ở giữa lak dấu nhân ak??????

=-1/2 nha bn!!!!!!

2 câu mà bạn

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

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

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

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

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

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

      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

bang 0 het nhetk cho minh di

a)    \(\frac{x+14}{86}+\frac{x+15}{85}+\frac{x+16}{84}+\frac{x+17}{83}+\frac{x+116}{4}=0\)

\(\Leftrightarrow\)\(\frac{x+14}{86}+1+\frac{x+15}{85}+1+\frac{x+16}{84}+1+\frac{x+17}{83}+1+\frac{x+116}{4}-4=0\)

\(\Leftrightarrow\)\(\frac{x+100}{86}+\frac{x+100}{85}+\frac{x+100}{84}+\frac{x+100}{83}+\frac{x+100}{4}=0\)

\(\Leftrightarrow\)\(\left(x+100\right)\left(\frac{1}{86}+\frac{1}{85}+\frac{1}{84}+\frac{1}{83}+\frac{1}{4}\right)=0\)

\(\Leftrightarrow\)\(x+100=0\)

\(\Leftrightarrow\)\(x=-100\)

Vậy...

câu 1:=\(\frac{x+14}{86}+1+\frac{x+15}{85}+1+\frac{x+16}{84}+1+\frac{x+17}{83}+1+\frac{x+116}{4}-4=0\)

=\(\frac{x+100}{86}+\frac{x+100}{85}+\frac{x+100}{84}+\frac{x+100}{83}+\frac{x+100}{4}=0\)

=\(^{(x+100).(\frac{1}{86}+\frac{1}{85}+\frac{1}{84}+\frac{1}{83}+\frac{1}{4})=0}\)

=\(\orbr{\begin{cases}100+x=0\\\frac{1}{86}+\frac{1}{85}+\frac{1}{84}+\frac{1}{83}+\frac{1}{4}=0\left(voly\right)\end{cases}}\Leftrightarrow100+x=0\Leftrightarrow x=-100\)

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= \(\frac{7^{ }}{8}-\frac{7}{8}=0\)

a)(\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{11.12}\)). x=\(\frac{1}{3}\)

(1-\(\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...-\frac{1}{11}_{ }+\frac{1}{12}\)).x=\(\frac{1}{3}\)

(1+\(\frac{1}{12}\)).x=\(\frac{1}{3}\)

x=\(\frac{1}{3}:\frac{13}{12}\)

x=\(\frac{4}{13}\)

b)( \(2-\frac{2}{3}+\frac{2}{3}-\frac{2}{5}+...+\frac{2}{9}-\frac{2}{11}_{ }\)):x =\(\frac{2}{3}\)

Giống câu a

a) \(\frac{6}{2.5}+\frac{6}{5.8}+\frac{6}{8.11}+.......+\frac{6}{44.47}+\frac{6}{47.50}\)

\(=2\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+......+\frac{3}{44.47}+\frac{3}{47.50}\right)\)

\(=2\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+......+\frac{1}{44}-\frac{1}{47}+\frac{1}{47}-\frac{1}{50}\right)\)

\(=2\left(\frac{1}{2}-\frac{1}{50}\right)\)

\(=1-\frac{1}{25}\)

\(=\frac{24}{25}\)

đặt \(A=\frac{1}{9.11}+\frac{1}{11.13}+........+\frac{1}{41.43}+\frac{1}{43.45}\)

\(2A=\frac{2}{9.11}+\frac{2}{11.13}+.......+\frac{2}{41.43}+\frac{2}{43.45}\)

\(2A=\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+......+\frac{1}{41}-\frac{1}{43}+\frac{1}{43}-\frac{1}{45}\)

\(2A=\frac{1}{9}-\frac{1}{45}\)

\(2A=\frac{4}{45}\)

\(A=\frac{4}{45}\div2\)

\(A=\frac{2}{45}\)