Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
A = \(\dfrac{5.4^{15}.9^9-4.3^{20}.8^9}{5.2^9.6^{19}-7.2^{29}.27^6}=\dfrac{5.2^{30}.3^{18}-2^{29}.3^{20}}{5.2^{28}.3.3^{19}-7.2^{29}.3^{18}}=\dfrac{2^{29}.3^{18}\left(5.2-3^2\right)}{2^{18}.3^{18}\left(5.3-7.2\right)}=2\)
B = \(\dfrac{8020}{2004.2006-2003.2005}\)
Đặt x = 2004, ta có:
\(\dfrac{4x+2}{x\left(x+2\right)-\left(x-1\right)\left(x+1\right)}=\dfrac{4x+2}{2x+1}=\dfrac{2\left(2x+1\right)}{2x+1}=2\)
Tử số= 2^19.9^3.3^3 + 5.3.2^9.2^9.9^4
= 2^19.9^3.3^3 + 5.3.2^18.9.9^3
= 2^19.9^3.3.3^2 + 5.3.2^18.3^2.9^3
= 2^18.9^3.3^2(2 + 5.) (đặt nhân tử chung)
=7.2^18.9^3.3^2
=7.2^18.9.9.9.3^2
=7.2^18.3^2.3^2.3^2.3^2
=7.2^18.3^8
Mẫu số= 6^9.2^10 + 6^10.2^10
= 6^9.2^10 + 6^10.2^10
=6^9.2^10(1+6)
=7.6^9.2^10.
=7.2^9.3^9.2^10
=7.2^19.3^9
Lấy tử số chia mẫu số ta được : 1/2.3 = 1/6
\(a.C=\dfrac{x^4+x^8+x^{12}+x^{16}+x^{20}+x^{24}+x^{28}+1}{x^3+x^7+x^{11}+x^{15}+x^{19}+x^{23}+x^{27}+x^{31}}=\dfrac{x^{28}+x^{24}+...+x^8+x^4+1}{x^3\left(x^{28}+x^{24}+...+x^8+x^4+1\right)}=\dfrac{1}{x^3}\) Tại x = 2015 thì : \(C=\dfrac{1}{x^3}=\dfrac{1}{2015^3}\)
\(b.F=\dfrac{1}{1.2.3.4}+\dfrac{1}{2.3.4.5}+\dfrac{1}{3.4.5.6}+...+\dfrac{1}{2011.2012.2013.2014}\)
\(3F=\dfrac{3}{1.2.3.4}+\dfrac{3}{2.3.4.5}+\dfrac{3}{3.4.5.6}+...+\dfrac{3}{2011.2012.2013.2014}\)
\(3F=\dfrac{1}{1.2.3}-\dfrac{1}{2.3.4}+\dfrac{1}{2.3.4}-\dfrac{1}{3.4.5}+\dfrac{1}{3.4.5}-\dfrac{1}{4.5.6}+...+\dfrac{1}{2011.2012.2013}-\dfrac{1}{2012.2013.2014}\)
\(3F=\dfrac{1}{1.2.3}-\dfrac{1}{2012.2013.2014}\)
Tới đây dễ rồi , bạn tự tính nốt .
Vì làm vậy để triệt tiêu dần mà ( dang bài kiểu ... này thường là phải triệt tiêu ) Triệu Tử Dương
\(\frac{x^{24}+x^{18}+x^{12}+x^6+1}{x^{27}+x^{24}+x^{21}+x^{18}+x^{15}+x^{12}+x^9+x^6+x^3+1}=\frac{x^{24}+x^{18}+x^{12}+x^6+1}{x^{24}\left(x^3+1\right)+x^{18}\left(x^3+1\right)+x^{12}\left(x^3+1\right)+x^6\left(x^3+1\right)+\left(x^3+1\right)}\)
=\(\frac{x^{24}+x^{18}+x^{12}+x^6+1}{\left(x^3+1\right)\left(x^{24}+x^{18}+x^{12}+x^6+1\right)}=\frac{1}{x^3+1}\)
a: \(=\dfrac{2^{19}\cdot3^9+2^{20}\cdot3^{10}}{2^{19}\cdot3^9+2^{18}\cdot3^9\cdot5}=\dfrac{2^{19}\cdot3^9\left(1+2\cdot3\right)}{2^{18}\cdot3^9\left(2+5\right)}=2\)
\(A=\dfrac{5.4^{15}.9^9-4.3^{20}.8^9}{5.2^9.6^{19}-7.2^{29}.27^6}\)
\(=\dfrac{5.2^{30}.3^{18}-2^{29}.3^{20}}{5.2^{28}.3^{19}-7.2^{29}.3^{18}}\)
\(=\dfrac{2^2.3^{18}\left(5.2-3^2\right)}{2^{18}.3^{18}\left(5.2-3^2\right)}\)
\(=2\)