• Bài 1 = 32/13
• Bài 2 ≈ -332.667
• Bài 3 ≈ 0
• Nha bạn,sai thì cho mình xin lỗi

\(\frac{44.2.60}{11.13.15}=\frac{\left(4.11\right).2.\left(4.15\right)}{11.13.15}=\frac{4.11.2.4.15}{11.13.15}=\frac{4.2.4}{13}=\frac{8.4}{13}=\frac{32}{13}\)

\(123.456456 - 456.123123\) \(=\) \(123.1001.456-456.1001.123=0\)

\(\frac{98.7676-9898.76}{2001.2002.2003...2010}=\frac{0}{2001.2002.2003...2010}=0\)

Đặt \(A=\frac{2}{10\cdot12}+\frac{2}{12\cdot14}+\frac{2}{14\cdot16}+...+\frac{2}{48\cdot50}\)

\(A=\frac{1}{10}-\frac{1}{12}+\frac{1}{12}-\frac{1}{14}+...+\frac{1}{48}-\frac{1}{50}\)

\(A=\frac{1}{10}-\frac{1}{50}=\frac{5}{50}-\frac{1}{50}=\frac{4}{50}=\frac{2}{25}\)

Vậy \(A=\frac{2}{25}\)

= \(\frac{1}{10}-\frac{1}{12}+\frac{1}{12}-\frac{1}{14}+...+\frac{1}{48}-\frac{1}{50}\)
=  \(\frac{1}{10}-\frac{1}{50}\)=  \(\frac{2}{25}\)

giúp tớ vs

17.(29-(-111))+29.(-17)

=17⋅(29+111)+(−17)⋅29

=17⋅140−17⋅29

=17(140-29)

=17.111

=1887

17.[29 - (-111)] + 29.(-17)

= 17.29 + 17.111 + 29.(-17)

= (17.29 - 29.17) + 17.111

= 0 + 1887

= 1887

ko

31.15.49.4-31.49.40

=15.4-40

=60-40

=20

a, \(A=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}...\frac{999^2}{998.1000}\)

\(=\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}...\frac{999.999}{998.1000}\)

\(=\frac{2.3.4...999}{1.2.3...998}.\frac{2.3.4...999}{3.4.5...1000}\)

\(=\frac{999}{1}.\frac{2}{1000}\)

\(=\frac{999.2}{1000.1}=\frac{999.2}{500.2.1}\)

\(=\frac{999}{500}\)

Vậy \(A=\frac{999}{500}\)

chúc bạn học giỏi

cảm ơn bạn nhiều nha

a,\(\frac{5}{3}.\frac{3}{7}+\frac{5}{3}.\frac{5}{7}-\frac{5}{3}\)

 =\(\frac{5}{3}.\left(\frac{3}{7}+\frac{5}{7}\right)-\frac{5}{3}\)

 = \(\frac{5}{21}\)

\(A=\left(9\frac{30303}{80808}+7\frac{303030}{484848}\right)+4,03\)

\(A=\left(9,375+7,625\right)+4,03\)

\(A=17+4,03\)

\(A=21,03\)

367+(-30)+1672+(-337)

= [367 + (-337) ] + (-30) + 1672

= 30 + (-30) + 1672

= 0 + 1672 = 1672

367 + (-30) + 1672 + (-337)

=367 + 1672  +[ (-30) + (-337)]

=367 + 1672  + (-367)

=[367 + (-367)] + 1672

=0 + 1672

=1672

#Học tốt

&YOUTUBER&