{37}\)
= 8