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\(A=\frac{2.\left(1-\frac{1}{19}+\frac{1}{43}-\frac{1}{1943}\right)}{3.\left(1-\frac{1}{19}+\frac{1}{43}-\frac{1}{1943}\right)}:\frac{4\left(1-\frac{1}{29}+\frac{1}{41}-\frac{1}{2941}\right)}{5.\left(1-\frac{1}{29}+\frac{1}{41}-\frac{1}{2941}\right)}=\frac{2}{3}:\frac{4}{5}=\frac{2}{3}.\frac{5}{4}=\frac{5}{6}\)
Xin lỗi nha, mình chỉ biết làm bài này nhõn bằng cách qui đồng nhưng số lớn lắm!
2-2/19+2/43-2/1943
3-3/19+3/43-3/1943
=2(1-1/19+1/43-1/1943)
3(1-1/19+1/43-1/1943)
=2/3
Chúc bạn học tốt!!!!!
\(3\frac{14}{19}+\frac{13}{17}+\frac{35}{43}+6\frac{5}{19}+\frac{8}{43}\)
\(=3+6+\left(\frac{14}{19}+\frac{5}{19}\right)+\left(\frac{35}{43}+\frac{8}{43}\right)+\frac{13}{17}\)
\(=9+1+1+\frac{13}{17}\)
\(=11+\frac{13}{17}\)
\(=\frac{200}{17}\)
a, 3 \(\frac{14}{19}\)+ \(\frac{13}{17}\)+ \(\frac{35}{43}\)+ 6\(\frac{5}{19}\)+ \(\frac{8}{43}\)= \(\left(3\frac{14}{19}+6\frac{5}{19}\right)+\left(\frac{35}{43}+\frac{8}{43}\right)+\frac{13}{17}=\)\(9+1+\frac{13}{17}=8+\frac{13}{17}=8\frac{13}{17}\)
b, \(\frac{-5}{7}.\frac{2}{11}+\frac{-5}{7}.\frac{9}{11}+1\frac{5}{7}\)\(=\frac{-5}{7}\left(\frac{2}{11}+\frac{9}{11}\right)+1\frac{5}{7}\)\(=\frac{-5}{7}.1+1\frac{5}{7}\)\(=\frac{-5}{7}+\frac{12}{7}=\frac{7}{7}=1\)
Chúc bn học tốt
\(3\frac{14}{19}+\frac{13}{17}+\frac{35}{43}+6\frac{5}{19}+\frac{8}{43}\)
\(=\left(3\frac{14}{19}+6\frac{5}{19}\right)+\left(\frac{35}{43}+\frac{8}{43}\right)+\frac{13}{17}\)
\(=10+1+\frac{13}{17}=11+\frac{13}{17}=11\frac{13}{17}\)
\(b)\) Đặt \(B=\frac{3}{10}+\frac{3}{11}+\frac{3}{12}+\frac{3}{13}+\frac{3}{14}\) ta có :
\(B>\frac{3}{15}+\frac{3}{15}+\frac{3}{15}+\frac{3}{15}+\frac{3}{15}=\frac{3+3+3+3+3}{15}=\frac{3.5}{15}=\frac{15}{15}=1\)
\(\Rightarrow\)\(B>1\) \(\left(1\right)\)
Lại có :
\(B< \frac{3}{10}+\frac{3}{10}+\frac{3}{10}+\frac{3}{10}+\frac{3}{10}=\frac{3+3+3+3+3}{10}=\frac{3.5}{10}=\frac{15}{10}< \frac{20}{10}=2\)
\(\Rightarrow\)\(B< 2\) \(\left(2\right)\)
Từ (1) và (2) suy ra :
\(1< B< 2\) ( đpcm )
Vậy \(1< B< 2\)
Chúc bạn học tốt ~
\(1-\frac{\frac{2}{19}+\frac{2}{43}-\frac{2}{2013}}{\frac{3}{19}+\frac{3}{43}-\frac{3}{2013}}\)= \(1-\frac{2\left(\frac{1}{19}+\frac{1}{43}-\frac{1}{2013}\right)}{3\left(\frac{1}{19}+\frac{1}{43}-\frac{1}{2013}\right)}\)= 1 - \(\frac{2}{3}\)=\(\frac{1}{3}\)
\(1-\frac{\frac{2}{19}+\frac{2}{43}-\frac{2}{2013}}{\frac{3}{19}+\frac{3}{43}-\frac{3}{2013}}=1-\frac{2\left(\frac{1}{19}+\frac{1}{43}-\frac{1}{2013}\right)}{3\left(\frac{1}{19}+\frac{1}{43}-\frac{1}{2013}\right)}\)=\(1-\frac{2}{3}=\frac{1}{3}\)