Số lẻ lắm bn!!!!!!!\(\frac{1492992}{7}\)

\(\frac{9.11+19.9}{27.5-27.15}=\frac{9.\left(11+19\right)}{27.\left(5-15\right)}=\frac{9.30}{3.9.\left(-10\right)}=\frac{30}{3.\left(-10\right)}=\frac{30}{-30}=-1\)

Rút gọn các biểu thức nha

Câu 1: \(\frac{9.\left(11+9\right)}{27.\left(5-15\right)}=\frac{9.20}{27.\left(-10\right)}=-\frac{2}{3}\)

Câu 2: Làm biếng giải quá: =\(\frac{1492992}{7}\)

\(A=\frac{49^2\cdot3^{11}}{81^2\cdot21^5}\)

\(=\frac{\left(7^2\right)^2\cdot3^{11}}{\left(3^4\right)^2\cdot\left(3\cdot7\right)^5}\)

\(=\frac{7^4\cdot3^{11}}{3^8\cdot3^5\cdot7^5}\)

\(=\frac{7^4\cdot3^{11}}{3^{13}\cdot7^5}\)

\(=\frac{1}{3^2\cdot7}=\frac{1}{63}\)

      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

\(A=\frac{49^2.3^{11}}{81^2.21^5}\)

\(=\frac{\left(7^2\right)^2.3^{11}}{\left(3^4\right)^2.\left(3.7\right)^5}\)

\(=\frac{7^4.3^{11}}{3^8.3^5.7^5}\)

\(=\frac{1}{3^2.7}\)

\(=\frac{1}{63}\)

Học tốt

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

ko sao

A= \(\frac{49^5.8^{10}}{14^7.49.4^{13}}-\frac{\frac{7}{10}-\frac{7}{12}+\frac{7}{5}}{0,8-\frac{2}{3}+\frac{8}{5}}\)

A= \(\frac{7^{10}.2^{30}}{2^7.7^9.2^{26}}-\frac{7\left(\frac{1}{10}-\frac{1}{12}+\frac{1}{5}\right)}{\frac{8}{10}-\frac{2.4}{3.4}+\frac{8}{5}}\)=\(\frac{7^{10}.2^{30}}{7^9.2^{33}}-\frac{7\left(\frac{1}{10}-\frac{1}{12}+\frac{1}{5}\right)}{\frac{8}{10}-\frac{2.4}{3.4}+\frac{8}{5}}\)

A= \(\frac{7^{ }}{2^3}-\frac{7\left(\frac{1}{10}-\frac{1}{12}+\frac{1}{5}\right)}{8.\left(\frac{1}{10}-\frac{1}{12}+\frac{1}{5}\right)}\)

A= \(\frac{7^{ }}{8}-\frac{7}{8}=0\)

\(S=\frac{6}{2.5}+\frac{6}{5.8}+.......+\frac{6}{29.32}\)

\(S=2\left(\frac{3}{2.5}+\frac{3}{5.8}+......+\frac{3}{29.32}\right)\)

\(S=2\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+......+\frac{1}{29}-\frac{1}{32}\right)\)

\(S=2\left(\frac{1}{2}-\frac{1}{32}\right)\)

\(S=2.\frac{15}{32}\)

\(S=\frac{15}{16}< 1\RightarrowĐPCM\)

Vậy \(S=\frac{15}{16}\)

\(=2.\left(\frac{3}{5.8}+\frac{11}{8.19}+...+\frac{99}{101.200}\right)\)

\(=2.\left(\frac{1}{5}-\frac{1}{200}\right)\)

\(=2.\frac{39}{200}=\frac{39}{100}\)

\(E=\frac{6}{5.8}+\frac{22}{8.19}+\frac{24}{19.31}+\frac{140}{31.101}+\frac{198}{101.200}\)

    \(=2.\left(\frac{3}{5.8}+\frac{11}{8.19}+\frac{12}{19.31}+\frac{70}{31.101}+\frac{99}{101.200}\right)\)

    \(=2\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{19}+\frac{1}{19}-\frac{1}{31}+\frac{1}{31}-\frac{1}{101}+\frac{1}{101}-\frac{1}{200}\right)\)

    \(=2\left(\frac{1}{5}-\frac{1}{200}\right)\)

    \(=2.\frac{39}{200}\)

      \(=\frac{39}{100}\)

\(\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+...+\frac{1}{x\left(x+3\right)}=\frac{1}{6}\)

\(\Leftrightarrow\frac{1}{3}\left(\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{x\left(x+3\right)}\right)=\frac{1}{6}\)

\(\Leftrightarrow\frac{1}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{1}{6}\)

\(\Leftrightarrow\frac{1}{3}\left(\frac{1}{5}-\frac{1}{x+3}\right)=\frac{1}{6}\)

\(\Leftrightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{1}{2}\)

\(\Leftrightarrow\frac{1}{x+3}=-\frac{3}{10}\)

\(\Leftrightarrow1\cdot10=-3\left(x+3\right)\)

\(\Leftrightarrow10=-3x-9\)

\(\Leftrightarrow10+9=-3x\)

\(\Leftrightarrow19=-3x\)

\(\Leftrightarrow x=-\frac{19}{3}\)

Đề sai à -.- 

\(\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+...+\frac{1}{x\left(x+3\right)}=\frac{1}{6}\)

=> \(\frac{1}{3}\left(\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{x\left(x+3\right)}\right)=\frac{1}{6}\)

=> \(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{1}{6}:\frac{1}{3}\)

=> \(\frac{1}{5}-\frac{1}{x+3}=\frac{1}{6}\cdot3=\frac{1}{2}\)

=> \(\frac{1}{x+3}=\frac{1}{5}-\frac{1}{2}=-\frac{3}{10}\)

=> \(10=-3\left(x+3\right)\)

=> 10 = -9x - 9

=> 10 + 9x + 9 = 0

=> 19 + 9x = 0

=> 9x = -19

=> x = -19/9

Nhầm khúc cuối 

=> 10 = -3x - 9

=> 10 + 3x + 9 = 0

=> 19 + 3x = 0

=> 3x = -19

=> x = -19/3

=11,4215495

\(=\frac{39}{100}=0,39\)

\(\frac{2^7.3^6}{6^5.8^2}=\frac{2^7.3^6}{\left(2.3\right)^5.\left(2^3\right)^2}=\frac{2^7.3^6}{2^5.3^5.2^6}=\frac{2^7.3^6}{2^{11}.3^5}=\frac{3}{2^4}=\frac{3}{16}\)