\(\frac{3\cdot4\cdot2^4\cdot3\cdot4\cdot2^4}{5\cdot2^5\cdot4^2-16^2}\)
Tìm C :
\(C=\frac{3\cdot4\cdot2^4\cdot3\cdot4\cdot2^4}{5\cdot2^5\cdot4^2-16^2}\)
[ Trình bày lời giải nha ]
\(C=\frac{3.4.2^4.3.4.2^4}{5.2^5.4^2-16^2}\)
\(C=\frac{12.12.2^8}{5.2^5.\left(2^2\right)^2-256}\)
\(C=\frac{144.256}{5.32.16-256}\)
\(C=\frac{36864}{2560-256}\)
\(C=\frac{36864}{2304}\)
\(C=16\)
Để câu trả lời của bạn nhanh chóng được duyệt và hiển thị, hãy gửi câu trả lời đầy đủ và không nên:
Yêu cầu, gợi ý các bạn khác chọn (k) đúng cho mìnhChỉ ghi đáp số mà không có lời giải, hoặc nội dung không liên quan đến câu hỏi.Tìm C biết :
\(C=\frac{3\cdot4\cdot2^4\cdot3\cdot4\cdot2^4}{5\cdot2^5\cdot4^2-16^2}\)
( Trình bay lời giải nha )
Tính hợp lý:
\(H=\frac{\left(3\cdot4\cdot2^{16}\right)}{11\cdot2^{13}\cdot4^{11}-16^9}\)
\(I=\frac{5\cdot4^{15}\cdot9^9-4\cdot3^{20}\cdot8^9}{5\cdot2^9\cdot6^{19}-7\cdot2^{29}\cdot27^6}\)
\(I=\frac{5.4^{15}.9^9-4.3^{20}.8^9}{5.2^9.6^{19}-7.2^{29}.27^6}=\frac{5.2^{30}.3^{27}-2^2.3^{20}.2^{27}}{5.2^9.2^{19}.3^{19}-7.2^{29}.3^{18}}\)
\(=\frac{5.2^{30}.3^{27}-3^{30}.2^{29}}{5.2^{28}.3^{19}-7.2^{29}.3^{18}}\)
\(=\frac{2^{29}.3^{27}.\left(5.2-3^3\right)}{2^{28}.3^{18}.\left(5.3-2.7\right)}\)
\(=\frac{2^{29}.3^{27}.-17}{2^{18}.3^{18}}\)
\(=\frac{2^9.3^9.-17}{1}\)
Ta có \(H=\frac{\left(3.4.2^{16}\right)}{11.2^{13}.4^{11}-16^9}\)
\(=\frac{3.4.2^{16}}{11.2^{13}.2^{22}-2^{36}}\)
\(=\frac{3.2^{18}}{11.2^{35}-2^{36}}\)
\(=\frac{3.2^{18}}{2^{35}.\left(11-2\right)}\)
\(=\frac{3.2^{18}}{2^{35}.3^2}\)
\(=\frac{1}{2^{17}.3}\)
\(A=\frac{1\cdot2}{2\cdot2}\cdot\frac{2\cdot3}{3\cdot3}\cdot\frac{3\cdot4}{4\cdot4}\cdot\frac{4\cdot5}{5\cdot5}\cdot.................\cdot\frac{2012\cdot2013}{2013\cdot2013}\)với
\(B=\frac{2012\cdot2013-2012\cdot2012}{2012\cdot2011+2012\cdot2}\)
A=\(\frac{1}{2}\).\(\frac{2}{3}\)....\(\frac{2012}{2013}\)=\(\frac{1}{2013}\)
B=\(\frac{2012}{2012.2013}\)=\(\frac{1}{2013}\)
vậy A=B
\(A=\frac{1.2}{2.2}.\frac{2.3}{3.3}.\frac{3.4}{4.4}.\frac{4.5}{5.5}.....\frac{2012.2013}{2013.2013}=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}....\frac{2012}{2013}=\frac{1.2.3.4.5....2012}{2.3.4.5....2013}=\frac{1}{2013}\)
\(B=\frac{2012.2013-2012.2012}{2012.2011+2012.2}=\frac{2012.\left(2013-2012\right)}{2012.\left(2011+2\right)}=\frac{2012}{2012.2013}=\frac{1}{2013}\)
\(\Rightarrow A=B\)
tính giá trị biểu thức ( tính nhanh nếu có thể )
c, \(1\cdot2\cdot3....9-1\cdot2\cdot3....8-1\cdot2\cdot3....7\cdot8^2\)
d,\(\frac{\left(3\cdot4\cdot2^{16}\right)^2}{11\cdot2^{13}\cdot4^{11}-16^9}\)
\(c,1.2.3...9-1.2.3...8-1.2.3...7.8^2\)
\(=1.2.3...8\left(9-1-8\right)\)
\(=1.2.3...8.0\)
\(=0\)
\(d,\frac{\left(3.4.2^{16}\right)^2}{11.2^{13}.4^{11}-16^9}\)
\(=\frac{3^2.4^2.2^{32}}{11.2^{13}.\left(2^2\right)^{11}-\left(2^4\right)^9}\)
\(=\frac{3^2.2^4.2^{32}}{11.2^{13}.2^{22}-2^{36}}\)
\(=\frac{3^2.2^{36}}{11.2^{35}-2^{36}}\)
\(=\frac{3^2.2^{36}}{2^{35}\left(11-2\right)}\)
\(=\frac{3^2.2^{36}}{2^{35}.9}\)
\(=\frac{3^2.2^{36}}{2^{35}.3^2}\)
\(=2\)
\(A=\frac{2^2\cdot3\cdot2^{16}\cdot81}{3^2\cdot2^{15}\cdot4+4^9;27}\)
A=\(\frac{1\cdot2\cdot3+2\cdot4\cdot6+4\cdot8\cdot12+8\cdot16\cdot24}{2\cdot2\cdot4+4\cdot6\cdot8+8\cdot12\cdot16+16\cdot24\cdot32}\)
A=
A=\(\frac{15\cdot3^{11}+4.27^4}{9^7}\)
B=\(\frac{2^{19}\cdot2^{73}+15\cdot4^9\cdot9^4}{6^9\cdot2^{10}+12^{10}}\)
C=\(\frac{5\cdot12^3\cdot4^{11}-16^8}{\left(3\cdot2^{17}\right)^2}\)
D=\(\frac{4^7\cdot2^8}{3\cdot2^{15}\cdot16^2-5\cdot2^2\cdot\left(2^{10}\right)^2}\)
tính bằng cách hợp lí
a) \(\left(10^2+11^2+12^2\right):\left(13^2+14^2\right)\)
b) \(1\cdot2\cdot3\cdot4\cdot...\cdot8\cdot9-1\cdot2\cdot3\cdot4\cdot...\cdot7\cdot8-1\cdot2\cdot3\cdot...\cdot7\cdot8^2\)
c) \(\frac{\left(3\cdot4\cdot2^{16}\right)^2}{11\cdot2^{13}\cdot4^{11}-16^9}\)
giúp em với ạ
c) \(\frac{\left(3\cdot4\cdot2^{16}\right)}{11\cdot2^{13}\cdot4^{11}-16^9}=\frac{\left(3\cdot2^2\cdot2^{16}\right)^2}{11\cdot2^{13}\cdot2^{22}-2^{36}}\)
\(=\frac{9\cdot2^4\cdot2^{32}}{11\cdot2^{35}-2^{26}}\)
\(=\frac{9\cdot2^4\cdot2^{32}2^{ }}{\left(11-2\right)\cdot2^{35}}\)
\(=\frac{9\cdot2^4\cdot2^{32}}{9\cdot2^{35}}\)
\(=\frac{9\cdot1\cdot2^{32}}{9\cdot2^{31}}=\frac{2^{32}}{2^{31}}=2\)