Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A=\dfrac{4}{7.31}+\dfrac{6}{7.41}+\dfrac{9}{10.41}+\dfrac{7}{10.57}\Rightarrow\dfrac{A}{5}=\dfrac{4}{35.31}+\dfrac{6}{35.41}+\dfrac{9}{50.41}+\dfrac{7}{50.57}\)
\(\dfrac{A}{5}=\dfrac{1}{31}-\dfrac{1}{35}+\dfrac{1}{35}-\dfrac{1}{41}+\dfrac{1}{41}-\dfrac{1}{50}+\dfrac{1}{50}-\dfrac{1}{57}\)
\(\dfrac{A}{5}=\dfrac{1}{31}-\dfrac{1}{57}=\dfrac{26}{31.57}\Rightarrow A=\dfrac{130}{31.57}\)
\(\dfrac{B}{2}=\dfrac{7}{38.31}+\dfrac{5}{38.41}+\dfrac{3}{46.43}+\dfrac{11}{46.57}\Rightarrow\dfrac{B}{2}=\dfrac{1}{31}-\dfrac{1}{57}=\dfrac{26}{31.57}\)
\(\Rightarrow B=\dfrac{52}{31.57}\)
\(\dfrac{A}{B}=\dfrac{130}{31.51}:\dfrac{52}{31.57}=\dfrac{5}{2}\)
Ta có : `A:5=4/35.31+6/35.41++9/50.41+7/50.57`
`A/5 = 1/31 - 1/35 +1/35-1/41+1/41-1/50+1/50-1/57`
`A/5 =1/31-1/57` `(1)`
Lại có : `B:2=7/38.31 +5/38.43+3/46.43+11/46.57`
`B/2 =1/31-1/38+1/38-1/43+1/43-1/46+1/46-1/57`
`B/2 =1/31-1/57` `(2)`
Từ `(1)` và `(2)` `=>A/5 =B/2`
`=>A/B=5/2`
a) Ta có: \(A=\dfrac{4}{7\cdot31}+\dfrac{6}{7\cdot41}+\dfrac{9}{10\cdot41}+\dfrac{7}{10\cdot57}\)
\(=\dfrac{20}{31\cdot35}+\dfrac{30}{35\cdot41}+\dfrac{45}{41\cdot50}+\dfrac{35}{50\cdot57}\)
\(=5\left(\dfrac{4}{31\cdot35}+\dfrac{6}{35\cdot41}+\dfrac{9}{41\cdot50}+\dfrac{7}{50\cdot57}\right)\)
\(=5\left(\dfrac{1}{31}-\dfrac{1}{35}+\dfrac{1}{35}-\dfrac{1}{41}+\dfrac{1}{41}-\dfrac{1}{50}+\dfrac{1}{50}-\dfrac{1}{57}\right)\)
\(=5\left(\dfrac{1}{31}-\dfrac{1}{57}\right)\)
Ta có: \(B=\dfrac{7}{19\cdot31}+\dfrac{5}{19\cdot43}+\dfrac{3}{23\cdot43}+\dfrac{11}{23\cdot57}\)
\(=\dfrac{14}{31\cdot38}+\dfrac{10}{38\cdot43}+\dfrac{6}{43\cdot46}+\dfrac{22}{46\cdot57}\)
\(=2\left(\dfrac{7}{31\cdot38}+\dfrac{5}{38\cdot43}+\dfrac{3}{43\cdot46}+\dfrac{11}{46\cdot57}\right)\)
\(=2\left(\dfrac{1}{31}-\dfrac{1}{38}+\dfrac{1}{38}-\dfrac{1}{43}+\dfrac{1}{43}-\dfrac{1}{46}+\dfrac{1}{46}-\dfrac{1}{57}\right)\)
\(=2\left(\dfrac{1}{31}-\dfrac{1}{57}\right)\)
Suy ra: \(\dfrac{A}{B}=\dfrac{5\left(\dfrac{1}{31}-\dfrac{1}{57}\right)}{2\left(\dfrac{1}{31}-\dfrac{1}{57}\right)}=\dfrac{5}{2}\)
\(B=\left(\dfrac{7}{19\cdot31}+\dfrac{5}{19\cdot43}+\dfrac{3}{23\cdot43}+\dfrac{11}{23\cdot57}\right):\left(\dfrac{1}{31}-\dfrac{1}{57}\right)\)
\(\Rightarrow\dfrac{1}{2}B=\left(\dfrac{7}{38\cdot31}+\dfrac{5}{38\cdot43}+\dfrac{3}{46\cdot43}+\dfrac{11}{46\cdot57}\right):\left(\dfrac{1}{31}-\dfrac{1}{57}\right)\)
\(\Rightarrow\dfrac{1}{2}B=\left(\dfrac{1}{31}-\dfrac{1}{38}+\dfrac{1}{38}-\dfrac{1}{43}+\dfrac{1}{43}-\dfrac{1}{46}+\dfrac{1}{46}-\dfrac{1}{57}\right):\left(\dfrac{1}{31}-\dfrac{1}{57}\right)\)
\(\Rightarrow\dfrac{1}{2}B=\left(\dfrac{1}{31}-\dfrac{1}{57}\right):\left(\dfrac{1}{31}-\dfrac{1}{57}\right)\)
\(\Rightarrow\dfrac{1}{2}B=1\)
\(\Rightarrow B=1:\dfrac{1}{2}=2\)
\(C=\frac{7}{19.31}+\frac{5}{19.43}+\frac{3}{23.43}+\frac{11}{23.57}\)
\(\frac{C}{2}=\frac{7}{31.38}+\frac{5}{38.43}+\frac{3}{43.46}+\frac{11}{46.57}\)
\(\frac{C}{2}=\frac{1}{31}-\frac{1}{38}+\frac{1}{38}-\frac{1}{43}+\frac{1}{43}-\frac{1}{46}+\frac{1}{46}-\frac{1}{57}\)
\(C=2\left(\frac{1}{31}-\frac{1}{57}\right)=2\cdot\frac{26}{1767}=\frac{52}{1767}\)
\(\left(\frac{7}{19.31}+\frac{5}{19.43}+\frac{3}{23.43}+\frac{11}{23.57}\right):\left(\frac{1}{31}-\frac{1}{57}\right)\)
\(=\frac{7}{19}.\frac{7}{31}+\frac{5}{19}.\frac{5}{43}+\frac{3}{23}.\frac{3}{43}+\frac{11}{23}.\frac{11}{57}:\frac{1}{31}-\frac{1}{57}\)
\(=\frac{7}{19}+\frac{5}{19}.\frac{7}{31}+\frac{1}{31}:\frac{5}{43}-\frac{3}{43}.\frac{3}{23}+\frac{11}{23}.\frac{11}{57}+\frac{1}{57}\)
\(=\frac{12}{19}.\frac{8}{31}:\frac{2}{43}.\frac{7}{16}.\frac{4}{19}\)
\(=\frac{1204}{1767}\)
\(\dfrac{7.43+5.31}{19.31.43}+\dfrac{3.57+11.43}{23.43.57}=\dfrac{456}{19.31.43}+\dfrac{644}{23.43.57}=\dfrac{24.19}{19.31.43}+\dfrac{23.28}{23.43.57}=\dfrac{24}{31.43}+\dfrac{28}{43.57}=\dfrac{24.57+28.31}{43.57.31}=\dfrac{2236}{75981}=\dfrac{52}{1767}\)
Thanks cô Huyền ạ(っ- • – ς)
ヾ(=`ω´=)ノ”