\(\frac{6^5.8^2.21}{49}\)
K
Khách
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N
5
14 tháng 2 2016
\(\frac{9.11+19.9}{27.5-27.15}=\frac{9.\left(11+19\right)}{27.\left(5-15\right)}=\frac{9.30}{3.9.\left(-10\right)}=\frac{30}{3.\left(-10\right)}=\frac{30}{-30}=-1\)
BP
4
3 tháng 9 2020
\(A=\frac{49^2\cdot3^{11}}{81^2\cdot21^5}\)
\(=\frac{\left(7^2\right)^2\cdot3^{11}}{\left(3^4\right)^2\cdot\left(3\cdot7\right)^5}\)
\(=\frac{7^4\cdot3^{11}}{3^8\cdot3^5\cdot7^5}\)
\(=\frac{7^4\cdot3^{11}}{3^{13}\cdot7^5}\)
\(=\frac{1}{3^2\cdot7}=\frac{1}{63}\)
B
3 tháng 9 2020
Bài làm :
Ta có :
\(A=\frac{49^2\cdot3^{11}}{81^2\cdot21^5}\)
\(A=\frac{\left(7^2\right)^2\cdot3^{11}}{\left(3^4\right)^2\cdot\left(3\cdot7\right)^5}\)
\(A=\frac{7^4\cdot3^{11}}{3^8\cdot3^5\cdot7^5}\)
\(A=\frac{7^4\cdot3^{11}}{3^{13}\cdot7^5}\)
\(A=\frac{1}{3^2\cdot7}\)
\(A=\frac{1}{63}\)
Vậy A=1/63
27 tháng 2 2020
Có:495.810/147.49.413=710.230/233.79=7/8 7/10-7/12+7/5 / 8/10-8/12+8/5=7(1/10-1/12+1/5) / 8(1/10-1/12+1/5)=7/8 =>A=7/8-7/8=0 xin lỗi nha mik ghi hơi khó hiểu chút vì mik mới dùng online math.HOK TỐT
LN
31 tháng 1 2019
\(S=\frac{6}{2.5}+\frac{6}{5.8}+.......+\frac{6}{29.32}\)
\(S=2\left(\frac{3}{2.5}+\frac{3}{5.8}+......+\frac{3}{29.32}\right)\)
\(S=2\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+......+\frac{1}{29}-\frac{1}{32}\right)\)
\(S=2\left(\frac{1}{2}-\frac{1}{32}\right)\)
\(S=2.\frac{15}{32}\)
\(S=\frac{15}{16}< 1\RightarrowĐPCM\)
Vậy \(S=\frac{15}{16}\)
10 tháng 4 2019
\(=2.\left(\frac{3}{5.8}+\frac{11}{8.19}+...+\frac{99}{101.200}\right)\)
\(=2.\left(\frac{1}{5}-\frac{1}{200}\right)\)
\(=2.\frac{39}{200}=\frac{39}{100}\)
TH
10 tháng 4 2019
\(E=\frac{6}{5.8}+\frac{22}{8.19}+\frac{24}{19.31}+\frac{140}{31.101}+\frac{198}{101.200}\)
\(=2.\left(\frac{3}{5.8}+\frac{11}{8.19}+\frac{12}{19.31}+\frac{70}{31.101}+\frac{99}{101.200}\right)\)
\(=2\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{19}+\frac{1}{19}-\frac{1}{31}+\frac{1}{31}-\frac{1}{101}+\frac{1}{101}-\frac{1}{200}\right)\)
\(=2\left(\frac{1}{5}-\frac{1}{200}\right)\)
\(=2.\frac{39}{200}\)
\(=\frac{39}{100}\)
TN
4
19 tháng 8 2020
\(\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+...+\frac{1}{x\left(x+3\right)}=\frac{1}{6}\)
\(\Leftrightarrow\frac{1}{3}\left(\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{x\left(x+3\right)}\right)=\frac{1}{6}\)
\(\Leftrightarrow\frac{1}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{1}{6}\)
\(\Leftrightarrow\frac{1}{3}\left(\frac{1}{5}-\frac{1}{x+3}\right)=\frac{1}{6}\)
\(\Leftrightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{1}{2}\)
\(\Leftrightarrow\frac{1}{x+3}=-\frac{3}{10}\)
\(\Leftrightarrow1\cdot10=-3\left(x+3\right)\)
\(\Leftrightarrow10=-3x-9\)
\(\Leftrightarrow10+9=-3x\)
\(\Leftrightarrow19=-3x\)
\(\Leftrightarrow x=-\frac{19}{3}\)
Đề sai à -.-
19 tháng 8 2020
\(\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+...+\frac{1}{x\left(x+3\right)}=\frac{1}{6}\)
=> \(\frac{1}{3}\left(\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{x\left(x+3\right)}\right)=\frac{1}{6}\)
=> \(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{1}{6}:\frac{1}{3}\)
=> \(\frac{1}{5}-\frac{1}{x+3}=\frac{1}{6}\cdot3=\frac{1}{2}\)
=> \(\frac{1}{x+3}=\frac{1}{5}-\frac{1}{2}=-\frac{3}{10}\)
=> \(10=-3\left(x+3\right)\)
=> 10 = -9x - 9
=> 10 + 9x + 9 = 0
=> 19 + 9x = 0
=> 9x = -19
=> x = -19/9
6 tháng 7 2017
\(\frac{2^7.3^6}{6^5.8^2}=\frac{2^7.3^6}{\left(2.3\right)^5.\left(2^3\right)^2}=\frac{2^7.3^6}{2^5.3^5.2^6}=\frac{2^7.3^6}{2^{11}.3^5}=\frac{3}{2^4}=\frac{3}{16}\)